Why does $1+2+3+cdots = -frac{1}{12}$?












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$displaystylesum_{n=1}^infty frac{1}{n^s}$ only converges to $zeta(s)$ if $text{Re}(s) > 1$.



Why should analytically continuing to $zeta(-1)$ give the right answer?










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migrated from physics.stackexchange.com May 18 '11 at 4:34


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  • 27




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    It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-frac{1}{12}$ is simply the answer that analytical continuation yields. See also en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· .
    $endgroup$
    – joriki
    May 18 '11 at 4:40






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    What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell.
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    – KCd
    May 18 '11 at 4:42






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    See: math.stackexchange.com/questions/37327/infinity-1-paradox, math.stackexchange.com/questions/27526/…, and terrytao.wordpress.com/2010/04/10/… for more information too.
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    – JavaMan
    May 18 '11 at 4:54








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    Someone should edit the Wikipedia section that says $1+2+3+... = frac{-1}{12}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values
    $endgroup$
    – zerosofthezeta
    Dec 5 '13 at 5:39






  • 8




    $begingroup$
    (For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) on (problems with) finding infinite sums via 'shift & add', ii) on analytic continuation, iii) on assigning values to divergent series
    $endgroup$
    – Grigory M
    Jan 12 '14 at 11:52


















368












$begingroup$


$displaystylesum_{n=1}^infty frac{1}{n^s}$ only converges to $zeta(s)$ if $text{Re}(s) > 1$.



Why should analytically continuing to $zeta(-1)$ give the right answer?










share|cite|improve this question











$endgroup$



migrated from physics.stackexchange.com May 18 '11 at 4:34


This question came from our site for active researchers, academics and students of physics.











  • 27




    $begingroup$
    It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-frac{1}{12}$ is simply the answer that analytical continuation yields. See also en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· .
    $endgroup$
    – joriki
    May 18 '11 at 4:40






  • 61




    $begingroup$
    What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell.
    $endgroup$
    – KCd
    May 18 '11 at 4:42






  • 19




    $begingroup$
    See: math.stackexchange.com/questions/37327/infinity-1-paradox, math.stackexchange.com/questions/27526/…, and terrytao.wordpress.com/2010/04/10/… for more information too.
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    – JavaMan
    May 18 '11 at 4:54








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    Someone should edit the Wikipedia section that says $1+2+3+... = frac{-1}{12}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values
    $endgroup$
    – zerosofthezeta
    Dec 5 '13 at 5:39






  • 8




    $begingroup$
    (For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) on (problems with) finding infinite sums via 'shift & add', ii) on analytic continuation, iii) on assigning values to divergent series
    $endgroup$
    – Grigory M
    Jan 12 '14 at 11:52
















368












368








368


206



$begingroup$


$displaystylesum_{n=1}^infty frac{1}{n^s}$ only converges to $zeta(s)$ if $text{Re}(s) > 1$.



Why should analytically continuing to $zeta(-1)$ give the right answer?










share|cite|improve this question











$endgroup$




$displaystylesum_{n=1}^infty frac{1}{n^s}$ only converges to $zeta(s)$ if $text{Re}(s) > 1$.



Why should analytically continuing to $zeta(-1)$ give the right answer?







analysis complex-analysis riemann-zeta divergent-series






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share|cite|improve this question













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edited Oct 29 '18 at 6:44









Alexander Gruber

19.9k25102172




19.9k25102172










asked May 18 '11 at 4:26







perplexed











migrated from physics.stackexchange.com May 18 '11 at 4:34


This question came from our site for active researchers, academics and students of physics.






migrated from physics.stackexchange.com May 18 '11 at 4:34


This question came from our site for active researchers, academics and students of physics.










  • 27




    $begingroup$
    It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-frac{1}{12}$ is simply the answer that analytical continuation yields. See also en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· .
    $endgroup$
    – joriki
    May 18 '11 at 4:40






  • 61




    $begingroup$
    What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell.
    $endgroup$
    – KCd
    May 18 '11 at 4:42






  • 19




    $begingroup$
    See: math.stackexchange.com/questions/37327/infinity-1-paradox, math.stackexchange.com/questions/27526/…, and terrytao.wordpress.com/2010/04/10/… for more information too.
    $endgroup$
    – JavaMan
    May 18 '11 at 4:54








  • 7




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    Someone should edit the Wikipedia section that says $1+2+3+... = frac{-1}{12}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values
    $endgroup$
    – zerosofthezeta
    Dec 5 '13 at 5:39






  • 8




    $begingroup$
    (For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) on (problems with) finding infinite sums via 'shift & add', ii) on analytic continuation, iii) on assigning values to divergent series
    $endgroup$
    – Grigory M
    Jan 12 '14 at 11:52
















  • 27




    $begingroup$
    It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-frac{1}{12}$ is simply the answer that analytical continuation yields. See also en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· .
    $endgroup$
    – joriki
    May 18 '11 at 4:40






  • 61




    $begingroup$
    What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell.
    $endgroup$
    – KCd
    May 18 '11 at 4:42






  • 19




    $begingroup$
    See: math.stackexchange.com/questions/37327/infinity-1-paradox, math.stackexchange.com/questions/27526/…, and terrytao.wordpress.com/2010/04/10/… for more information too.
    $endgroup$
    – JavaMan
    May 18 '11 at 4:54








  • 7




    $begingroup$
    Someone should edit the Wikipedia section that says $1+2+3+... = frac{-1}{12}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values
    $endgroup$
    – zerosofthezeta
    Dec 5 '13 at 5:39






  • 8




    $begingroup$
    (For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) on (problems with) finding infinite sums via 'shift & add', ii) on analytic continuation, iii) on assigning values to divergent series
    $endgroup$
    – Grigory M
    Jan 12 '14 at 11:52










27




27




$begingroup$
It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-frac{1}{12}$ is simply the answer that analytical continuation yields. See also en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· .
$endgroup$
– joriki
May 18 '11 at 4:40




$begingroup$
It's not clear (to me) what you mean by "the right answer". Since this was migrated from physics, I suspect that you have in mind the use of this summation in string theory. However, mathematically speaking, there is no "right answer" for summation techniques for divergent series; $-frac{1}{12}$ is simply the answer that analytical continuation yields. See also en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_·_·_· .
$endgroup$
– joriki
May 18 '11 at 4:40




61




61




$begingroup$
What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell.
$endgroup$
– KCd
May 18 '11 at 4:42




$begingroup$
What does the phrase "right answer" mean? If you define zeta(s) in the usual way for s with real part greater than 1, it has an analytic continuation to s = -1 and the value there is -1/12. That does not mean the series defining zeta(s) for Re(s) > 1 converges at s = -1 with value -1/12 except in a hand-waving Euler kind of way. This question doesn't really have a mathematical meaning as far as I can tell.
$endgroup$
– KCd
May 18 '11 at 4:42




19




19




$begingroup$
See: math.stackexchange.com/questions/37327/infinity-1-paradox, math.stackexchange.com/questions/27526/…, and terrytao.wordpress.com/2010/04/10/… for more information too.
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– JavaMan
May 18 '11 at 4:54






$begingroup$
See: math.stackexchange.com/questions/37327/infinity-1-paradox, math.stackexchange.com/questions/27526/…, and terrytao.wordpress.com/2010/04/10/… for more information too.
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– JavaMan
May 18 '11 at 4:54






7




7




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Someone should edit the Wikipedia section that says $1+2+3+... = frac{-1}{12}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values
$endgroup$
– zerosofthezeta
Dec 5 '13 at 5:39




$begingroup$
Someone should edit the Wikipedia section that says $1+2+3+... = frac{-1}{12}$ en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values
$endgroup$
– zerosofthezeta
Dec 5 '13 at 5:39




8




8




$begingroup$
(For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) on (problems with) finding infinite sums via 'shift & add', ii) on analytic continuation, iii) on assigning values to divergent series
$endgroup$
– Grigory M
Jan 12 '14 at 11:52






$begingroup$
(For newcomers) there are a lot of linked questions now, but most relevant, IMO: i) on (problems with) finding infinite sums via 'shift & add', ii) on analytic continuation, iii) on assigning values to divergent series
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– Grigory M
Jan 12 '14 at 11:52












16 Answers
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there are many ways to see that your result is the right one. What does the right one mean?



It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.



It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.



Exponential regulator method



Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes
$$ S = sum_{n=1}^{infty} n e^{-epsilon n} $$
Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $epsilonto 0$ limit of the formula for $S$. In physics, $epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.



The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself)
$$ S = frac{e^epsilon}{(e^epsilon-1)^2} $$
We will only need some Laurent expansion around $epsilon = 0$.
$$ S = frac{1+epsilon+epsilon^2/2 + O(epsilon^3)}{(epsilon+epsilon^2/2+epsilon^3/6+O(epsilon^4))^2} $$
We have
$$ S = frac{1}{epsilon^2} frac{1+epsilon+epsilon^2/2+O(epsilon^3)}{(1+epsilon/2+epsilon^2/6+O(epsilon^3))^2} $$
You see that the $1/epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]



and the result is
$$ frac{1}{epsilon^2} - frac{1}{12} + frac{epsilon^2}{240} + O(epsilon^4) $$
In the $epsilonto 0$ limit we were interested in, the $epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.



In physics applications, $epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $exp(-(epsilon+epsilon^2)n)$.



There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.



Euler's method



Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.



We will work with two sums,
$$ S = 1+2+3+4+5+dots, quad T = 1-2+3-4+5-dots $$
Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion
$$ frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + dots $$
Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now:
$$ T = (1+2+3+dots) - 2times (2+4+6+dots) = (1+2+3+dots) (1 - 4) = -3S$$
so $S=-T/3=-1/12$.



A zeta-function calculation



A somewhat unusual calculation of $zeta(-1)=-1/12$ of mine may be found here:




http://www.kolej.mff.cuni.cz/~lmotm275/RUZE/09/node7.html




The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.






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    The fact that other methods also yield the result $-frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is.
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    – joriki
    May 18 '11 at 6:04






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    @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics.
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    – Luboš Motl
    May 18 '11 at 6:43






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    @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't.
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    – joriki
    May 18 '11 at 6:47








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    @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong.
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    – KCd
    May 18 '11 at 7:59






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    @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+dots=-1/12$ and (2) $1+1+1+dots=-1/2$ and (3) is additive must also assign $0+1+2+dots=5/12$ and thus is not (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up?
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    – Mario Carneiro
    Feb 19 '15 at 7:44



















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Here is a variant on Lubos Motl's answer:



Let $S = sum_{n=1}^{infty} n$. Then $S - 4 S = sum_{n = 1}^{infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's.



Namely, consider $$sum_{n=1}^{infty} (-1)^{n-1} n t^n
= -t dfrac{d}{dt} sum_{n=1}^{infty} (-t)^n = -t dfrac{d}{dt} dfrac{1}{1+t}
= dfrac{t}{(1+t)^2}.$$

Letting $t to 1,$ we find that $-3 S = dfrac{1}{4}$, and hence that
$S = dfrac{-1}{12}.$





To see the relationship between this approach and Lubos Motl's, note that if
we write $t = e^{epsilon},$ then $tdfrac{d}{dt} = dfrac{d}{depsilon},$
so in fact the arguments are essentially the same, except that Lubos doesn't
perform the initial step of replacing $S$ by $S - 4S$, which means that
he has the pole $dfrac{1}{epsilon^2}$ which he then subtracts away.





As far as I know, this trick of replacing $zeta(s)$ by $(1-2^{-s+1})^{-1}zeta(s)$ is due to Euler, and it is a now standard method for replacing
$zeta(s)$ by a function which carries the same information, but
does not have a pole at $s = 1$. The evaluation of $zeta(s)$ at negative integers by passing to $(1-2^{-s+1})zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see
the Bernoulli numbers appearing in this way, for example.





Of course, taken literally, the series $sum_{n=1}^{infty} n$ diverges to $+infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $sum_{n = 1}^{infty} n,$
I'm reasonably confident that any sensible regularization will necessarily yield the same value of $dfrac{-1}{12}$. (Lubos Motl makes the same assertion in his answer.)






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    @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+dotso$ or $-1-1-dotso$ First, these should be different. Also, $1+1+dotso=zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$.
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    – joriki
    May 18 '11 at 6:57






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    @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.)
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    – joriki
    May 18 '11 at 7:31








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    @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.)
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    – joriki
    May 18 '11 at 9:55






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    @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards,
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    – Matt E
    May 18 '11 at 11:58






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    I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+cdots$ and $0+1+0+2+0+3+0+cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+cdots$ has a different value (namely $1/24$).
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    – Axel Boldt
    Mar 9 '14 at 3:10



















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What a great method! I tried a few...
$$
sum_{n = 1}^{infty} left(frac{1}{(2 n)^{s}} - frac{1}{(2 n - 1)^{s}}right) = zeta (s) Bigl(2^{(1 - s)} - 1Bigr)
$$
put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{1}{4}$



or
$$
sum_{n = 1}^{infty} left(frac{1}{(2 n + 1)^{s}} - frac{1}{(2 n)^{s}}right) = zeta (s) - frac{2 zeta (s)}{2^{s}} - 1
$$
put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{3}{4}$






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    You should read Terence Tao's article on this "oddities".
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    – Pedro Tamaroff
    Jun 11 '12 at 17:37






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    For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question.
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    – Marnix Klooster
    Jan 16 '14 at 20:04






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    Well this is exactly about an answer I just posted
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    – reuns
    Nov 1 '16 at 6:36





















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Here is a direct computation of $zeta(-1)$ taken from this answer:



Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get
$$
begin{align}
(1-2^{1-x})zeta(x)Gamma(x+2)
&=int_0^inftyfrac{(x+1)xt^{x-1}}{e^t+1}mathrm{d}t\
&=int_0^inftyfrac{(x+1)t^xe^t}{(e^t+1)^2}mathrm{d}t\
&=int_0^inftyfrac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}mathrm{d}ttag{1}
end{align}
$$
Now we can plug in $x=-1$ into $(1)$ to get
$$
begin{align}
(1-2^2)zeta(-1)Gamma(1)
&=int_0^inftyfrac{e^{2t}-e^t}{(e^t+1)^3}mathrm{d}t\
&=int_1^inftyfrac{u-1}{(u+1)^3}mathrm{d}u\
&=int_1^inftyleft(frac1{(u+1)^2}-frac2{(u+1)^3}right)mathrm{d}u\
&=frac14tag{2}
end{align}
$$
Since $(1-2^2)Gamma(1)=-3$, $(2)$ says that
$$
zeta(-1)=-frac1{12}tag{3}
$$
Naively plugging $n=-1$ into
$$
zeta(n)=frac1{1^n}+frac1{2^n}+frac1{3^n}+dotstag{4}
$$
yields
$$
-frac1{12}=1+2+3+4+dotstag{5}
$$
However, the series on the right of $(5)$ is obviously divergent by the Term Test.






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    44












    $begingroup$

    A good explanation of how this is true and why and where it is useful and significant is beautifully summarized in 24 by John Baez (as part of the 19/09/08 Rankin Lectures) and yes, I do agree that this is wierd.





    The following is the same Euler's method described by Luboš Motl and Matt E in the other answers.



    Let us consider the infinite geometric progression
    " $1, x, x^2, x^3, x^4, dots$ "



    The sum of this progression can be found by the formula $S_infty = large frac{a}{1-r}$,



    $$implies 1 + x + x^2 + x^3 + x^4 + dots = frac{1}{1-x}$$
    Differentiating with respect to $x$,
    $$ implies 0 + 1 + 2x + 3x^2 + 4x^3 + dots = frac{1}{(x-1)^2}$$
    Let $x = -1$, (Note: In the LHS, you get the alternating counterpart of the natural numbers series)
    $$implies 1-2+3-4+5-6+...= frac{1}{4} $$



    $$ bbox[5px,border:2px solid green]{therefore sumlimits_{i=1}^infty n(-1)^{n-1} = frac{1}{4}} $$



    Subtracting this from $sumlimits_{i=1}^infty n$,
    $$sumlimits_{i=1}^infty n -sumlimits_{i=1}^infty n(-1)^{n-1} =4+8+12+...=4sumlimits_{i=1}^infty n$$
    $$implies 3sumlimits_{i=1}^infty n =-sumlimits_{i=1}^infty n(-1)^{n-1}=-1/4$$
    $$bbox[5px,border:2px solid lime]{therefore sumlimits_{i=1}^infty n=-frac{1}{12}}$$




    This intuitive reasoning was made famous in a video by the physicist Phil Plait of Numberphile. It is heavily criticized for being sloppy but it shows a simple method that has fooled a large population of its high school audience (as noted in the comments).


    Funfact: Googling -1/12 doesn't give you any results




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    • 17




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      That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit)
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      – Mariano Suárez-Álvarez
      Jan 18 '14 at 2:37








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      This proof was featured today in a video posted to Reddit.
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      – Steven Gubkin
      Jan 18 '14 at 3:31






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      Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up.
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      – Jeyekomon
      Jan 18 '14 at 17:05






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      @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature...
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      – Mariano Suárez-Álvarez
      Jan 19 '14 at 6:27






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      You manipulate that infinite sum as if it were finite, like when you assume that $sum_{n=1}^infty n-4sum_{n=1}^infty n=-3sum_{n=1}^infty n.$ But that's impossible, since $infty-infty$ is undefined.
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      – Hakim
      May 16 '14 at 20:52





















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    If the following were true: $$sum_{n=1}^infty{n}=-frac1{12}tag{hypothesis}$$ then we would expect the following: $$lim_{ntoinfty}frac{n(n+1)}2=-frac1{12}tag{expectation}$$ which is the formula for the infinite triangular number limit. Unfortunately this is a result that we do not get when the limit is correctly taken. The correct value is $$lim_{ntoinfty}frac{n(n+1)}2=frac{infty(infty+1)}2=frac{infty^2+infty}2=fracinfty2=inftyneq-frac1{12}$$ This sort of mathematical sleight of hand, smoke and mirrors, pulling a finite negative rabbit out of an empty positively infinite hat does not impress me; worse yet, it gives legitimate, observable, repeatable mathematics a bad name.






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    • 12




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      Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity...
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      – zerosofthezeta
      Mar 13 '14 at 3:33






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      Not sure that is expectable. I made a question related to it.
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      – JMCF125
      Apr 27 '14 at 20:10






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      I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch.
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      – primo
      May 29 '14 at 10:24












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      Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth.
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      – Sara Tancredi
      Jun 5 '14 at 23:47






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      @GottfriedHelms, did you just divide by 0 ?
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      – GinKin
      Jun 17 '14 at 19:44



















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    The notation "$1+2+3+cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$.






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    • 21




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      Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards,
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      – Matt E
      Sep 8 '12 at 19:11






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      Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent.
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      – John Bentin
      Sep 9 '12 at 12:03






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      The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon.
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      – Steven Gubkin
      Jan 18 '14 at 3:29






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      @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st.
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      – Did
      Jan 18 '14 at 21:50








    • 1




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      I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.)
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      – Jesse Madnick
      Jan 22 '14 at 1:36



















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    This infinite series is ultimately divergent because
    $$
    1+2+3+4+cdots=sumlimits_{k=1}^{infty} k$$
    $$ = limlimits_{ntoinfty} sumlimits_{k=1}^{n} k = limlimits_{ntoinfty} frac{n(n+1)}{2} = infty
    $$
    The value of $-frac{1}{12}$ is assigned to the infinite series by the methods of zeta function regularization and several others. If you'd like to understand why someone would assign a value to a divergent series, then have a look at this question.






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      I am missing here that



      $$1+2+3+cdots rightarrow infty$$



      $$zeta(-1) neq 1+2+3+cdots$$



      As you say, we only define $zeta$ using the infinite sum if $Re(s)>1$.





      It is just as defining $$f(x) = begin{cases} frac{1}{x} &mathrm{if} ; x neq 0 \ 0 & mathrm{if} ; x = 0 end{cases}$$



      And asking why $f(0)=0$.






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      • $begingroup$
        your example is not correct. In first place the zeta function defined at values where $Re(z)le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation.
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        – Masacroso
        Jan 17 '18 at 11:34



















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      A nice alternative representation, by a double sum.



      Consider the infinite square array where the rowsums and the (formal) column-sums are also given in closed forms
      $$ small begin{array} {r|rrrrr|rr}
      & & & & & & text{rowsums} \
      hline
      & 1/0! & log(1)/1! & log(1)^2/2! &log(1)^3/3! & cdots & = e^{log(1)}&=1 \
      & 1/0! & log(2)/1! & log(2)^2/2! &log(2)^3/3! & cdots & = e^{log(2)}&=2 \
      & 1/0! & log(3)/1! & log(3)^2/2! &log(3)^3/3! & cdots & = e^{log(3)}&=3 \
      & vdots & & & & ddots & vdots & vdots\
      hline
      text{colsums:}
      & zeta(0) & -zeta(0)'/1! & zeta(0)''/2!& -zeta(0)^{(3)}/3! & cdots &&=zeta(-1)
      end{array} $$
      and the row-sum of the derivatives of the $zeta()$ at $0$ in the bottom row can numerically be written as
      $$ small -0.5 +0.9189... -1.003178...+1.000785...-0.999879...+1.0000019... pm cdots $$
      If we split the terms and do two series we find
      $$ Tiny begin{array} {}
      -1 &+1 &-1 &+1 &-1 &+1 &pm &cdots &=-1/2\
      +0.5 &-0.0810... &-0.003178...&+0.000785...&+0.000120...&+0.0000019... &pm &cdots &= 5/12\
      end{array}$$
      Here we must now resort to the earlier definition of the divergent sum of the alternating units (in the first row), but the second row is convergent and can conventionally be summed. The sum of the two rowsums $small -1/2 + 5/12 = -1/12$ gives the expected result.



      However, that fiddling with double-sums, when non-convergent-series are involved (as are the columnsums in the above matrix and the splitting in the numerical expression) must explicitely be justified as "legal"/algebraically consistent operation.

      But because I encounter it not frequently, that such non-convergent double-sum schemes come out with the expected result without further ado, I think this is a specific nice observation here.



      And the more algebraically manipulations come out to be consistent with the assumed value of a divergent series, the more is the hypothese acceptable, that this should be taken as the canonical numerical replacement for the series-expression ( like we do it with the rational fraction for the geometric series even in the divergent case (except for that with quotient $q=1$)) .






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        Suppose a rigorous way of doing a computation yields a well defined real number as the answer. But with shortcut formal manipulations that are not allowed (e.g. interchanging summations and integrations when that isn't allowed) one ends up with a divergent series and then the question is that given only the divergent series, can one guess what real number is most likely the answer to what the unknown original problem was.



        This is similar to guessing that the next term of the integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 is probably 16 rather than e.g. 32431, even though the latter possibility cannot be ruled out. Strictly speaking this problem is not well defined, what one really is doing here is assuming that the sequence that is specified by the least amount of information is the most likely answer. The person who invented the puzzle had some simple algorithm in mind, therefore the much more complicated algorithm that would yield 32431 as the next number in the sequence is not the likely answer.



        Similarly, the formal manipulations that yield the divergent series won't have introduced a lot of additional information, these are just generic mathematical manipulations that would have been correct when used in a wide class of problems, but not the one it actually has been applied to. This means that almost all the information about the unknown real number is present in the divergent series, it can be extracted from it by applying certain formal manipulations to the series that, like the unknown manipulations that led to the divergent series, are formally correct for a wide class of convergent series, but not in case of this divergent series.






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          13












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          Euler's approach, but using algebra instead of subtraction of infinite series: Let $f(x)=(1-x)^{-2}$. Then $f(-x) = f(x) -4xf(x^2)$. Plug $x=1$ to get $f(-1)=-3f(1)$, whence $f(1)=-1/12$.



          EDIT: This argument is a motivation for the mysterious value $-1/12$. Yes, $f(x) :=(1-x)^{-2}$ is undefined at $x=1$. If we define $f(1)=-1/12$, then the above identity holds for all $x$.






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          • 4




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            So... $0^{-2}$ is well defined? And all these years I thought you can't divide by zero.
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            – Asaf Karagila
            Feb 21 '15 at 22:44






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            $frac14=infty-4cdotinfty$ So $-frac1{12}=infty$
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            – robjohn
            Feb 22 '15 at 4:18





















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          I like this simple graphical explanation.



          There are first partial sums of the series 1 + 2 + 3 + 4 + ⋯ on picture. The parabola is their smoothed asymptote; its y-intercept is −1/12.



          enter image description here






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          • 6




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            I'm afraid I don't follow this at all.
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            – Cheerful Parsnip
            Sep 11 '15 at 21:02










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            @GrumpyParsnip It's explained at en.wikipedia.org/wiki/… . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself?
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            – Chris Culter
            Sep 13 '15 at 21:01










          • $begingroup$
            @ChrisCulter: thanks!
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            – Cheerful Parsnip
            Sep 13 '15 at 21:12










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            @ChrisCulter, yep, it's my additions. I updated the answer.
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            – Ivan Kochurkin
            Sep 27 '15 at 18:12






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            It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the colors in the picture?
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            – Did
            Dec 4 '16 at 9:42



















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          As I recently showed in another answer, we have the wonderful pattern:



          $$sum_{k=1}^n1=nimpliesint_{-1}^0x~mathrm dx=zeta(0)\sum_{k=1}^nk=frac{n(n+1)}2impliesint_{-1}^0frac{x(x+1)}2~mathrm dx=zeta(-1)\sum_{k=1}^nk^2=frac{n(n+1)(2n+1)}6impliesint_{-1}^0frac{x(x+1)(2x+1)}6~mathrm dx=zeta(-2)\sum_{k=1}^nk^3=left[frac{n(n+1)}2right]^2impliesint_{-1}^0left[frac{x(x+1)}2right]^2~mathrm dx=zeta(-3)\vdots$$



          This pattern works for all $zeta(-k)$.






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            8












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            Here is a useful place for Euler's transform and the Dirichlet eta function:



            $$zeta(s)=sum_{n=1}^inftyfrac1{n^s}$$



            $$eta(s)=sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$



            $$zeta(s)-eta(s)=sum_{n=1}^inftyfrac{1+(-1)^n}{n^s}=sum_{n=1}^inftyfrac2{(2n)^s}=2^{1-s}sum_{n=1}^inftyfrac1{n^s}=2^{1-s}zeta(s)$$



            $$zeta(s)-eta(s)=2^{1-s}zeta(s)implies (1-2^{1-s})zeta(s)=eta(s)$$




            $$zeta(s)=frac1{1-2^{1-s}}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$




            After a quick application of Euler's transform, we get a nice analytic continuation to the entire complex plane.



            $${small E_1}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}=sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$



            Finally giving us




            $$zeta(s)=frac1{1-2^{1-s}}sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$




            And at $s=-1$,



            $$zeta(-1)=-frac1{12}$$






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              Full disclosure: I saw a different question ask a similar question to the OP and it got closed before I could explain it. I'm mostly focusing on why people say the sum of the natural numbers is $-1/12$, and why that's wrong, in as simple a way as I can communicate and assuming as little knowledge as possible, save for perhaps a bit of basic calculus, since the OP of that question was only in high school.






              The notion that the sum of the natural numbers, in any sense, is $-1/12$ is actually completely untrue and it results from a misconception some people have ... I actually saw a professional mathematician make this mistake in a lecture yesterday so I guess it penetrates even academia.



              I'll try to give you the rough idea of what this, and what the misconception is.



              We define the Riemann zeta function $zeta(s)$ by the summation



              $$zeta(s) = sum_{n=1}^infty frac{1}{n^s} = 1 + frac{1}{2^s} + frac{1}{3^s} + frac{1}{4^s} + frac{1}{5^s} + ...$$



              This $s$ can be any complex or real number, within a specific constant. If $s$ is real, then we must have $s>1$. If $s$ is complex, the real part of $s$, denoted $Re(s)$, must satisfy $Re(s)>1$. (Of course, both fall under this latter condition but we're going to focus on the former here since the complex inputs for the function aren't important to describe the phenomenon.)



              Why do we impose this constraint? You might recall $p$-series from Calculus I or II in school: the convergence of this sum diverges if $sleq 1$ as a result. For example, if $s=1$:



              $$zeta(1) "=" sum_{n=1}^infty frac{1}{n} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + ...$$



              This is a well-known divergent sum, the harmonic series. As an example of a convergent sum, consider $s=2$:



              $$zeta(2) = sum_{n=1}^infty frac{1}{n^2} = 1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + ... = frac{pi^2}{6}$$



              This notion - the sum of the reciprocals of the perfect squares - is known as the Basel problem, and the result was first proved by Euler. 3Blue1Brown has a great video showing (somewhat intuitively) how the sum converges to this value if you're curious - https://www.youtube.com/watch?v=d-o3eB9sfls



              So it should be relatively clear that the sum itself diverges when $s>1$. Okay, so how does this relate to the sum of the natural numbers, i.e. $1+2+3+4+...$?



              Well, notice:



              $$zeta(-1) "=" sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$






              What you should be saying to yourself right now: "But wait! $zeta(s)$ isn't convergent for $s<1$! $-1$ is clearly less than $1$, what the hell are you doing?"






              This is exactly where the misconceptions begin to start.



              Flat out, the notion of $zeta(-1)$ is nonsensical right now. $zeta(s)$ only works for $s>1$ - it is a very bad idea to even consider messing with a clearly divergent sum like this.



              However, out of our interests in the field of complex analysis, we can define the $zeta(s)$ in such a way that it can take in inputs $s leq 1$ (or $Re(s)leq 1$ for complex $s$). This is a process known as "analytic continuation." The notion is a bit above even my head, but the general notion is to create a definition for $zeta(s)$ (or any function) for $s leq 1$ such that the function is differentiable everywhere. The notion of differentiability also gives us a unique definition in this sense.



              I'll skip over the details because I'm not qualified (it is discussed here), but what results is this definition for $zeta(s)$ for $s < 1$:



              $$zeta(s) = 2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s)$$



              Oof, what a messy definition. (Note: $s=1$ is still divergent for this!)



              So this gives us the "grander" definition of the $zeta$ function for all $s$, negative or positive or complex, except $s=1$:



              $$zeta(s) = )left{begin{matrix}
              sum_{n=1}^infty n^{-s} & Re(s)>1 \
              2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s) & Re(s)<1
              end{matrix}right.$$



              It can be shown, through this definition, $zeta(-1) = -1/12$.



              However, if you plug $-1$ naively into the original zeta function - the summation over the natural numbers - and ignore that the function doesn't converge there, you get $1+2+3+4+5+...$. Taken alongside the fact that under the analytic continuation of the zeta function $zeta(-1) = -1/12$, this leads one to naively conclude



              $$1+2+3+4+5+ ... = -frac{1}{12}$$



              Now, obviously, this is untrue. As stated, $zeta(s)$ in its summation form does not converge for $s leq 1$, so it's just outright wrong to say it's equal in this way.



              That this has almost become prolific as a sort of meme and spread to people not knowledgeable enough to no any better doesn't help matters.






              A digression on "convergence," Ramanujan, and partial sums:






              Interestingly, Ramanujan came up with a similar result without making use of the zeta function, by manipulating the infinite sum $1+2+3+4+5+...$, letting it equal some constant $c$, and then manipulations yielded $c = -frac{1}{12}$. I've seen people tempted to defend the result $zeta(-1) = -frac{1}{12}$ since they're basically the same thing. So before that notion comes up, I'd also like to note that while it depends on your notion of "convergence" (perhaps there's a context in which the sum can be said to converge?), most people use the definition of convergence as the limit of the partial sums.



              In which case, let us consider the partial sums of $1+2+3+4+5+...$. We let $P(n)$ denote the sum of the first $n$ terms in the sum. Clearly,




              • $P(1) = 1$

              • $P(2) = 3$

              • $P(3) = 6$

              • $P(4) = 10$

              • $P(5) = 15$

              • $P(6) = 21$

              • $P(7) = 28$

              • ...

              • $P(n) = frac{n(n+1)}{2}$


              Clearly,



              $$lim_{n to infty} P(n) = lim_{n to infty} frac{n(n+1)}{2} = infty$$



              i.e. the sum diverges.



              This is not the only way one can define convergence: for example, Cesaro convergence looks at the average of the partial sums and their limit. This is a context in which



              $$sum_{n=1}^infty (-1)^n = (-1) + 1 +(-1) + 1 +(-1) + 1 +... = frac{1}{2}$$



              It should be clear that the averages for the summation $1+2+3+4+5+...$ also diverge - and certainly don't approach $-1/12$!



              So in that sense we can't utilize Ramanujan's summation since the notion of manipulating the summation as he did (which involved multiplying the sum by scalars and adding/subtracting such things from the original sum) is generally forbidden for divergent sums.






              Thus, the end conclusion and what needs to be said:






              $$zeta(-1) = -frac{1}{12} ; ; ; ; ; text{BUT} ; ; ; ; ; zeta(-1) neq sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$



              And therefore,



              $$-frac{1}{12} neq 1+2+3+4+5+...$$






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                protected by Bruno Joyal Dec 6 '13 at 22:43



                Thank you for your interest in this question.
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                190












                $begingroup$

                there are many ways to see that your result is the right one. What does the right one mean?



                It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.



                It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.



                Exponential regulator method



                Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes
                $$ S = sum_{n=1}^{infty} n e^{-epsilon n} $$
                Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $epsilonto 0$ limit of the formula for $S$. In physics, $epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.



                The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself)
                $$ S = frac{e^epsilon}{(e^epsilon-1)^2} $$
                We will only need some Laurent expansion around $epsilon = 0$.
                $$ S = frac{1+epsilon+epsilon^2/2 + O(epsilon^3)}{(epsilon+epsilon^2/2+epsilon^3/6+O(epsilon^4))^2} $$
                We have
                $$ S = frac{1}{epsilon^2} frac{1+epsilon+epsilon^2/2+O(epsilon^3)}{(1+epsilon/2+epsilon^2/6+O(epsilon^3))^2} $$
                You see that the $1/epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
                1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]



                and the result is
                $$ frac{1}{epsilon^2} - frac{1}{12} + frac{epsilon^2}{240} + O(epsilon^4) $$
                In the $epsilonto 0$ limit we were interested in, the $epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.



                In physics applications, $epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $exp(-(epsilon+epsilon^2)n)$.



                There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.



                Euler's method



                Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.



                We will work with two sums,
                $$ S = 1+2+3+4+5+dots, quad T = 1-2+3-4+5-dots $$
                Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion
                $$ frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + dots $$
                Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now:
                $$ T = (1+2+3+dots) - 2times (2+4+6+dots) = (1+2+3+dots) (1 - 4) = -3S$$
                so $S=-T/3=-1/12$.



                A zeta-function calculation



                A somewhat unusual calculation of $zeta(-1)=-1/12$ of mine may be found here:




                http://www.kolej.mff.cuni.cz/~lmotm275/RUZE/09/node7.html




                The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.






                share|cite|improve this answer











                $endgroup$









                • 138




                  $begingroup$
                  The fact that other methods also yield the result $-frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:04






                • 12




                  $begingroup$
                  @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics.
                  $endgroup$
                  – Luboš Motl
                  May 18 '11 at 6:43






                • 61




                  $begingroup$
                  @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:47








                • 58




                  $begingroup$
                  @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong.
                  $endgroup$
                  – KCd
                  May 18 '11 at 7:59






                • 8




                  $begingroup$
                  @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+dots=-1/12$ and (2) $1+1+1+dots=-1/2$ and (3) is additive must also assign $0+1+2+dots=5/12$ and thus is not (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up?
                  $endgroup$
                  – Mario Carneiro
                  Feb 19 '15 at 7:44
















                190












                $begingroup$

                there are many ways to see that your result is the right one. What does the right one mean?



                It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.



                It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.



                Exponential regulator method



                Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes
                $$ S = sum_{n=1}^{infty} n e^{-epsilon n} $$
                Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $epsilonto 0$ limit of the formula for $S$. In physics, $epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.



                The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself)
                $$ S = frac{e^epsilon}{(e^epsilon-1)^2} $$
                We will only need some Laurent expansion around $epsilon = 0$.
                $$ S = frac{1+epsilon+epsilon^2/2 + O(epsilon^3)}{(epsilon+epsilon^2/2+epsilon^3/6+O(epsilon^4))^2} $$
                We have
                $$ S = frac{1}{epsilon^2} frac{1+epsilon+epsilon^2/2+O(epsilon^3)}{(1+epsilon/2+epsilon^2/6+O(epsilon^3))^2} $$
                You see that the $1/epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
                1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]



                and the result is
                $$ frac{1}{epsilon^2} - frac{1}{12} + frac{epsilon^2}{240} + O(epsilon^4) $$
                In the $epsilonto 0$ limit we were interested in, the $epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.



                In physics applications, $epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $exp(-(epsilon+epsilon^2)n)$.



                There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.



                Euler's method



                Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.



                We will work with two sums,
                $$ S = 1+2+3+4+5+dots, quad T = 1-2+3-4+5-dots $$
                Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion
                $$ frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + dots $$
                Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now:
                $$ T = (1+2+3+dots) - 2times (2+4+6+dots) = (1+2+3+dots) (1 - 4) = -3S$$
                so $S=-T/3=-1/12$.



                A zeta-function calculation



                A somewhat unusual calculation of $zeta(-1)=-1/12$ of mine may be found here:




                http://www.kolej.mff.cuni.cz/~lmotm275/RUZE/09/node7.html




                The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.






                share|cite|improve this answer











                $endgroup$









                • 138




                  $begingroup$
                  The fact that other methods also yield the result $-frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:04






                • 12




                  $begingroup$
                  @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics.
                  $endgroup$
                  – Luboš Motl
                  May 18 '11 at 6:43






                • 61




                  $begingroup$
                  @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:47








                • 58




                  $begingroup$
                  @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong.
                  $endgroup$
                  – KCd
                  May 18 '11 at 7:59






                • 8




                  $begingroup$
                  @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+dots=-1/12$ and (2) $1+1+1+dots=-1/2$ and (3) is additive must also assign $0+1+2+dots=5/12$ and thus is not (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up?
                  $endgroup$
                  – Mario Carneiro
                  Feb 19 '15 at 7:44














                190












                190








                190





                $begingroup$

                there are many ways to see that your result is the right one. What does the right one mean?



                It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.



                It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.



                Exponential regulator method



                Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes
                $$ S = sum_{n=1}^{infty} n e^{-epsilon n} $$
                Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $epsilonto 0$ limit of the formula for $S$. In physics, $epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.



                The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself)
                $$ S = frac{e^epsilon}{(e^epsilon-1)^2} $$
                We will only need some Laurent expansion around $epsilon = 0$.
                $$ S = frac{1+epsilon+epsilon^2/2 + O(epsilon^3)}{(epsilon+epsilon^2/2+epsilon^3/6+O(epsilon^4))^2} $$
                We have
                $$ S = frac{1}{epsilon^2} frac{1+epsilon+epsilon^2/2+O(epsilon^3)}{(1+epsilon/2+epsilon^2/6+O(epsilon^3))^2} $$
                You see that the $1/epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
                1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]



                and the result is
                $$ frac{1}{epsilon^2} - frac{1}{12} + frac{epsilon^2}{240} + O(epsilon^4) $$
                In the $epsilonto 0$ limit we were interested in, the $epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.



                In physics applications, $epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $exp(-(epsilon+epsilon^2)n)$.



                There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.



                Euler's method



                Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.



                We will work with two sums,
                $$ S = 1+2+3+4+5+dots, quad T = 1-2+3-4+5-dots $$
                Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion
                $$ frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + dots $$
                Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now:
                $$ T = (1+2+3+dots) - 2times (2+4+6+dots) = (1+2+3+dots) (1 - 4) = -3S$$
                so $S=-T/3=-1/12$.



                A zeta-function calculation



                A somewhat unusual calculation of $zeta(-1)=-1/12$ of mine may be found here:




                http://www.kolej.mff.cuni.cz/~lmotm275/RUZE/09/node7.html




                The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.






                share|cite|improve this answer











                $endgroup$



                there are many ways to see that your result is the right one. What does the right one mean?



                It means that whenever such a sum appears anywhere in physics - I explicitly emphasize that not just in string theory, also in experimentally doable measurements of the Casimir force (between parallel metals resulting from quantized standing electromagnetic waves in between) - and one knows that the result is finite, the only possible finite part of the result that may be consistent with other symmetries of the problem (and that is actually confirmed experimentally whenever it is possible) is equal to $-1/12$.



                It's another widespread misconception (see all the incorrect comments right below your question) that the zeta-function regularization is the only way how to calculate the proper value. Let me show a completely different calculation - one that is a homework exercise in Joe Polchinski's "String Theory" textbook.



                Exponential regulator method



                Add an exponentially decreasing regulator to make the sum convergent - so that the sum becomes
                $$ S = sum_{n=1}^{infty} n e^{-epsilon n} $$
                Note that this is not equivalent to generalizing the sum to the zeta-function. In the zeta-function, the $n$ is the base that is exponentiated to the $s$th power. Here, the regulator has $n$ in the exponent. Obviously, the original sum of natural numbers is obtained in the $epsilonto 0$ limit of the formula for $S$. In physics, $epsilon$ would be viewed as a kind of "minimum distance" that can be resolved.



                The sum above may be exactly evaluated and the result is (use Mathematica if you don't want to do it yourself, but you can do it yourself)
                $$ S = frac{e^epsilon}{(e^epsilon-1)^2} $$
                We will only need some Laurent expansion around $epsilon = 0$.
                $$ S = frac{1+epsilon+epsilon^2/2 + O(epsilon^3)}{(epsilon+epsilon^2/2+epsilon^3/6+O(epsilon^4))^2} $$
                We have
                $$ S = frac{1}{epsilon^2} frac{1+epsilon+epsilon^2/2+O(epsilon^3)}{(1+epsilon/2+epsilon^2/6+O(epsilon^3))^2} $$
                You see that the $1/epsilon^2$ leading divergence survives and the next subleading term cancels. The resulting expansion may be calculated with this Mathematica command
                1/epsilon^2 * Series[epsilon^2 Sum[n Exp[-n epsilon], {n, 1, Infinity}], {epsilon, 0, 5}]



                and the result is
                $$ frac{1}{epsilon^2} - frac{1}{12} + frac{epsilon^2}{240} + O(epsilon^4) $$
                In the $epsilonto 0$ limit we were interested in, the $epsilon^2/240$ term as well as the smaller ones go to zero and may be erased. The leading divergence $1/epsilon^2$ may be and must be canceled by a local counterterm - a vacuum energy term. This is true for the Casimir effect in electromagnetism (in this case, the cancelled pole may be interpreted as the sum of the zero-point energies in the case that no metals were bounding the region), zero-point energies in string theory, and everywhere else. The cancellation of the leading divergence is needed for physics to be finite - but one may guarantee that the counterterm won't affect the finite term, $-1/12$, which is the correct result of the sum.



                In physics applications, $epsilon$ would be dimensionful and its different powers are sharply separated and may be treated individually. That's why the local counterterms may eliminate the leading divergence but don't affect the finite part. That's also why you couldn't have used a more complex regulator, like $exp(-(epsilon+epsilon^2)n)$.



                There are many other, apparently inequivalent ways to compute the right value of the sum. It is not just the zeta function.



                Euler's method



                Let me present one more, slightly less modern, method that was used by Leonhard Euler to calculate that the sum of natural numbers is $-1/12$. It's of course a bit more heuristic but his heuristic approach showed that he had a good intuition and the derivation could be turned into a modern physics derivation, too.



                We will work with two sums,
                $$ S = 1+2+3+4+5+dots, quad T = 1-2+3-4+5-dots $$
                Extrapolating the geometric and similar sums to the divergent (and, in this case, marginally divergent) domain of values of $x$, the expression $T$ may be summed according to the Taylor expansion
                $$ frac{1}{(1+x)^2} = 1 - 2x + 3x^2 -4x^3 + dots $$
                Substitute $x=1$ to see that $T=+1/4$. The value of $S$ is easily calculated now:
                $$ T = (1+2+3+dots) - 2times (2+4+6+dots) = (1+2+3+dots) (1 - 4) = -3S$$
                so $S=-T/3=-1/12$.



                A zeta-function calculation



                A somewhat unusual calculation of $zeta(-1)=-1/12$ of mine may be found here:




                http://www.kolej.mff.cuni.cz/~lmotm275/RUZE/09/node7.html




                The comments are in Czech but the equations represent bulk of the language that really matters, so the Czech comments shouldn't be a problem. A new argument (subscript) $s$ is added to the zeta function. The new function is the old zeta function for $s=0$ and for $s=1$, it only differs by one. We Taylor expand around $s=0$ to get to $s=1$ and we find out that only a finite number of terms survives if the main argument $x$ is a non-positive integer. The resulting recursive relations for the zeta function allow us to compute the values of the zeta-function at integers smaller than $1$, and prove that the function vanishes at negative even values of $x$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 29 '17 at 4:12









                Mehrdad

                6,62463778




                6,62463778










                answered May 18 '11 at 5:54









                Luboš MotlLuboš Motl

                7,02611625




                7,02611625








                • 138




                  $begingroup$
                  The fact that other methods also yield the result $-frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:04






                • 12




                  $begingroup$
                  @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics.
                  $endgroup$
                  – Luboš Motl
                  May 18 '11 at 6:43






                • 61




                  $begingroup$
                  @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:47








                • 58




                  $begingroup$
                  @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong.
                  $endgroup$
                  – KCd
                  May 18 '11 at 7:59






                • 8




                  $begingroup$
                  @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+dots=-1/12$ and (2) $1+1+1+dots=-1/2$ and (3) is additive must also assign $0+1+2+dots=5/12$ and thus is not (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up?
                  $endgroup$
                  – Mario Carneiro
                  Feb 19 '15 at 7:44














                • 138




                  $begingroup$
                  The fact that other methods also yield the result $-frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:04






                • 12




                  $begingroup$
                  @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics.
                  $endgroup$
                  – Luboš Motl
                  May 18 '11 at 6:43






                • 61




                  $begingroup$
                  @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:47








                • 58




                  $begingroup$
                  @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong.
                  $endgroup$
                  – KCd
                  May 18 '11 at 7:59






                • 8




                  $begingroup$
                  @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+dots=-1/12$ and (2) $1+1+1+dots=-1/2$ and (3) is additive must also assign $0+1+2+dots=5/12$ and thus is not (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up?
                  $endgroup$
                  – Mario Carneiro
                  Feb 19 '15 at 7:44








                138




                138




                $begingroup$
                The fact that other methods also yield the result $-frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is.
                $endgroup$
                – joriki
                May 18 '11 at 6:04




                $begingroup$
                The fact that other methods also yield the result $-frac{1}{12}$ doesn't make any of the comments under the question "incorrect". You interpreted "right answer" to mean the one that works in physical applications; that's fine (and I suggested in my comment that it was probably meant that way), but that doesn't make it the right answer in any mathematical sense of the term, and nor does the consistency of several summation methods. I'm not aware of any theory of summation of divergent series that provides a definition of what "the right answer" for a resummation is.
                $endgroup$
                – joriki
                May 18 '11 at 6:04




                12




                12




                $begingroup$
                @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics.
                $endgroup$
                – Luboš Motl
                May 18 '11 at 6:43




                $begingroup$
                @Joriki, the right answer -1/12 is a deep mathematical result that is relevant not only in physics but also in any branch of maths that cares about the deep relationships between structures such as sums, functions of complex variables, and many other things. The convergent result is not the right result given the most naive mathematical ways to interpret the sum - as a limit of partial sums but it is surely the result of all the most mathematically profound interpretations of the sum. That's also why Euler who was not quite silly knew that the right sum was $-1/12$ long before quantum physics.
                $endgroup$
                – Luboš Motl
                May 18 '11 at 6:43




                61




                61




                $begingroup$
                @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't.
                $endgroup$
                – joriki
                May 18 '11 at 6:47






                $begingroup$
                @Luboš: It seems we have a difference of opinion about the use of "right" in this context. I don't object to you using "right" in the sense that you're using it; what I object to is labelling other views as "incorrect". To call something "incorrect", you'd need a definition of what makes an answer the right answer. As long as you don't have one, it's a matter of taste that you call the consistent result of several different methods "the right answer" and I don't.
                $endgroup$
                – joriki
                May 18 '11 at 6:47






                58




                58




                $begingroup$
                @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong.
                $endgroup$
                – KCd
                May 18 '11 at 7:59




                $begingroup$
                @Luboš: I too object to your label of comments by others as "shallow". Look at the question: it asks specifically about analytically continuing and writes the value as $zeta(-1)$. That is what I was responding to in my comment. I am well aware that there are other methods of "deriving" the value -1/12 and I had sketched Euler's technique a while ago on Mathoverflow (see my 2nd answer to mathoverflow.net/questions/13130/…). You answered a better question than the one that was asked. It's not a reason to say more focused answers are wrong.
                $endgroup$
                – KCd
                May 18 '11 at 7:59




                8




                8




                $begingroup$
                @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+dots=-1/12$ and (2) $1+1+1+dots=-1/2$ and (3) is additive must also assign $0+1+2+dots=5/12$ and thus is not (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up?
                $endgroup$
                – Mario Carneiro
                Feb 19 '15 at 7:44




                $begingroup$
                @LubošMotl As @${}$Axel points out, any summability method that gives (1) $1+2+3+dots=-1/12$ and (2) $1+1+1+dots=-1/2$ and (3) is additive must also assign $0+1+2+dots=5/12$ and thus is not (4) shift invariant, so holding all beliefs (1,2,3,4) leads to contradiction. Which one are you giving up?
                $endgroup$
                – Mario Carneiro
                Feb 19 '15 at 7:44











                107












                $begingroup$

                Here is a variant on Lubos Motl's answer:



                Let $S = sum_{n=1}^{infty} n$. Then $S - 4 S = sum_{n = 1}^{infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's.



                Namely, consider $$sum_{n=1}^{infty} (-1)^{n-1} n t^n
                = -t dfrac{d}{dt} sum_{n=1}^{infty} (-t)^n = -t dfrac{d}{dt} dfrac{1}{1+t}
                = dfrac{t}{(1+t)^2}.$$

                Letting $t to 1,$ we find that $-3 S = dfrac{1}{4}$, and hence that
                $S = dfrac{-1}{12}.$





                To see the relationship between this approach and Lubos Motl's, note that if
                we write $t = e^{epsilon},$ then $tdfrac{d}{dt} = dfrac{d}{depsilon},$
                so in fact the arguments are essentially the same, except that Lubos doesn't
                perform the initial step of replacing $S$ by $S - 4S$, which means that
                he has the pole $dfrac{1}{epsilon^2}$ which he then subtracts away.





                As far as I know, this trick of replacing $zeta(s)$ by $(1-2^{-s+1})^{-1}zeta(s)$ is due to Euler, and it is a now standard method for replacing
                $zeta(s)$ by a function which carries the same information, but
                does not have a pole at $s = 1$. The evaluation of $zeta(s)$ at negative integers by passing to $(1-2^{-s+1})zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see
                the Bernoulli numbers appearing in this way, for example.





                Of course, taken literally, the series $sum_{n=1}^{infty} n$ diverges to $+infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $sum_{n = 1}^{infty} n,$
                I'm reasonably confident that any sensible regularization will necessarily yield the same value of $dfrac{-1}{12}$. (Lubos Motl makes the same assertion in his answer.)






                share|cite|improve this answer











                $endgroup$









                • 8




                  $begingroup$
                  @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+dotso$ or $-1-1-dotso$ First, these should be different. Also, $1+1+dotso=zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:57






                • 5




                  $begingroup$
                  @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 7:31








                • 37




                  $begingroup$
                  @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 9:55






                • 10




                  $begingroup$
                  @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards,
                  $endgroup$
                  – Matt E
                  May 18 '11 at 11:58






                • 5




                  $begingroup$
                  I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+cdots$ and $0+1+0+2+0+3+0+cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+cdots$ has a different value (namely $1/24$).
                  $endgroup$
                  – Axel Boldt
                  Mar 9 '14 at 3:10
















                107












                $begingroup$

                Here is a variant on Lubos Motl's answer:



                Let $S = sum_{n=1}^{infty} n$. Then $S - 4 S = sum_{n = 1}^{infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's.



                Namely, consider $$sum_{n=1}^{infty} (-1)^{n-1} n t^n
                = -t dfrac{d}{dt} sum_{n=1}^{infty} (-t)^n = -t dfrac{d}{dt} dfrac{1}{1+t}
                = dfrac{t}{(1+t)^2}.$$

                Letting $t to 1,$ we find that $-3 S = dfrac{1}{4}$, and hence that
                $S = dfrac{-1}{12}.$





                To see the relationship between this approach and Lubos Motl's, note that if
                we write $t = e^{epsilon},$ then $tdfrac{d}{dt} = dfrac{d}{depsilon},$
                so in fact the arguments are essentially the same, except that Lubos doesn't
                perform the initial step of replacing $S$ by $S - 4S$, which means that
                he has the pole $dfrac{1}{epsilon^2}$ which he then subtracts away.





                As far as I know, this trick of replacing $zeta(s)$ by $(1-2^{-s+1})^{-1}zeta(s)$ is due to Euler, and it is a now standard method for replacing
                $zeta(s)$ by a function which carries the same information, but
                does not have a pole at $s = 1$. The evaluation of $zeta(s)$ at negative integers by passing to $(1-2^{-s+1})zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see
                the Bernoulli numbers appearing in this way, for example.





                Of course, taken literally, the series $sum_{n=1}^{infty} n$ diverges to $+infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $sum_{n = 1}^{infty} n,$
                I'm reasonably confident that any sensible regularization will necessarily yield the same value of $dfrac{-1}{12}$. (Lubos Motl makes the same assertion in his answer.)






                share|cite|improve this answer











                $endgroup$









                • 8




                  $begingroup$
                  @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+dotso$ or $-1-1-dotso$ First, these should be different. Also, $1+1+dotso=zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:57






                • 5




                  $begingroup$
                  @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 7:31








                • 37




                  $begingroup$
                  @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 9:55






                • 10




                  $begingroup$
                  @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards,
                  $endgroup$
                  – Matt E
                  May 18 '11 at 11:58






                • 5




                  $begingroup$
                  I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+cdots$ and $0+1+0+2+0+3+0+cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+cdots$ has a different value (namely $1/24$).
                  $endgroup$
                  – Axel Boldt
                  Mar 9 '14 at 3:10














                107












                107








                107





                $begingroup$

                Here is a variant on Lubos Motl's answer:



                Let $S = sum_{n=1}^{infty} n$. Then $S - 4 S = sum_{n = 1}^{infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's.



                Namely, consider $$sum_{n=1}^{infty} (-1)^{n-1} n t^n
                = -t dfrac{d}{dt} sum_{n=1}^{infty} (-t)^n = -t dfrac{d}{dt} dfrac{1}{1+t}
                = dfrac{t}{(1+t)^2}.$$

                Letting $t to 1,$ we find that $-3 S = dfrac{1}{4}$, and hence that
                $S = dfrac{-1}{12}.$





                To see the relationship between this approach and Lubos Motl's, note that if
                we write $t = e^{epsilon},$ then $tdfrac{d}{dt} = dfrac{d}{depsilon},$
                so in fact the arguments are essentially the same, except that Lubos doesn't
                perform the initial step of replacing $S$ by $S - 4S$, which means that
                he has the pole $dfrac{1}{epsilon^2}$ which he then subtracts away.





                As far as I know, this trick of replacing $zeta(s)$ by $(1-2^{-s+1})^{-1}zeta(s)$ is due to Euler, and it is a now standard method for replacing
                $zeta(s)$ by a function which carries the same information, but
                does not have a pole at $s = 1$. The evaluation of $zeta(s)$ at negative integers by passing to $(1-2^{-s+1})zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see
                the Bernoulli numbers appearing in this way, for example.





                Of course, taken literally, the series $sum_{n=1}^{infty} n$ diverges to $+infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $sum_{n = 1}^{infty} n,$
                I'm reasonably confident that any sensible regularization will necessarily yield the same value of $dfrac{-1}{12}$. (Lubos Motl makes the same assertion in his answer.)






                share|cite|improve this answer











                $endgroup$



                Here is a variant on Lubos Motl's answer:



                Let $S = sum_{n=1}^{infty} n$. Then $S - 4 S = sum_{n = 1}^{infty} (-1)^{n-1} n.$ We will evaluate this latter expression with a regularization similar to Lubos Motl's.



                Namely, consider $$sum_{n=1}^{infty} (-1)^{n-1} n t^n
                = -t dfrac{d}{dt} sum_{n=1}^{infty} (-t)^n = -t dfrac{d}{dt} dfrac{1}{1+t}
                = dfrac{t}{(1+t)^2}.$$

                Letting $t to 1,$ we find that $-3 S = dfrac{1}{4}$, and hence that
                $S = dfrac{-1}{12}.$





                To see the relationship between this approach and Lubos Motl's, note that if
                we write $t = e^{epsilon},$ then $tdfrac{d}{dt} = dfrac{d}{depsilon},$
                so in fact the arguments are essentially the same, except that Lubos doesn't
                perform the initial step of replacing $S$ by $S - 4S$, which means that
                he has the pole $dfrac{1}{epsilon^2}$ which he then subtracts away.





                As far as I know, this trick of replacing $zeta(s)$ by $(1-2^{-s+1})^{-1}zeta(s)$ is due to Euler, and it is a now standard method for replacing
                $zeta(s)$ by a function which carries the same information, but
                does not have a pole at $s = 1$. The evaluation of $zeta(s)$ at negative integers by passing to $(1-2^{-s+1})zeta(s)$ and then performing Abelian regularization as above is also due to Euler, I believe. It is easy to see
                the Bernoulli numbers appearing in this way, for example.





                Of course, taken literally, the series $sum_{n=1}^{infty} n$ diverges to $+infty$, so any attempt to assign it a finite value will involve some form of regularization. Analytic continuation of the $zeta$-function is one form of regularization, and the Abelian regularization that Lubos Motl and I are making is another. I can't quote a precise theorem to this effect (although maybe others can), but with such a simple expression as $sum_{n = 1}^{infty} n,$
                I'm reasonably confident that any sensible regularization will necessarily yield the same value of $dfrac{-1}{12}$. (Lubos Motl makes the same assertion in his answer.)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 1 '15 at 16:17









                amWhy

                192k28225439




                192k28225439










                answered May 18 '11 at 6:10









                Matt EMatt E

                104k8217385




                104k8217385








                • 8




                  $begingroup$
                  @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+dotso$ or $-1-1-dotso$ First, these should be different. Also, $1+1+dotso=zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:57






                • 5




                  $begingroup$
                  @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 7:31








                • 37




                  $begingroup$
                  @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 9:55






                • 10




                  $begingroup$
                  @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards,
                  $endgroup$
                  – Matt E
                  May 18 '11 at 11:58






                • 5




                  $begingroup$
                  I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+cdots$ and $0+1+0+2+0+3+0+cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+cdots$ has a different value (namely $1/24$).
                  $endgroup$
                  – Axel Boldt
                  Mar 9 '14 at 3:10














                • 8




                  $begingroup$
                  @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+dotso$ or $-1-1-dotso$ First, these should be different. Also, $1+1+dotso=zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$.
                  $endgroup$
                  – joriki
                  May 18 '11 at 6:57






                • 5




                  $begingroup$
                  @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 7:31








                • 37




                  $begingroup$
                  @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.)
                  $endgroup$
                  – joriki
                  May 18 '11 at 9:55






                • 10




                  $begingroup$
                  @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards,
                  $endgroup$
                  – Matt E
                  May 18 '11 at 11:58






                • 5




                  $begingroup$
                  I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+cdots$ and $0+1+0+2+0+3+0+cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+cdots$ has a different value (namely $1/24$).
                  $endgroup$
                  – Axel Boldt
                  Mar 9 '14 at 3:10








                8




                8




                $begingroup$
                @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+dotso$ or $-1-1-dotso$ First, these should be different. Also, $1+1+dotso=zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$.
                $endgroup$
                – joriki
                May 18 '11 at 6:57




                $begingroup$
                @Matt: My first example was flawed, but I think my general point still stands that a satisfactory theory of summation of divergent series would have to explain not only why several seemingly different methods yield the same result, but also why other methods don't. Here's a new example (hopefully less flawed): $S-2S=1+3+5+dotso$ Now we can subtract $2S$ again, and depending on whether we start substracting the $2$ from the $3$ or the $1$, we get $1+1+dotso$ or $-1-1-dotso$ First, these should be different. Also, $1+1+dotso=zeta(0)=-1/2$, so that would yield $S-4S=-1/2$, $S=1/6$.
                $endgroup$
                – joriki
                May 18 '11 at 6:57




                5




                5




                $begingroup$
                @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.)
                $endgroup$
                – joriki
                May 18 '11 at 7:31






                $begingroup$
                @Matt: (I'm not saying that this can't be explained, or that the results that do coincide do so by chance; to the contrary, I find it intriguing how many of them coincide; all I'm saying is that the ones that don't coincide are also part of the picture, and a satisfactory theory (one that would in my opinion allow us to speak of "the right answer") would have to have something to say about which of these methods are "valid", which aren't and why.)
                $endgroup$
                – joriki
                May 18 '11 at 7:31






                37




                37




                $begingroup$
                @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.)
                $endgroup$
                – joriki
                May 18 '11 at 9:55




                $begingroup$
                @Luboš: Instead of insisting on using terms like "simple errors" "in reality" and "wrong interpretations", it would be more helpful if you could specifically point out in which sense you believe that your calculation of $S-2S-2S$ has more merit than mine. As I emphasized, it seems quite likely to me that there is such a difference; but I was looking for an explanation. How did I "randomly clump" terms in my calculation? In which framework does your combination of the terms appear more systematic than mine? (Not rhetorical questions.)
                $endgroup$
                – joriki
                May 18 '11 at 9:55




                10




                10




                $begingroup$
                @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards,
                $endgroup$
                – Matt E
                May 18 '11 at 11:58




                $begingroup$
                @joriki: Dear Joriki, I deleted my comment that was made in reference to your deleted comment. As for your revised example (obtaining $zeta(0)$), there is a "graded" aspect to $S$, which is being disturbed in your computation. (This is what Lubos is referring to when he writes that you "randomly clump" terms.) In terms of Lubos's answer, this disturbing of the grading is not allowed because of dimensional considerations. Mathematically, all I can think to say right now is that both $zeta$-regularization and Euler regularization involve the graded aspect of $S$, and preserve it. Regards,
                $endgroup$
                – Matt E
                May 18 '11 at 11:58




                5




                5




                $begingroup$
                I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+cdots$ and $0+1+0+2+0+3+0+cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+cdots$ has a different value (namely $1/24$).
                $endgroup$
                – Axel Boldt
                Mar 9 '14 at 3:10




                $begingroup$
                I don't like the first step of your derivation very much, where you compute $S-4S$. Implicitly, this uses the fact that the sums $1+2+3+cdots$ and $0+1+0+2+0+3+0+cdots$ have the same value. While true, this is not immediately obvious. For instance, the sum $1+0+2+0+3+0+cdots$ has a different value (namely $1/24$).
                $endgroup$
                – Axel Boldt
                Mar 9 '14 at 3:10











                71












                $begingroup$

                What a great method! I tried a few...
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n)^{s}} - frac{1}{(2 n - 1)^{s}}right) = zeta (s) Bigl(2^{(1 - s)} - 1Bigr)
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{1}{4}$



                or
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n + 1)^{s}} - frac{1}{(2 n)^{s}}right) = zeta (s) - frac{2 zeta (s)}{2^{s}} - 1
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{3}{4}$






                share|cite|improve this answer









                $endgroup$









                • 7




                  $begingroup$
                  You should read Terence Tao's article on this "oddities".
                  $endgroup$
                  – Pedro Tamaroff
                  Jun 11 '12 at 17:37






                • 11




                  $begingroup$
                  For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question.
                  $endgroup$
                  – Marnix Klooster
                  Jan 16 '14 at 20:04






                • 1




                  $begingroup$
                  Well this is exactly about an answer I just posted
                  $endgroup$
                  – reuns
                  Nov 1 '16 at 6:36


















                71












                $begingroup$

                What a great method! I tried a few...
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n)^{s}} - frac{1}{(2 n - 1)^{s}}right) = zeta (s) Bigl(2^{(1 - s)} - 1Bigr)
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{1}{4}$



                or
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n + 1)^{s}} - frac{1}{(2 n)^{s}}right) = zeta (s) - frac{2 zeta (s)}{2^{s}} - 1
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{3}{4}$






                share|cite|improve this answer









                $endgroup$









                • 7




                  $begingroup$
                  You should read Terence Tao's article on this "oddities".
                  $endgroup$
                  – Pedro Tamaroff
                  Jun 11 '12 at 17:37






                • 11




                  $begingroup$
                  For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question.
                  $endgroup$
                  – Marnix Klooster
                  Jan 16 '14 at 20:04






                • 1




                  $begingroup$
                  Well this is exactly about an answer I just posted
                  $endgroup$
                  – reuns
                  Nov 1 '16 at 6:36
















                71












                71








                71





                $begingroup$

                What a great method! I tried a few...
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n)^{s}} - frac{1}{(2 n - 1)^{s}}right) = zeta (s) Bigl(2^{(1 - s)} - 1Bigr)
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{1}{4}$



                or
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n + 1)^{s}} - frac{1}{(2 n)^{s}}right) = zeta (s) - frac{2 zeta (s)}{2^{s}} - 1
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{3}{4}$






                share|cite|improve this answer









                $endgroup$



                What a great method! I tried a few...
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n)^{s}} - frac{1}{(2 n - 1)^{s}}right) = zeta (s) Bigl(2^{(1 - s)} - 1Bigr)
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{1}{4}$



                or
                $$
                sum_{n = 1}^{infty} left(frac{1}{(2 n + 1)^{s}} - frac{1}{(2 n)^{s}}right) = zeta (s) - frac{2 zeta (s)}{2^{s}} - 1
                $$
                put $s=-1$ to get $sum_{n = 1}^{infty} 1 = -frac{3}{4}$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered May 18 '11 at 12:58









                GEdgarGEdgar

                62k267168




                62k267168








                • 7




                  $begingroup$
                  You should read Terence Tao's article on this "oddities".
                  $endgroup$
                  – Pedro Tamaroff
                  Jun 11 '12 at 17:37






                • 11




                  $begingroup$
                  For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question.
                  $endgroup$
                  – Marnix Klooster
                  Jan 16 '14 at 20:04






                • 1




                  $begingroup$
                  Well this is exactly about an answer I just posted
                  $endgroup$
                  – reuns
                  Nov 1 '16 at 6:36
















                • 7




                  $begingroup$
                  You should read Terence Tao's article on this "oddities".
                  $endgroup$
                  – Pedro Tamaroff
                  Jun 11 '12 at 17:37






                • 11




                  $begingroup$
                  For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question.
                  $endgroup$
                  – Marnix Klooster
                  Jan 16 '14 at 20:04






                • 1




                  $begingroup$
                  Well this is exactly about an answer I just posted
                  $endgroup$
                  – reuns
                  Nov 1 '16 at 6:36










                7




                7




                $begingroup$
                You should read Terence Tao's article on this "oddities".
                $endgroup$
                – Pedro Tamaroff
                Jun 11 '12 at 17:37




                $begingroup$
                You should read Terence Tao's article on this "oddities".
                $endgroup$
                – Pedro Tamaroff
                Jun 11 '12 at 17:37




                11




                11




                $begingroup$
                For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question.
                $endgroup$
                – Marnix Klooster
                Jan 16 '14 at 20:04




                $begingroup$
                For clarity: I'm assuming you mean terrytao.wordpress.com/2010/04/10/… (which I found through JavaMan's comment on the question.
                $endgroup$
                – Marnix Klooster
                Jan 16 '14 at 20:04




                1




                1




                $begingroup$
                Well this is exactly about an answer I just posted
                $endgroup$
                – reuns
                Nov 1 '16 at 6:36






                $begingroup$
                Well this is exactly about an answer I just posted
                $endgroup$
                – reuns
                Nov 1 '16 at 6:36













                52












                $begingroup$

                Here is a direct computation of $zeta(-1)$ taken from this answer:



                Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get
                $$
                begin{align}
                (1-2^{1-x})zeta(x)Gamma(x+2)
                &=int_0^inftyfrac{(x+1)xt^{x-1}}{e^t+1}mathrm{d}t\
                &=int_0^inftyfrac{(x+1)t^xe^t}{(e^t+1)^2}mathrm{d}t\
                &=int_0^inftyfrac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}mathrm{d}ttag{1}
                end{align}
                $$
                Now we can plug in $x=-1$ into $(1)$ to get
                $$
                begin{align}
                (1-2^2)zeta(-1)Gamma(1)
                &=int_0^inftyfrac{e^{2t}-e^t}{(e^t+1)^3}mathrm{d}t\
                &=int_1^inftyfrac{u-1}{(u+1)^3}mathrm{d}u\
                &=int_1^inftyleft(frac1{(u+1)^2}-frac2{(u+1)^3}right)mathrm{d}u\
                &=frac14tag{2}
                end{align}
                $$
                Since $(1-2^2)Gamma(1)=-3$, $(2)$ says that
                $$
                zeta(-1)=-frac1{12}tag{3}
                $$
                Naively plugging $n=-1$ into
                $$
                zeta(n)=frac1{1^n}+frac1{2^n}+frac1{3^n}+dotstag{4}
                $$
                yields
                $$
                -frac1{12}=1+2+3+4+dotstag{5}
                $$
                However, the series on the right of $(5)$ is obviously divergent by the Term Test.






                share|cite|improve this answer











                $endgroup$


















                  52












                  $begingroup$

                  Here is a direct computation of $zeta(-1)$ taken from this answer:



                  Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get
                  $$
                  begin{align}
                  (1-2^{1-x})zeta(x)Gamma(x+2)
                  &=int_0^inftyfrac{(x+1)xt^{x-1}}{e^t+1}mathrm{d}t\
                  &=int_0^inftyfrac{(x+1)t^xe^t}{(e^t+1)^2}mathrm{d}t\
                  &=int_0^inftyfrac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}mathrm{d}ttag{1}
                  end{align}
                  $$
                  Now we can plug in $x=-1$ into $(1)$ to get
                  $$
                  begin{align}
                  (1-2^2)zeta(-1)Gamma(1)
                  &=int_0^inftyfrac{e^{2t}-e^t}{(e^t+1)^3}mathrm{d}t\
                  &=int_1^inftyfrac{u-1}{(u+1)^3}mathrm{d}u\
                  &=int_1^inftyleft(frac1{(u+1)^2}-frac2{(u+1)^3}right)mathrm{d}u\
                  &=frac14tag{2}
                  end{align}
                  $$
                  Since $(1-2^2)Gamma(1)=-3$, $(2)$ says that
                  $$
                  zeta(-1)=-frac1{12}tag{3}
                  $$
                  Naively plugging $n=-1$ into
                  $$
                  zeta(n)=frac1{1^n}+frac1{2^n}+frac1{3^n}+dotstag{4}
                  $$
                  yields
                  $$
                  -frac1{12}=1+2+3+4+dotstag{5}
                  $$
                  However, the series on the right of $(5)$ is obviously divergent by the Term Test.






                  share|cite|improve this answer











                  $endgroup$
















                    52












                    52








                    52





                    $begingroup$

                    Here is a direct computation of $zeta(-1)$ taken from this answer:



                    Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get
                    $$
                    begin{align}
                    (1-2^{1-x})zeta(x)Gamma(x+2)
                    &=int_0^inftyfrac{(x+1)xt^{x-1}}{e^t+1}mathrm{d}t\
                    &=int_0^inftyfrac{(x+1)t^xe^t}{(e^t+1)^2}mathrm{d}t\
                    &=int_0^inftyfrac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}mathrm{d}ttag{1}
                    end{align}
                    $$
                    Now we can plug in $x=-1$ into $(1)$ to get
                    $$
                    begin{align}
                    (1-2^2)zeta(-1)Gamma(1)
                    &=int_0^inftyfrac{e^{2t}-e^t}{(e^t+1)^3}mathrm{d}t\
                    &=int_1^inftyfrac{u-1}{(u+1)^3}mathrm{d}u\
                    &=int_1^inftyleft(frac1{(u+1)^2}-frac2{(u+1)^3}right)mathrm{d}u\
                    &=frac14tag{2}
                    end{align}
                    $$
                    Since $(1-2^2)Gamma(1)=-3$, $(2)$ says that
                    $$
                    zeta(-1)=-frac1{12}tag{3}
                    $$
                    Naively plugging $n=-1$ into
                    $$
                    zeta(n)=frac1{1^n}+frac1{2^n}+frac1{3^n}+dotstag{4}
                    $$
                    yields
                    $$
                    -frac1{12}=1+2+3+4+dotstag{5}
                    $$
                    However, the series on the right of $(5)$ is obviously divergent by the Term Test.






                    share|cite|improve this answer











                    $endgroup$



                    Here is a direct computation of $zeta(-1)$ taken from this answer:



                    Multiply equation $(1)$ from this answer by $x+1$, then integrate by parts twice, to get
                    $$
                    begin{align}
                    (1-2^{1-x})zeta(x)Gamma(x+2)
                    &=int_0^inftyfrac{(x+1)xt^{x-1}}{e^t+1}mathrm{d}t\
                    &=int_0^inftyfrac{(x+1)t^xe^t}{(e^t+1)^2}mathrm{d}t\
                    &=int_0^inftyfrac{t^{x+1}(e^{2t}-e^t)}{(e^t+1)^3}mathrm{d}ttag{1}
                    end{align}
                    $$
                    Now we can plug in $x=-1$ into $(1)$ to get
                    $$
                    begin{align}
                    (1-2^2)zeta(-1)Gamma(1)
                    &=int_0^inftyfrac{e^{2t}-e^t}{(e^t+1)^3}mathrm{d}t\
                    &=int_1^inftyfrac{u-1}{(u+1)^3}mathrm{d}u\
                    &=int_1^inftyleft(frac1{(u+1)^2}-frac2{(u+1)^3}right)mathrm{d}u\
                    &=frac14tag{2}
                    end{align}
                    $$
                    Since $(1-2^2)Gamma(1)=-3$, $(2)$ says that
                    $$
                    zeta(-1)=-frac1{12}tag{3}
                    $$
                    Naively plugging $n=-1$ into
                    $$
                    zeta(n)=frac1{1^n}+frac1{2^n}+frac1{3^n}+dotstag{4}
                    $$
                    yields
                    $$
                    -frac1{12}=1+2+3+4+dotstag{5}
                    $$
                    However, the series on the right of $(5)$ is obviously divergent by the Term Test.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Apr 13 '17 at 12:19









                    Community

                    1




                    1










                    answered Jan 9 '14 at 18:14









                    robjohnrobjohn

                    265k27304626




                    265k27304626























                        44












                        $begingroup$

                        A good explanation of how this is true and why and where it is useful and significant is beautifully summarized in 24 by John Baez (as part of the 19/09/08 Rankin Lectures) and yes, I do agree that this is wierd.





                        The following is the same Euler's method described by Luboš Motl and Matt E in the other answers.



                        Let us consider the infinite geometric progression
                        " $1, x, x^2, x^3, x^4, dots$ "



                        The sum of this progression can be found by the formula $S_infty = large frac{a}{1-r}$,



                        $$implies 1 + x + x^2 + x^3 + x^4 + dots = frac{1}{1-x}$$
                        Differentiating with respect to $x$,
                        $$ implies 0 + 1 + 2x + 3x^2 + 4x^3 + dots = frac{1}{(x-1)^2}$$
                        Let $x = -1$, (Note: In the LHS, you get the alternating counterpart of the natural numbers series)
                        $$implies 1-2+3-4+5-6+...= frac{1}{4} $$



                        $$ bbox[5px,border:2px solid green]{therefore sumlimits_{i=1}^infty n(-1)^{n-1} = frac{1}{4}} $$



                        Subtracting this from $sumlimits_{i=1}^infty n$,
                        $$sumlimits_{i=1}^infty n -sumlimits_{i=1}^infty n(-1)^{n-1} =4+8+12+...=4sumlimits_{i=1}^infty n$$
                        $$implies 3sumlimits_{i=1}^infty n =-sumlimits_{i=1}^infty n(-1)^{n-1}=-1/4$$
                        $$bbox[5px,border:2px solid lime]{therefore sumlimits_{i=1}^infty n=-frac{1}{12}}$$




                        This intuitive reasoning was made famous in a video by the physicist Phil Plait of Numberphile. It is heavily criticized for being sloppy but it shows a simple method that has fooled a large population of its high school audience (as noted in the comments).


                        Funfact: Googling -1/12 doesn't give you any results




                        share|cite|improve this answer











                        $endgroup$









                        • 17




                          $begingroup$
                          That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit)
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 18 '14 at 2:37








                        • 3




                          $begingroup$
                          This proof was featured today in a video posted to Reddit.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:31






                        • 8




                          $begingroup$
                          Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up.
                          $endgroup$
                          – Jeyekomon
                          Jan 18 '14 at 17:05






                        • 7




                          $begingroup$
                          @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature...
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 19 '14 at 6:27






                        • 4




                          $begingroup$
                          You manipulate that infinite sum as if it were finite, like when you assume that $sum_{n=1}^infty n-4sum_{n=1}^infty n=-3sum_{n=1}^infty n.$ But that's impossible, since $infty-infty$ is undefined.
                          $endgroup$
                          – Hakim
                          May 16 '14 at 20:52


















                        44












                        $begingroup$

                        A good explanation of how this is true and why and where it is useful and significant is beautifully summarized in 24 by John Baez (as part of the 19/09/08 Rankin Lectures) and yes, I do agree that this is wierd.





                        The following is the same Euler's method described by Luboš Motl and Matt E in the other answers.



                        Let us consider the infinite geometric progression
                        " $1, x, x^2, x^3, x^4, dots$ "



                        The sum of this progression can be found by the formula $S_infty = large frac{a}{1-r}$,



                        $$implies 1 + x + x^2 + x^3 + x^4 + dots = frac{1}{1-x}$$
                        Differentiating with respect to $x$,
                        $$ implies 0 + 1 + 2x + 3x^2 + 4x^3 + dots = frac{1}{(x-1)^2}$$
                        Let $x = -1$, (Note: In the LHS, you get the alternating counterpart of the natural numbers series)
                        $$implies 1-2+3-4+5-6+...= frac{1}{4} $$



                        $$ bbox[5px,border:2px solid green]{therefore sumlimits_{i=1}^infty n(-1)^{n-1} = frac{1}{4}} $$



                        Subtracting this from $sumlimits_{i=1}^infty n$,
                        $$sumlimits_{i=1}^infty n -sumlimits_{i=1}^infty n(-1)^{n-1} =4+8+12+...=4sumlimits_{i=1}^infty n$$
                        $$implies 3sumlimits_{i=1}^infty n =-sumlimits_{i=1}^infty n(-1)^{n-1}=-1/4$$
                        $$bbox[5px,border:2px solid lime]{therefore sumlimits_{i=1}^infty n=-frac{1}{12}}$$




                        This intuitive reasoning was made famous in a video by the physicist Phil Plait of Numberphile. It is heavily criticized for being sloppy but it shows a simple method that has fooled a large population of its high school audience (as noted in the comments).


                        Funfact: Googling -1/12 doesn't give you any results




                        share|cite|improve this answer











                        $endgroup$









                        • 17




                          $begingroup$
                          That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit)
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 18 '14 at 2:37








                        • 3




                          $begingroup$
                          This proof was featured today in a video posted to Reddit.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:31






                        • 8




                          $begingroup$
                          Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up.
                          $endgroup$
                          – Jeyekomon
                          Jan 18 '14 at 17:05






                        • 7




                          $begingroup$
                          @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature...
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 19 '14 at 6:27






                        • 4




                          $begingroup$
                          You manipulate that infinite sum as if it were finite, like when you assume that $sum_{n=1}^infty n-4sum_{n=1}^infty n=-3sum_{n=1}^infty n.$ But that's impossible, since $infty-infty$ is undefined.
                          $endgroup$
                          – Hakim
                          May 16 '14 at 20:52
















                        44












                        44








                        44





                        $begingroup$

                        A good explanation of how this is true and why and where it is useful and significant is beautifully summarized in 24 by John Baez (as part of the 19/09/08 Rankin Lectures) and yes, I do agree that this is wierd.





                        The following is the same Euler's method described by Luboš Motl and Matt E in the other answers.



                        Let us consider the infinite geometric progression
                        " $1, x, x^2, x^3, x^4, dots$ "



                        The sum of this progression can be found by the formula $S_infty = large frac{a}{1-r}$,



                        $$implies 1 + x + x^2 + x^3 + x^4 + dots = frac{1}{1-x}$$
                        Differentiating with respect to $x$,
                        $$ implies 0 + 1 + 2x + 3x^2 + 4x^3 + dots = frac{1}{(x-1)^2}$$
                        Let $x = -1$, (Note: In the LHS, you get the alternating counterpart of the natural numbers series)
                        $$implies 1-2+3-4+5-6+...= frac{1}{4} $$



                        $$ bbox[5px,border:2px solid green]{therefore sumlimits_{i=1}^infty n(-1)^{n-1} = frac{1}{4}} $$



                        Subtracting this from $sumlimits_{i=1}^infty n$,
                        $$sumlimits_{i=1}^infty n -sumlimits_{i=1}^infty n(-1)^{n-1} =4+8+12+...=4sumlimits_{i=1}^infty n$$
                        $$implies 3sumlimits_{i=1}^infty n =-sumlimits_{i=1}^infty n(-1)^{n-1}=-1/4$$
                        $$bbox[5px,border:2px solid lime]{therefore sumlimits_{i=1}^infty n=-frac{1}{12}}$$




                        This intuitive reasoning was made famous in a video by the physicist Phil Plait of Numberphile. It is heavily criticized for being sloppy but it shows a simple method that has fooled a large population of its high school audience (as noted in the comments).


                        Funfact: Googling -1/12 doesn't give you any results




                        share|cite|improve this answer











                        $endgroup$



                        A good explanation of how this is true and why and where it is useful and significant is beautifully summarized in 24 by John Baez (as part of the 19/09/08 Rankin Lectures) and yes, I do agree that this is wierd.





                        The following is the same Euler's method described by Luboš Motl and Matt E in the other answers.



                        Let us consider the infinite geometric progression
                        " $1, x, x^2, x^3, x^4, dots$ "



                        The sum of this progression can be found by the formula $S_infty = large frac{a}{1-r}$,



                        $$implies 1 + x + x^2 + x^3 + x^4 + dots = frac{1}{1-x}$$
                        Differentiating with respect to $x$,
                        $$ implies 0 + 1 + 2x + 3x^2 + 4x^3 + dots = frac{1}{(x-1)^2}$$
                        Let $x = -1$, (Note: In the LHS, you get the alternating counterpart of the natural numbers series)
                        $$implies 1-2+3-4+5-6+...= frac{1}{4} $$



                        $$ bbox[5px,border:2px solid green]{therefore sumlimits_{i=1}^infty n(-1)^{n-1} = frac{1}{4}} $$



                        Subtracting this from $sumlimits_{i=1}^infty n$,
                        $$sumlimits_{i=1}^infty n -sumlimits_{i=1}^infty n(-1)^{n-1} =4+8+12+...=4sumlimits_{i=1}^infty n$$
                        $$implies 3sumlimits_{i=1}^infty n =-sumlimits_{i=1}^infty n(-1)^{n-1}=-1/4$$
                        $$bbox[5px,border:2px solid lime]{therefore sumlimits_{i=1}^infty n=-frac{1}{12}}$$




                        This intuitive reasoning was made famous in a video by the physicist Phil Plait of Numberphile. It is heavily criticized for being sloppy but it shows a simple method that has fooled a large population of its high school audience (as noted in the comments).


                        Funfact: Googling -1/12 doesn't give you any results





                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Feb 28 '14 at 3:34









                        Brian J. Fink

                        954616




                        954616










                        answered Jan 18 '14 at 2:32









                        NickNick

                        3,70673461




                        3,70673461








                        • 17




                          $begingroup$
                          That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit)
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 18 '14 at 2:37








                        • 3




                          $begingroup$
                          This proof was featured today in a video posted to Reddit.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:31






                        • 8




                          $begingroup$
                          Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up.
                          $endgroup$
                          – Jeyekomon
                          Jan 18 '14 at 17:05






                        • 7




                          $begingroup$
                          @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature...
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 19 '14 at 6:27






                        • 4




                          $begingroup$
                          You manipulate that infinite sum as if it were finite, like when you assume that $sum_{n=1}^infty n-4sum_{n=1}^infty n=-3sum_{n=1}^infty n.$ But that's impossible, since $infty-infty$ is undefined.
                          $endgroup$
                          – Hakim
                          May 16 '14 at 20:52
















                        • 17




                          $begingroup$
                          That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit)
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 18 '14 at 2:37








                        • 3




                          $begingroup$
                          This proof was featured today in a video posted to Reddit.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:31






                        • 8




                          $begingroup$
                          Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up.
                          $endgroup$
                          – Jeyekomon
                          Jan 18 '14 at 17:05






                        • 7




                          $begingroup$
                          @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature...
                          $endgroup$
                          – Mariano Suárez-Álvarez
                          Jan 19 '14 at 6:27






                        • 4




                          $begingroup$
                          You manipulate that infinite sum as if it were finite, like when you assume that $sum_{n=1}^infty n-4sum_{n=1}^infty n=-3sum_{n=1}^infty n.$ But that's impossible, since $infty-infty$ is undefined.
                          $endgroup$
                          – Hakim
                          May 16 '14 at 20:52










                        17




                        17




                        $begingroup$
                        That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit)
                        $endgroup$
                        – Mariano Suárez-Álvarez
                        Jan 18 '14 at 2:37






                        $begingroup$
                        That is not a proof, since you are manipulating expressions to which no meaning is attached. One can attach meaning to (some of) them, but that is a somewhat technical thing, and quite non-obvious. (It may be the case that you do know how to cope with these technical difficulties, and that when you write, say, «$1+2+3+4+cdots$» you h ave in mind something concrete and meaningful, but in that case you should make it explicit)
                        $endgroup$
                        – Mariano Suárez-Álvarez
                        Jan 18 '14 at 2:37






                        3




                        3




                        $begingroup$
                        This proof was featured today in a video posted to Reddit.
                        $endgroup$
                        – Steven Gubkin
                        Jan 18 '14 at 3:31




                        $begingroup$
                        This proof was featured today in a video posted to Reddit.
                        $endgroup$
                        – Steven Gubkin
                        Jan 18 '14 at 3:31




                        8




                        8




                        $begingroup$
                        Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up.
                        $endgroup$
                        – Jeyekomon
                        Jan 18 '14 at 17:05




                        $begingroup$
                        Maybe not a proof but still a nice and simple way how to fool a random high-school audience. Thumbs up.
                        $endgroup$
                        – Jeyekomon
                        Jan 18 '14 at 17:05




                        7




                        7




                        $begingroup$
                        @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature...
                        $endgroup$
                        – Mariano Suárez-Álvarez
                        Jan 19 '14 at 6:27




                        $begingroup$
                        @BalarkaSen, if you take the trouble of looking at history of edits in this post, you will notice that at the time I made the comment above the answer was of a somewhat different nature...
                        $endgroup$
                        – Mariano Suárez-Álvarez
                        Jan 19 '14 at 6:27




                        4




                        4




                        $begingroup$
                        You manipulate that infinite sum as if it were finite, like when you assume that $sum_{n=1}^infty n-4sum_{n=1}^infty n=-3sum_{n=1}^infty n.$ But that's impossible, since $infty-infty$ is undefined.
                        $endgroup$
                        – Hakim
                        May 16 '14 at 20:52






                        $begingroup$
                        You manipulate that infinite sum as if it were finite, like when you assume that $sum_{n=1}^infty n-4sum_{n=1}^infty n=-3sum_{n=1}^infty n.$ But that's impossible, since $infty-infty$ is undefined.
                        $endgroup$
                        – Hakim
                        May 16 '14 at 20:52













                        38












                        $begingroup$

                        If the following were true: $$sum_{n=1}^infty{n}=-frac1{12}tag{hypothesis}$$ then we would expect the following: $$lim_{ntoinfty}frac{n(n+1)}2=-frac1{12}tag{expectation}$$ which is the formula for the infinite triangular number limit. Unfortunately this is a result that we do not get when the limit is correctly taken. The correct value is $$lim_{ntoinfty}frac{n(n+1)}2=frac{infty(infty+1)}2=frac{infty^2+infty}2=fracinfty2=inftyneq-frac1{12}$$ This sort of mathematical sleight of hand, smoke and mirrors, pulling a finite negative rabbit out of an empty positively infinite hat does not impress me; worse yet, it gives legitimate, observable, repeatable mathematics a bad name.






                        share|cite|improve this answer









                        $endgroup$









                        • 12




                          $begingroup$
                          Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity...
                          $endgroup$
                          – zerosofthezeta
                          Mar 13 '14 at 3:33






                        • 1




                          $begingroup$
                          Not sure that is expectable. I made a question related to it.
                          $endgroup$
                          – JMCF125
                          Apr 27 '14 at 20:10






                        • 6




                          $begingroup$
                          I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch.
                          $endgroup$
                          – primo
                          May 29 '14 at 10:24












                        • $begingroup$
                          Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth.
                          $endgroup$
                          – Sara Tancredi
                          Jun 5 '14 at 23:47






                        • 5




                          $begingroup$
                          @GottfriedHelms, did you just divide by 0 ?
                          $endgroup$
                          – GinKin
                          Jun 17 '14 at 19:44
















                        38












                        $begingroup$

                        If the following were true: $$sum_{n=1}^infty{n}=-frac1{12}tag{hypothesis}$$ then we would expect the following: $$lim_{ntoinfty}frac{n(n+1)}2=-frac1{12}tag{expectation}$$ which is the formula for the infinite triangular number limit. Unfortunately this is a result that we do not get when the limit is correctly taken. The correct value is $$lim_{ntoinfty}frac{n(n+1)}2=frac{infty(infty+1)}2=frac{infty^2+infty}2=fracinfty2=inftyneq-frac1{12}$$ This sort of mathematical sleight of hand, smoke and mirrors, pulling a finite negative rabbit out of an empty positively infinite hat does not impress me; worse yet, it gives legitimate, observable, repeatable mathematics a bad name.






                        share|cite|improve this answer









                        $endgroup$









                        • 12




                          $begingroup$
                          Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity...
                          $endgroup$
                          – zerosofthezeta
                          Mar 13 '14 at 3:33






                        • 1




                          $begingroup$
                          Not sure that is expectable. I made a question related to it.
                          $endgroup$
                          – JMCF125
                          Apr 27 '14 at 20:10






                        • 6




                          $begingroup$
                          I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch.
                          $endgroup$
                          – primo
                          May 29 '14 at 10:24












                        • $begingroup$
                          Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth.
                          $endgroup$
                          – Sara Tancredi
                          Jun 5 '14 at 23:47






                        • 5




                          $begingroup$
                          @GottfriedHelms, did you just divide by 0 ?
                          $endgroup$
                          – GinKin
                          Jun 17 '14 at 19:44














                        38












                        38








                        38





                        $begingroup$

                        If the following were true: $$sum_{n=1}^infty{n}=-frac1{12}tag{hypothesis}$$ then we would expect the following: $$lim_{ntoinfty}frac{n(n+1)}2=-frac1{12}tag{expectation}$$ which is the formula for the infinite triangular number limit. Unfortunately this is a result that we do not get when the limit is correctly taken. The correct value is $$lim_{ntoinfty}frac{n(n+1)}2=frac{infty(infty+1)}2=frac{infty^2+infty}2=fracinfty2=inftyneq-frac1{12}$$ This sort of mathematical sleight of hand, smoke and mirrors, pulling a finite negative rabbit out of an empty positively infinite hat does not impress me; worse yet, it gives legitimate, observable, repeatable mathematics a bad name.






                        share|cite|improve this answer









                        $endgroup$



                        If the following were true: $$sum_{n=1}^infty{n}=-frac1{12}tag{hypothesis}$$ then we would expect the following: $$lim_{ntoinfty}frac{n(n+1)}2=-frac1{12}tag{expectation}$$ which is the formula for the infinite triangular number limit. Unfortunately this is a result that we do not get when the limit is correctly taken. The correct value is $$lim_{ntoinfty}frac{n(n+1)}2=frac{infty(infty+1)}2=frac{infty^2+infty}2=fracinfty2=inftyneq-frac1{12}$$ This sort of mathematical sleight of hand, smoke and mirrors, pulling a finite negative rabbit out of an empty positively infinite hat does not impress me; worse yet, it gives legitimate, observable, repeatable mathematics a bad name.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Feb 25 '14 at 21:14









                        Brian J. FinkBrian J. Fink

                        954616




                        954616








                        • 12




                          $begingroup$
                          Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity...
                          $endgroup$
                          – zerosofthezeta
                          Mar 13 '14 at 3:33






                        • 1




                          $begingroup$
                          Not sure that is expectable. I made a question related to it.
                          $endgroup$
                          – JMCF125
                          Apr 27 '14 at 20:10






                        • 6




                          $begingroup$
                          I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch.
                          $endgroup$
                          – primo
                          May 29 '14 at 10:24












                        • $begingroup$
                          Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth.
                          $endgroup$
                          – Sara Tancredi
                          Jun 5 '14 at 23:47






                        • 5




                          $begingroup$
                          @GottfriedHelms, did you just divide by 0 ?
                          $endgroup$
                          – GinKin
                          Jun 17 '14 at 19:44














                        • 12




                          $begingroup$
                          Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity...
                          $endgroup$
                          – zerosofthezeta
                          Mar 13 '14 at 3:33






                        • 1




                          $begingroup$
                          Not sure that is expectable. I made a question related to it.
                          $endgroup$
                          – JMCF125
                          Apr 27 '14 at 20:10






                        • 6




                          $begingroup$
                          I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch.
                          $endgroup$
                          – primo
                          May 29 '14 at 10:24












                        • $begingroup$
                          Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth.
                          $endgroup$
                          – Sara Tancredi
                          Jun 5 '14 at 23:47






                        • 5




                          $begingroup$
                          @GottfriedHelms, did you just divide by 0 ?
                          $endgroup$
                          – GinKin
                          Jun 17 '14 at 19:44








                        12




                        12




                        $begingroup$
                        Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity...
                        $endgroup$
                        – zerosofthezeta
                        Mar 13 '14 at 3:33




                        $begingroup$
                        Great answer! You should keep the "lim n to infinity" operator in front of every expression until the end instead of simply plugging in infinity...
                        $endgroup$
                        – zerosofthezeta
                        Mar 13 '14 at 3:33




                        1




                        1




                        $begingroup$
                        Not sure that is expectable. I made a question related to it.
                        $endgroup$
                        – JMCF125
                        Apr 27 '14 at 20:10




                        $begingroup$
                        Not sure that is expectable. I made a question related to it.
                        $endgroup$
                        – JMCF125
                        Apr 27 '14 at 20:10




                        6




                        6




                        $begingroup$
                        I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch.
                        $endgroup$
                        – primo
                        May 29 '14 at 10:24






                        $begingroup$
                        I can't tell if this was meant to be a joke, or not. $∞²$ is a nice touch.
                        $endgroup$
                        – primo
                        May 29 '14 at 10:24














                        $begingroup$
                        Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth.
                        $endgroup$
                        – Sara Tancredi
                        Jun 5 '14 at 23:47




                        $begingroup$
                        Sum of all natural numbers=negative fraction.This mathematical truth is too bitter. Isn't it violating the basic rules of mathematics(i.e. sum of positive natural numbers is positive). This mathematical phenomenon or behaviour is weird and incomprehensible. Doesn't this kind of situation tell us that human mind is insufficient to understand everything; may be thought is not capable enough to grasp the nature of every truth.
                        $endgroup$
                        – Sara Tancredi
                        Jun 5 '14 at 23:47




                        5




                        5




                        $begingroup$
                        @GottfriedHelms, did you just divide by 0 ?
                        $endgroup$
                        – GinKin
                        Jun 17 '14 at 19:44




                        $begingroup$
                        @GottfriedHelms, did you just divide by 0 ?
                        $endgroup$
                        – GinKin
                        Jun 17 '14 at 19:44











                        31












                        $begingroup$

                        The notation "$1+2+3+cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$.






                        share|cite|improve this answer









                        $endgroup$









                        • 21




                          $begingroup$
                          Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards,
                          $endgroup$
                          – Matt E
                          Sep 8 '12 at 19:11






                        • 8




                          $begingroup$
                          Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent.
                          $endgroup$
                          – John Bentin
                          Sep 9 '12 at 12:03






                        • 4




                          $begingroup$
                          The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:29






                        • 11




                          $begingroup$
                          @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st.
                          $endgroup$
                          – Did
                          Jan 18 '14 at 21:50








                        • 1




                          $begingroup$
                          I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.)
                          $endgroup$
                          – Jesse Madnick
                          Jan 22 '14 at 1:36
















                        31












                        $begingroup$

                        The notation "$1+2+3+cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$.






                        share|cite|improve this answer









                        $endgroup$









                        • 21




                          $begingroup$
                          Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards,
                          $endgroup$
                          – Matt E
                          Sep 8 '12 at 19:11






                        • 8




                          $begingroup$
                          Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent.
                          $endgroup$
                          – John Bentin
                          Sep 9 '12 at 12:03






                        • 4




                          $begingroup$
                          The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:29






                        • 11




                          $begingroup$
                          @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st.
                          $endgroup$
                          – Did
                          Jan 18 '14 at 21:50








                        • 1




                          $begingroup$
                          I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.)
                          $endgroup$
                          – Jesse Madnick
                          Jan 22 '14 at 1:36














                        31












                        31








                        31





                        $begingroup$

                        The notation "$1+2+3+cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$.






                        share|cite|improve this answer









                        $endgroup$



                        The notation "$1+2+3+cdots$" is as meaningless as "$1/0$". If you treat such notation as though it defined a real number and conformed in its syntax to the rules of formation for genuine real numbers, you can easily "prove" it to equal any number you like, including $-1/12$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Sep 8 '12 at 18:05









                        John BentinJohn Bentin

                        11.2k22554




                        11.2k22554








                        • 21




                          $begingroup$
                          Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards,
                          $endgroup$
                          – Matt E
                          Sep 8 '12 at 19:11






                        • 8




                          $begingroup$
                          Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent.
                          $endgroup$
                          – John Bentin
                          Sep 9 '12 at 12:03






                        • 4




                          $begingroup$
                          The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:29






                        • 11




                          $begingroup$
                          @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st.
                          $endgroup$
                          – Did
                          Jan 18 '14 at 21:50








                        • 1




                          $begingroup$
                          I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.)
                          $endgroup$
                          – Jesse Madnick
                          Jan 22 '14 at 1:36














                        • 21




                          $begingroup$
                          Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards,
                          $endgroup$
                          – Matt E
                          Sep 8 '12 at 19:11






                        • 8




                          $begingroup$
                          Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent.
                          $endgroup$
                          – John Bentin
                          Sep 9 '12 at 12:03






                        • 4




                          $begingroup$
                          The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon.
                          $endgroup$
                          – Steven Gubkin
                          Jan 18 '14 at 3:29






                        • 11




                          $begingroup$
                          @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st.
                          $endgroup$
                          – Did
                          Jan 18 '14 at 21:50








                        • 1




                          $begingroup$
                          I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.)
                          $endgroup$
                          – Jesse Madnick
                          Jan 22 '14 at 1:36








                        21




                        21




                        $begingroup$
                        Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards,
                        $endgroup$
                        – Matt E
                        Sep 8 '12 at 19:11




                        $begingroup$
                        Dear John, Your claim that this notation is "meanlingless" seems unnecessarily absolute, in light of the answers explaining that it does in fact admit meaningful interpretations. Regards,
                        $endgroup$
                        – Matt E
                        Sep 8 '12 at 19:11




                        8




                        8




                        $begingroup$
                        Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent.
                        $endgroup$
                        – John Bentin
                        Sep 9 '12 at 12:03




                        $begingroup$
                        Dear Matt: Sure, it admits meaningful interpretations, for some people. But they are not consistent.
                        $endgroup$
                        – John Bentin
                        Sep 9 '12 at 12:03




                        4




                        4




                        $begingroup$
                        The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon.
                        $endgroup$
                        – Steven Gubkin
                        Jan 18 '14 at 3:29




                        $begingroup$
                        The fact that so many different methods which connect to deeper waters in mathematics all lead to the same sum suggest that, even though there is currently no way to make it precise, there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet. To me, these are some of the most exciting things in mathematics: having only a glimpse of something great just beyond the horizon.
                        $endgroup$
                        – Steven Gubkin
                        Jan 18 '14 at 3:29




                        11




                        11




                        $begingroup$
                        @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st.
                        $endgroup$
                        – Did
                        Jan 18 '14 at 21:50






                        $begingroup$
                        @StevenGubkin Replace "there might be a rigorous consistent theory of divergent sums which no one has made fully precise yet" by "there exists a rigorous consistent theory of divergent sums, made fully precise more than 80 years ago". As Littlewood explains in the preface of Hardy's treatise, "[I]n the early years of the century the subject [Divergent Series], while in no way mystical or unrigorous, was regarded as sensational, and about the present title, now colourless, there hung an aroma of paradox and audacity"... only these were the first years of the 20th century, not the 21st.
                        $endgroup$
                        – Did
                        Jan 18 '14 at 21:50






                        1




                        1




                        $begingroup$
                        I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.)
                        $endgroup$
                        – Jesse Madnick
                        Jan 22 '14 at 1:36




                        $begingroup$
                        I think everything said in this post is true... but none of it explains why $-1/12$ is a better choice of "sum" than any other real number. (If you think $-1/12$ is no better than any other real number here, well, just read the other answers.)
                        $endgroup$
                        – Jesse Madnick
                        Jan 22 '14 at 1:36











                        28












                        $begingroup$

                        This infinite series is ultimately divergent because
                        $$
                        1+2+3+4+cdots=sumlimits_{k=1}^{infty} k$$
                        $$ = limlimits_{ntoinfty} sumlimits_{k=1}^{n} k = limlimits_{ntoinfty} frac{n(n+1)}{2} = infty
                        $$
                        The value of $-frac{1}{12}$ is assigned to the infinite series by the methods of zeta function regularization and several others. If you'd like to understand why someone would assign a value to a divergent series, then have a look at this question.






                        share|cite|improve this answer











                        $endgroup$


















                          28












                          $begingroup$

                          This infinite series is ultimately divergent because
                          $$
                          1+2+3+4+cdots=sumlimits_{k=1}^{infty} k$$
                          $$ = limlimits_{ntoinfty} sumlimits_{k=1}^{n} k = limlimits_{ntoinfty} frac{n(n+1)}{2} = infty
                          $$
                          The value of $-frac{1}{12}$ is assigned to the infinite series by the methods of zeta function regularization and several others. If you'd like to understand why someone would assign a value to a divergent series, then have a look at this question.






                          share|cite|improve this answer











                          $endgroup$
















                            28












                            28








                            28





                            $begingroup$

                            This infinite series is ultimately divergent because
                            $$
                            1+2+3+4+cdots=sumlimits_{k=1}^{infty} k$$
                            $$ = limlimits_{ntoinfty} sumlimits_{k=1}^{n} k = limlimits_{ntoinfty} frac{n(n+1)}{2} = infty
                            $$
                            The value of $-frac{1}{12}$ is assigned to the infinite series by the methods of zeta function regularization and several others. If you'd like to understand why someone would assign a value to a divergent series, then have a look at this question.






                            share|cite|improve this answer











                            $endgroup$



                            This infinite series is ultimately divergent because
                            $$
                            1+2+3+4+cdots=sumlimits_{k=1}^{infty} k$$
                            $$ = limlimits_{ntoinfty} sumlimits_{k=1}^{n} k = limlimits_{ntoinfty} frac{n(n+1)}{2} = infty
                            $$
                            The value of $-frac{1}{12}$ is assigned to the infinite series by the methods of zeta function regularization and several others. If you'd like to understand why someone would assign a value to a divergent series, then have a look at this question.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Apr 13 '17 at 12:21









                            Community

                            1




                            1










                            answered Jan 20 '15 at 3:59









                            k170k170

                            7,46131540




                            7,46131540























                                22












                                $begingroup$

                                I am missing here that



                                $$1+2+3+cdots rightarrow infty$$



                                $$zeta(-1) neq 1+2+3+cdots$$



                                As you say, we only define $zeta$ using the infinite sum if $Re(s)>1$.





                                It is just as defining $$f(x) = begin{cases} frac{1}{x} &mathrm{if} ; x neq 0 \ 0 & mathrm{if} ; x = 0 end{cases}$$



                                And asking why $f(0)=0$.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  your example is not correct. In first place the zeta function defined at values where $Re(z)le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation.
                                  $endgroup$
                                  – Masacroso
                                  Jan 17 '18 at 11:34
















                                22












                                $begingroup$

                                I am missing here that



                                $$1+2+3+cdots rightarrow infty$$



                                $$zeta(-1) neq 1+2+3+cdots$$



                                As you say, we only define $zeta$ using the infinite sum if $Re(s)>1$.





                                It is just as defining $$f(x) = begin{cases} frac{1}{x} &mathrm{if} ; x neq 0 \ 0 & mathrm{if} ; x = 0 end{cases}$$



                                And asking why $f(0)=0$.






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  your example is not correct. In first place the zeta function defined at values where $Re(z)le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation.
                                  $endgroup$
                                  – Masacroso
                                  Jan 17 '18 at 11:34














                                22












                                22








                                22





                                $begingroup$

                                I am missing here that



                                $$1+2+3+cdots rightarrow infty$$



                                $$zeta(-1) neq 1+2+3+cdots$$



                                As you say, we only define $zeta$ using the infinite sum if $Re(s)>1$.





                                It is just as defining $$f(x) = begin{cases} frac{1}{x} &mathrm{if} ; x neq 0 \ 0 & mathrm{if} ; x = 0 end{cases}$$



                                And asking why $f(0)=0$.






                                share|cite|improve this answer









                                $endgroup$



                                I am missing here that



                                $$1+2+3+cdots rightarrow infty$$



                                $$zeta(-1) neq 1+2+3+cdots$$



                                As you say, we only define $zeta$ using the infinite sum if $Re(s)>1$.





                                It is just as defining $$f(x) = begin{cases} frac{1}{x} &mathrm{if} ; x neq 0 \ 0 & mathrm{if} ; x = 0 end{cases}$$



                                And asking why $f(0)=0$.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jun 8 '15 at 7:28









                                wythagoraswythagoras

                                21.6k444104




                                21.6k444104












                                • $begingroup$
                                  your example is not correct. In first place the zeta function defined at values where $Re(z)le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation.
                                  $endgroup$
                                  – Masacroso
                                  Jan 17 '18 at 11:34


















                                • $begingroup$
                                  your example is not correct. In first place the zeta function defined at values where $Re(z)le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation.
                                  $endgroup$
                                  – Masacroso
                                  Jan 17 '18 at 11:34
















                                $begingroup$
                                your example is not correct. In first place the zeta function defined at values where $Re(z)le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation.
                                $endgroup$
                                – Masacroso
                                Jan 17 '18 at 11:34




                                $begingroup$
                                your example is not correct. In first place the zeta function defined at values where $Re(z)le1$ is defined by analytic continuation. However your function $f$ is not defined by analytic continuation.
                                $endgroup$
                                – Masacroso
                                Jan 17 '18 at 11:34











                                14












                                $begingroup$

                                A nice alternative representation, by a double sum.



                                Consider the infinite square array where the rowsums and the (formal) column-sums are also given in closed forms
                                $$ small begin{array} {r|rrrrr|rr}
                                & & & & & & text{rowsums} \
                                hline
                                & 1/0! & log(1)/1! & log(1)^2/2! &log(1)^3/3! & cdots & = e^{log(1)}&=1 \
                                & 1/0! & log(2)/1! & log(2)^2/2! &log(2)^3/3! & cdots & = e^{log(2)}&=2 \
                                & 1/0! & log(3)/1! & log(3)^2/2! &log(3)^3/3! & cdots & = e^{log(3)}&=3 \
                                & vdots & & & & ddots & vdots & vdots\
                                hline
                                text{colsums:}
                                & zeta(0) & -zeta(0)'/1! & zeta(0)''/2!& -zeta(0)^{(3)}/3! & cdots &&=zeta(-1)
                                end{array} $$
                                and the row-sum of the derivatives of the $zeta()$ at $0$ in the bottom row can numerically be written as
                                $$ small -0.5 +0.9189... -1.003178...+1.000785...-0.999879...+1.0000019... pm cdots $$
                                If we split the terms and do two series we find
                                $$ Tiny begin{array} {}
                                -1 &+1 &-1 &+1 &-1 &+1 &pm &cdots &=-1/2\
                                +0.5 &-0.0810... &-0.003178...&+0.000785...&+0.000120...&+0.0000019... &pm &cdots &= 5/12\
                                end{array}$$
                                Here we must now resort to the earlier definition of the divergent sum of the alternating units (in the first row), but the second row is convergent and can conventionally be summed. The sum of the two rowsums $small -1/2 + 5/12 = -1/12$ gives the expected result.



                                However, that fiddling with double-sums, when non-convergent-series are involved (as are the columnsums in the above matrix and the splitting in the numerical expression) must explicitely be justified as "legal"/algebraically consistent operation.

                                But because I encounter it not frequently, that such non-convergent double-sum schemes come out with the expected result without further ado, I think this is a specific nice observation here.



                                And the more algebraically manipulations come out to be consistent with the assumed value of a divergent series, the more is the hypothese acceptable, that this should be taken as the canonical numerical replacement for the series-expression ( like we do it with the rational fraction for the geometric series even in the divergent case (except for that with quotient $q=1$)) .






                                share|cite|improve this answer











                                $endgroup$


















                                  14












                                  $begingroup$

                                  A nice alternative representation, by a double sum.



                                  Consider the infinite square array where the rowsums and the (formal) column-sums are also given in closed forms
                                  $$ small begin{array} {r|rrrrr|rr}
                                  & & & & & & text{rowsums} \
                                  hline
                                  & 1/0! & log(1)/1! & log(1)^2/2! &log(1)^3/3! & cdots & = e^{log(1)}&=1 \
                                  & 1/0! & log(2)/1! & log(2)^2/2! &log(2)^3/3! & cdots & = e^{log(2)}&=2 \
                                  & 1/0! & log(3)/1! & log(3)^2/2! &log(3)^3/3! & cdots & = e^{log(3)}&=3 \
                                  & vdots & & & & ddots & vdots & vdots\
                                  hline
                                  text{colsums:}
                                  & zeta(0) & -zeta(0)'/1! & zeta(0)''/2!& -zeta(0)^{(3)}/3! & cdots &&=zeta(-1)
                                  end{array} $$
                                  and the row-sum of the derivatives of the $zeta()$ at $0$ in the bottom row can numerically be written as
                                  $$ small -0.5 +0.9189... -1.003178...+1.000785...-0.999879...+1.0000019... pm cdots $$
                                  If we split the terms and do two series we find
                                  $$ Tiny begin{array} {}
                                  -1 &+1 &-1 &+1 &-1 &+1 &pm &cdots &=-1/2\
                                  +0.5 &-0.0810... &-0.003178...&+0.000785...&+0.000120...&+0.0000019... &pm &cdots &= 5/12\
                                  end{array}$$
                                  Here we must now resort to the earlier definition of the divergent sum of the alternating units (in the first row), but the second row is convergent and can conventionally be summed. The sum of the two rowsums $small -1/2 + 5/12 = -1/12$ gives the expected result.



                                  However, that fiddling with double-sums, when non-convergent-series are involved (as are the columnsums in the above matrix and the splitting in the numerical expression) must explicitely be justified as "legal"/algebraically consistent operation.

                                  But because I encounter it not frequently, that such non-convergent double-sum schemes come out with the expected result without further ado, I think this is a specific nice observation here.



                                  And the more algebraically manipulations come out to be consistent with the assumed value of a divergent series, the more is the hypothese acceptable, that this should be taken as the canonical numerical replacement for the series-expression ( like we do it with the rational fraction for the geometric series even in the divergent case (except for that with quotient $q=1$)) .






                                  share|cite|improve this answer











                                  $endgroup$
















                                    14












                                    14








                                    14





                                    $begingroup$

                                    A nice alternative representation, by a double sum.



                                    Consider the infinite square array where the rowsums and the (formal) column-sums are also given in closed forms
                                    $$ small begin{array} {r|rrrrr|rr}
                                    & & & & & & text{rowsums} \
                                    hline
                                    & 1/0! & log(1)/1! & log(1)^2/2! &log(1)^3/3! & cdots & = e^{log(1)}&=1 \
                                    & 1/0! & log(2)/1! & log(2)^2/2! &log(2)^3/3! & cdots & = e^{log(2)}&=2 \
                                    & 1/0! & log(3)/1! & log(3)^2/2! &log(3)^3/3! & cdots & = e^{log(3)}&=3 \
                                    & vdots & & & & ddots & vdots & vdots\
                                    hline
                                    text{colsums:}
                                    & zeta(0) & -zeta(0)'/1! & zeta(0)''/2!& -zeta(0)^{(3)}/3! & cdots &&=zeta(-1)
                                    end{array} $$
                                    and the row-sum of the derivatives of the $zeta()$ at $0$ in the bottom row can numerically be written as
                                    $$ small -0.5 +0.9189... -1.003178...+1.000785...-0.999879...+1.0000019... pm cdots $$
                                    If we split the terms and do two series we find
                                    $$ Tiny begin{array} {}
                                    -1 &+1 &-1 &+1 &-1 &+1 &pm &cdots &=-1/2\
                                    +0.5 &-0.0810... &-0.003178...&+0.000785...&+0.000120...&+0.0000019... &pm &cdots &= 5/12\
                                    end{array}$$
                                    Here we must now resort to the earlier definition of the divergent sum of the alternating units (in the first row), but the second row is convergent and can conventionally be summed. The sum of the two rowsums $small -1/2 + 5/12 = -1/12$ gives the expected result.



                                    However, that fiddling with double-sums, when non-convergent-series are involved (as are the columnsums in the above matrix and the splitting in the numerical expression) must explicitely be justified as "legal"/algebraically consistent operation.

                                    But because I encounter it not frequently, that such non-convergent double-sum schemes come out with the expected result without further ado, I think this is a specific nice observation here.



                                    And the more algebraically manipulations come out to be consistent with the assumed value of a divergent series, the more is the hypothese acceptable, that this should be taken as the canonical numerical replacement for the series-expression ( like we do it with the rational fraction for the geometric series even in the divergent case (except for that with quotient $q=1$)) .






                                    share|cite|improve this answer











                                    $endgroup$



                                    A nice alternative representation, by a double sum.



                                    Consider the infinite square array where the rowsums and the (formal) column-sums are also given in closed forms
                                    $$ small begin{array} {r|rrrrr|rr}
                                    & & & & & & text{rowsums} \
                                    hline
                                    & 1/0! & log(1)/1! & log(1)^2/2! &log(1)^3/3! & cdots & = e^{log(1)}&=1 \
                                    & 1/0! & log(2)/1! & log(2)^2/2! &log(2)^3/3! & cdots & = e^{log(2)}&=2 \
                                    & 1/0! & log(3)/1! & log(3)^2/2! &log(3)^3/3! & cdots & = e^{log(3)}&=3 \
                                    & vdots & & & & ddots & vdots & vdots\
                                    hline
                                    text{colsums:}
                                    & zeta(0) & -zeta(0)'/1! & zeta(0)''/2!& -zeta(0)^{(3)}/3! & cdots &&=zeta(-1)
                                    end{array} $$
                                    and the row-sum of the derivatives of the $zeta()$ at $0$ in the bottom row can numerically be written as
                                    $$ small -0.5 +0.9189... -1.003178...+1.000785...-0.999879...+1.0000019... pm cdots $$
                                    If we split the terms and do two series we find
                                    $$ Tiny begin{array} {}
                                    -1 &+1 &-1 &+1 &-1 &+1 &pm &cdots &=-1/2\
                                    +0.5 &-0.0810... &-0.003178...&+0.000785...&+0.000120...&+0.0000019... &pm &cdots &= 5/12\
                                    end{array}$$
                                    Here we must now resort to the earlier definition of the divergent sum of the alternating units (in the first row), but the second row is convergent and can conventionally be summed. The sum of the two rowsums $small -1/2 + 5/12 = -1/12$ gives the expected result.



                                    However, that fiddling with double-sums, when non-convergent-series are involved (as are the columnsums in the above matrix and the splitting in the numerical expression) must explicitely be justified as "legal"/algebraically consistent operation.

                                    But because I encounter it not frequently, that such non-convergent double-sum schemes come out with the expected result without further ado, I think this is a specific nice observation here.



                                    And the more algebraically manipulations come out to be consistent with the assumed value of a divergent series, the more is the hypothese acceptable, that this should be taken as the canonical numerical replacement for the series-expression ( like we do it with the rational fraction for the geometric series even in the divergent case (except for that with quotient $q=1$)) .







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited Jan 10 '15 at 0:32

























                                    answered Jan 10 '15 at 0:17









                                    Gottfried HelmsGottfried Helms

                                    23.2k24398




                                    23.2k24398























                                        14












                                        $begingroup$

                                        Suppose a rigorous way of doing a computation yields a well defined real number as the answer. But with shortcut formal manipulations that are not allowed (e.g. interchanging summations and integrations when that isn't allowed) one ends up with a divergent series and then the question is that given only the divergent series, can one guess what real number is most likely the answer to what the unknown original problem was.



                                        This is similar to guessing that the next term of the integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 is probably 16 rather than e.g. 32431, even though the latter possibility cannot be ruled out. Strictly speaking this problem is not well defined, what one really is doing here is assuming that the sequence that is specified by the least amount of information is the most likely answer. The person who invented the puzzle had some simple algorithm in mind, therefore the much more complicated algorithm that would yield 32431 as the next number in the sequence is not the likely answer.



                                        Similarly, the formal manipulations that yield the divergent series won't have introduced a lot of additional information, these are just generic mathematical manipulations that would have been correct when used in a wide class of problems, but not the one it actually has been applied to. This means that almost all the information about the unknown real number is present in the divergent series, it can be extracted from it by applying certain formal manipulations to the series that, like the unknown manipulations that led to the divergent series, are formally correct for a wide class of convergent series, but not in case of this divergent series.






                                        share|cite|improve this answer











                                        $endgroup$


















                                          14












                                          $begingroup$

                                          Suppose a rigorous way of doing a computation yields a well defined real number as the answer. But with shortcut formal manipulations that are not allowed (e.g. interchanging summations and integrations when that isn't allowed) one ends up with a divergent series and then the question is that given only the divergent series, can one guess what real number is most likely the answer to what the unknown original problem was.



                                          This is similar to guessing that the next term of the integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 is probably 16 rather than e.g. 32431, even though the latter possibility cannot be ruled out. Strictly speaking this problem is not well defined, what one really is doing here is assuming that the sequence that is specified by the least amount of information is the most likely answer. The person who invented the puzzle had some simple algorithm in mind, therefore the much more complicated algorithm that would yield 32431 as the next number in the sequence is not the likely answer.



                                          Similarly, the formal manipulations that yield the divergent series won't have introduced a lot of additional information, these are just generic mathematical manipulations that would have been correct when used in a wide class of problems, but not the one it actually has been applied to. This means that almost all the information about the unknown real number is present in the divergent series, it can be extracted from it by applying certain formal manipulations to the series that, like the unknown manipulations that led to the divergent series, are formally correct for a wide class of convergent series, but not in case of this divergent series.






                                          share|cite|improve this answer











                                          $endgroup$
















                                            14












                                            14








                                            14





                                            $begingroup$

                                            Suppose a rigorous way of doing a computation yields a well defined real number as the answer. But with shortcut formal manipulations that are not allowed (e.g. interchanging summations and integrations when that isn't allowed) one ends up with a divergent series and then the question is that given only the divergent series, can one guess what real number is most likely the answer to what the unknown original problem was.



                                            This is similar to guessing that the next term of the integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 is probably 16 rather than e.g. 32431, even though the latter possibility cannot be ruled out. Strictly speaking this problem is not well defined, what one really is doing here is assuming that the sequence that is specified by the least amount of information is the most likely answer. The person who invented the puzzle had some simple algorithm in mind, therefore the much more complicated algorithm that would yield 32431 as the next number in the sequence is not the likely answer.



                                            Similarly, the formal manipulations that yield the divergent series won't have introduced a lot of additional information, these are just generic mathematical manipulations that would have been correct when used in a wide class of problems, but not the one it actually has been applied to. This means that almost all the information about the unknown real number is present in the divergent series, it can be extracted from it by applying certain formal manipulations to the series that, like the unknown manipulations that led to the divergent series, are formally correct for a wide class of convergent series, but not in case of this divergent series.






                                            share|cite|improve this answer











                                            $endgroup$



                                            Suppose a rigorous way of doing a computation yields a well defined real number as the answer. But with shortcut formal manipulations that are not allowed (e.g. interchanging summations and integrations when that isn't allowed) one ends up with a divergent series and then the question is that given only the divergent series, can one guess what real number is most likely the answer to what the unknown original problem was.



                                            This is similar to guessing that the next term of the integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 is probably 16 rather than e.g. 32431, even though the latter possibility cannot be ruled out. Strictly speaking this problem is not well defined, what one really is doing here is assuming that the sequence that is specified by the least amount of information is the most likely answer. The person who invented the puzzle had some simple algorithm in mind, therefore the much more complicated algorithm that would yield 32431 as the next number in the sequence is not the likely answer.



                                            Similarly, the formal manipulations that yield the divergent series won't have introduced a lot of additional information, these are just generic mathematical manipulations that would have been correct when used in a wide class of problems, but not the one it actually has been applied to. This means that almost all the information about the unknown real number is present in the divergent series, it can be extracted from it by applying certain formal manipulations to the series that, like the unknown manipulations that led to the divergent series, are formally correct for a wide class of convergent series, but not in case of this divergent series.







                                            share|cite|improve this answer














                                            share|cite|improve this answer



                                            share|cite|improve this answer








                                            edited Nov 27 '15 at 23:47

























                                            answered Aug 20 '15 at 16:13









                                            Count IblisCount Iblis

                                            8,23121533




                                            8,23121533























                                                13












                                                $begingroup$

                                                Euler's approach, but using algebra instead of subtraction of infinite series: Let $f(x)=(1-x)^{-2}$. Then $f(-x) = f(x) -4xf(x^2)$. Plug $x=1$ to get $f(-1)=-3f(1)$, whence $f(1)=-1/12$.



                                                EDIT: This argument is a motivation for the mysterious value $-1/12$. Yes, $f(x) :=(1-x)^{-2}$ is undefined at $x=1$. If we define $f(1)=-1/12$, then the above identity holds for all $x$.






                                                share|cite|improve this answer











                                                $endgroup$









                                                • 4




                                                  $begingroup$
                                                  So... $0^{-2}$ is well defined? And all these years I thought you can't divide by zero.
                                                  $endgroup$
                                                  – Asaf Karagila
                                                  Feb 21 '15 at 22:44






                                                • 3




                                                  $begingroup$
                                                  $frac14=infty-4cdotinfty$ So $-frac1{12}=infty$
                                                  $endgroup$
                                                  – robjohn
                                                  Feb 22 '15 at 4:18


















                                                13












                                                $begingroup$

                                                Euler's approach, but using algebra instead of subtraction of infinite series: Let $f(x)=(1-x)^{-2}$. Then $f(-x) = f(x) -4xf(x^2)$. Plug $x=1$ to get $f(-1)=-3f(1)$, whence $f(1)=-1/12$.



                                                EDIT: This argument is a motivation for the mysterious value $-1/12$. Yes, $f(x) :=(1-x)^{-2}$ is undefined at $x=1$. If we define $f(1)=-1/12$, then the above identity holds for all $x$.






                                                share|cite|improve this answer











                                                $endgroup$









                                                • 4




                                                  $begingroup$
                                                  So... $0^{-2}$ is well defined? And all these years I thought you can't divide by zero.
                                                  $endgroup$
                                                  – Asaf Karagila
                                                  Feb 21 '15 at 22:44






                                                • 3




                                                  $begingroup$
                                                  $frac14=infty-4cdotinfty$ So $-frac1{12}=infty$
                                                  $endgroup$
                                                  – robjohn
                                                  Feb 22 '15 at 4:18
















                                                13












                                                13








                                                13





                                                $begingroup$

                                                Euler's approach, but using algebra instead of subtraction of infinite series: Let $f(x)=(1-x)^{-2}$. Then $f(-x) = f(x) -4xf(x^2)$. Plug $x=1$ to get $f(-1)=-3f(1)$, whence $f(1)=-1/12$.



                                                EDIT: This argument is a motivation for the mysterious value $-1/12$. Yes, $f(x) :=(1-x)^{-2}$ is undefined at $x=1$. If we define $f(1)=-1/12$, then the above identity holds for all $x$.






                                                share|cite|improve this answer











                                                $endgroup$



                                                Euler's approach, but using algebra instead of subtraction of infinite series: Let $f(x)=(1-x)^{-2}$. Then $f(-x) = f(x) -4xf(x^2)$. Plug $x=1$ to get $f(-1)=-3f(1)$, whence $f(1)=-1/12$.



                                                EDIT: This argument is a motivation for the mysterious value $-1/12$. Yes, $f(x) :=(1-x)^{-2}$ is undefined at $x=1$. If we define $f(1)=-1/12$, then the above identity holds for all $x$.







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Feb 22 '15 at 16:52

























                                                answered Feb 21 '15 at 22:32









                                                grand_chatgrand_chat

                                                20.1k11226




                                                20.1k11226








                                                • 4




                                                  $begingroup$
                                                  So... $0^{-2}$ is well defined? And all these years I thought you can't divide by zero.
                                                  $endgroup$
                                                  – Asaf Karagila
                                                  Feb 21 '15 at 22:44






                                                • 3




                                                  $begingroup$
                                                  $frac14=infty-4cdotinfty$ So $-frac1{12}=infty$
                                                  $endgroup$
                                                  – robjohn
                                                  Feb 22 '15 at 4:18
















                                                • 4




                                                  $begingroup$
                                                  So... $0^{-2}$ is well defined? And all these years I thought you can't divide by zero.
                                                  $endgroup$
                                                  – Asaf Karagila
                                                  Feb 21 '15 at 22:44






                                                • 3




                                                  $begingroup$
                                                  $frac14=infty-4cdotinfty$ So $-frac1{12}=infty$
                                                  $endgroup$
                                                  – robjohn
                                                  Feb 22 '15 at 4:18










                                                4




                                                4




                                                $begingroup$
                                                So... $0^{-2}$ is well defined? And all these years I thought you can't divide by zero.
                                                $endgroup$
                                                – Asaf Karagila
                                                Feb 21 '15 at 22:44




                                                $begingroup$
                                                So... $0^{-2}$ is well defined? And all these years I thought you can't divide by zero.
                                                $endgroup$
                                                – Asaf Karagila
                                                Feb 21 '15 at 22:44




                                                3




                                                3




                                                $begingroup$
                                                $frac14=infty-4cdotinfty$ So $-frac1{12}=infty$
                                                $endgroup$
                                                – robjohn
                                                Feb 22 '15 at 4:18






                                                $begingroup$
                                                $frac14=infty-4cdotinfty$ So $-frac1{12}=infty$
                                                $endgroup$
                                                – robjohn
                                                Feb 22 '15 at 4:18













                                                13












                                                $begingroup$

                                                I like this simple graphical explanation.



                                                There are first partial sums of the series 1 + 2 + 3 + 4 + ⋯ on picture. The parabola is their smoothed asymptote; its y-intercept is −1/12.



                                                enter image description here






                                                share|cite|improve this answer











                                                $endgroup$









                                                • 6




                                                  $begingroup$
                                                  I'm afraid I don't follow this at all.
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 11 '15 at 21:02










                                                • $begingroup$
                                                  @GrumpyParsnip It's explained at en.wikipedia.org/wiki/… . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself?
                                                  $endgroup$
                                                  – Chris Culter
                                                  Sep 13 '15 at 21:01










                                                • $begingroup$
                                                  @ChrisCulter: thanks!
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 13 '15 at 21:12










                                                • $begingroup$
                                                  @ChrisCulter, yep, it's my additions. I updated the answer.
                                                  $endgroup$
                                                  – Ivan Kochurkin
                                                  Sep 27 '15 at 18:12






                                                • 4




                                                  $begingroup$
                                                  It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the colors in the picture?
                                                  $endgroup$
                                                  – Did
                                                  Dec 4 '16 at 9:42
















                                                13












                                                $begingroup$

                                                I like this simple graphical explanation.



                                                There are first partial sums of the series 1 + 2 + 3 + 4 + ⋯ on picture. The parabola is their smoothed asymptote; its y-intercept is −1/12.



                                                enter image description here






                                                share|cite|improve this answer











                                                $endgroup$









                                                • 6




                                                  $begingroup$
                                                  I'm afraid I don't follow this at all.
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 11 '15 at 21:02










                                                • $begingroup$
                                                  @GrumpyParsnip It's explained at en.wikipedia.org/wiki/… . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself?
                                                  $endgroup$
                                                  – Chris Culter
                                                  Sep 13 '15 at 21:01










                                                • $begingroup$
                                                  @ChrisCulter: thanks!
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 13 '15 at 21:12










                                                • $begingroup$
                                                  @ChrisCulter, yep, it's my additions. I updated the answer.
                                                  $endgroup$
                                                  – Ivan Kochurkin
                                                  Sep 27 '15 at 18:12






                                                • 4




                                                  $begingroup$
                                                  It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the colors in the picture?
                                                  $endgroup$
                                                  – Did
                                                  Dec 4 '16 at 9:42














                                                13












                                                13








                                                13





                                                $begingroup$

                                                I like this simple graphical explanation.



                                                There are first partial sums of the series 1 + 2 + 3 + 4 + ⋯ on picture. The parabola is their smoothed asymptote; its y-intercept is −1/12.



                                                enter image description here






                                                share|cite|improve this answer











                                                $endgroup$



                                                I like this simple graphical explanation.



                                                There are first partial sums of the series 1 + 2 + 3 + 4 + ⋯ on picture. The parabola is their smoothed asymptote; its y-intercept is −1/12.



                                                enter image description here







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Sep 27 '15 at 18:11

























                                                answered Sep 11 '15 at 20:54









                                                Ivan KochurkinIvan Kochurkin

                                                5641620




                                                5641620








                                                • 6




                                                  $begingroup$
                                                  I'm afraid I don't follow this at all.
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 11 '15 at 21:02










                                                • $begingroup$
                                                  @GrumpyParsnip It's explained at en.wikipedia.org/wiki/… . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself?
                                                  $endgroup$
                                                  – Chris Culter
                                                  Sep 13 '15 at 21:01










                                                • $begingroup$
                                                  @ChrisCulter: thanks!
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 13 '15 at 21:12










                                                • $begingroup$
                                                  @ChrisCulter, yep, it's my additions. I updated the answer.
                                                  $endgroup$
                                                  – Ivan Kochurkin
                                                  Sep 27 '15 at 18:12






                                                • 4




                                                  $begingroup$
                                                  It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the colors in the picture?
                                                  $endgroup$
                                                  – Did
                                                  Dec 4 '16 at 9:42














                                                • 6




                                                  $begingroup$
                                                  I'm afraid I don't follow this at all.
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 11 '15 at 21:02










                                                • $begingroup$
                                                  @GrumpyParsnip It's explained at en.wikipedia.org/wiki/… . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself?
                                                  $endgroup$
                                                  – Chris Culter
                                                  Sep 13 '15 at 21:01










                                                • $begingroup$
                                                  @ChrisCulter: thanks!
                                                  $endgroup$
                                                  – Cheerful Parsnip
                                                  Sep 13 '15 at 21:12










                                                • $begingroup$
                                                  @ChrisCulter, yep, it's my additions. I updated the answer.
                                                  $endgroup$
                                                  – Ivan Kochurkin
                                                  Sep 27 '15 at 18:12






                                                • 4




                                                  $begingroup$
                                                  It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the colors in the picture?
                                                  $endgroup$
                                                  – Did
                                                  Dec 4 '16 at 9:42








                                                6




                                                6




                                                $begingroup$
                                                I'm afraid I don't follow this at all.
                                                $endgroup$
                                                – Cheerful Parsnip
                                                Sep 11 '15 at 21:02




                                                $begingroup$
                                                I'm afraid I don't follow this at all.
                                                $endgroup$
                                                – Cheerful Parsnip
                                                Sep 11 '15 at 21:02












                                                $begingroup$
                                                @GrumpyParsnip It's explained at en.wikipedia.org/wiki/… . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself?
                                                $endgroup$
                                                – Chris Culter
                                                Sep 13 '15 at 21:01




                                                $begingroup$
                                                @GrumpyParsnip It's explained at en.wikipedia.org/wiki/… . I'm interested in where the version with the labeled axes came from, though. KvanTTT, did you add those yourself?
                                                $endgroup$
                                                – Chris Culter
                                                Sep 13 '15 at 21:01












                                                $begingroup$
                                                @ChrisCulter: thanks!
                                                $endgroup$
                                                – Cheerful Parsnip
                                                Sep 13 '15 at 21:12




                                                $begingroup$
                                                @ChrisCulter: thanks!
                                                $endgroup$
                                                – Cheerful Parsnip
                                                Sep 13 '15 at 21:12












                                                $begingroup$
                                                @ChrisCulter, yep, it's my additions. I updated the answer.
                                                $endgroup$
                                                – Ivan Kochurkin
                                                Sep 27 '15 at 18:12




                                                $begingroup$
                                                @ChrisCulter, yep, it's my additions. I updated the answer.
                                                $endgroup$
                                                – Ivan Kochurkin
                                                Sep 27 '15 at 18:12




                                                4




                                                4




                                                $begingroup$
                                                It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the colors in the picture?
                                                $endgroup$
                                                – Did
                                                Dec 4 '16 at 9:42




                                                $begingroup$
                                                It is somewhat odd that this completely opaque "answer", which the OP themselves concedes they do not know what it means or where it comes from, received 9 upvotes. Maybe for the colors in the picture?
                                                $endgroup$
                                                – Did
                                                Dec 4 '16 at 9:42











                                                9












                                                $begingroup$

                                                As I recently showed in another answer, we have the wonderful pattern:



                                                $$sum_{k=1}^n1=nimpliesint_{-1}^0x~mathrm dx=zeta(0)\sum_{k=1}^nk=frac{n(n+1)}2impliesint_{-1}^0frac{x(x+1)}2~mathrm dx=zeta(-1)\sum_{k=1}^nk^2=frac{n(n+1)(2n+1)}6impliesint_{-1}^0frac{x(x+1)(2x+1)}6~mathrm dx=zeta(-2)\sum_{k=1}^nk^3=left[frac{n(n+1)}2right]^2impliesint_{-1}^0left[frac{x(x+1)}2right]^2~mathrm dx=zeta(-3)\vdots$$



                                                This pattern works for all $zeta(-k)$.






                                                share|cite|improve this answer









                                                $endgroup$


















                                                  9












                                                  $begingroup$

                                                  As I recently showed in another answer, we have the wonderful pattern:



                                                  $$sum_{k=1}^n1=nimpliesint_{-1}^0x~mathrm dx=zeta(0)\sum_{k=1}^nk=frac{n(n+1)}2impliesint_{-1}^0frac{x(x+1)}2~mathrm dx=zeta(-1)\sum_{k=1}^nk^2=frac{n(n+1)(2n+1)}6impliesint_{-1}^0frac{x(x+1)(2x+1)}6~mathrm dx=zeta(-2)\sum_{k=1}^nk^3=left[frac{n(n+1)}2right]^2impliesint_{-1}^0left[frac{x(x+1)}2right]^2~mathrm dx=zeta(-3)\vdots$$



                                                  This pattern works for all $zeta(-k)$.






                                                  share|cite|improve this answer









                                                  $endgroup$
















                                                    9












                                                    9








                                                    9





                                                    $begingroup$

                                                    As I recently showed in another answer, we have the wonderful pattern:



                                                    $$sum_{k=1}^n1=nimpliesint_{-1}^0x~mathrm dx=zeta(0)\sum_{k=1}^nk=frac{n(n+1)}2impliesint_{-1}^0frac{x(x+1)}2~mathrm dx=zeta(-1)\sum_{k=1}^nk^2=frac{n(n+1)(2n+1)}6impliesint_{-1}^0frac{x(x+1)(2x+1)}6~mathrm dx=zeta(-2)\sum_{k=1}^nk^3=left[frac{n(n+1)}2right]^2impliesint_{-1}^0left[frac{x(x+1)}2right]^2~mathrm dx=zeta(-3)\vdots$$



                                                    This pattern works for all $zeta(-k)$.






                                                    share|cite|improve this answer









                                                    $endgroup$



                                                    As I recently showed in another answer, we have the wonderful pattern:



                                                    $$sum_{k=1}^n1=nimpliesint_{-1}^0x~mathrm dx=zeta(0)\sum_{k=1}^nk=frac{n(n+1)}2impliesint_{-1}^0frac{x(x+1)}2~mathrm dx=zeta(-1)\sum_{k=1}^nk^2=frac{n(n+1)(2n+1)}6impliesint_{-1}^0frac{x(x+1)(2x+1)}6~mathrm dx=zeta(-2)\sum_{k=1}^nk^3=left[frac{n(n+1)}2right]^2impliesint_{-1}^0left[frac{x(x+1)}2right]^2~mathrm dx=zeta(-3)\vdots$$



                                                    This pattern works for all $zeta(-k)$.







                                                    share|cite|improve this answer












                                                    share|cite|improve this answer



                                                    share|cite|improve this answer










                                                    answered Jul 8 '17 at 11:30









                                                    Simply Beautiful ArtSimply Beautiful Art

                                                    50.4k578181




                                                    50.4k578181























                                                        8












                                                        $begingroup$

                                                        Here is a useful place for Euler's transform and the Dirichlet eta function:



                                                        $$zeta(s)=sum_{n=1}^inftyfrac1{n^s}$$



                                                        $$eta(s)=sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$



                                                        $$zeta(s)-eta(s)=sum_{n=1}^inftyfrac{1+(-1)^n}{n^s}=sum_{n=1}^inftyfrac2{(2n)^s}=2^{1-s}sum_{n=1}^inftyfrac1{n^s}=2^{1-s}zeta(s)$$



                                                        $$zeta(s)-eta(s)=2^{1-s}zeta(s)implies (1-2^{1-s})zeta(s)=eta(s)$$




                                                        $$zeta(s)=frac1{1-2^{1-s}}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$




                                                        After a quick application of Euler's transform, we get a nice analytic continuation to the entire complex plane.



                                                        $${small E_1}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}=sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$



                                                        Finally giving us




                                                        $$zeta(s)=frac1{1-2^{1-s}}sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$




                                                        And at $s=-1$,



                                                        $$zeta(-1)=-frac1{12}$$






                                                        share|cite|improve this answer











                                                        $endgroup$


















                                                          8












                                                          $begingroup$

                                                          Here is a useful place for Euler's transform and the Dirichlet eta function:



                                                          $$zeta(s)=sum_{n=1}^inftyfrac1{n^s}$$



                                                          $$eta(s)=sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$



                                                          $$zeta(s)-eta(s)=sum_{n=1}^inftyfrac{1+(-1)^n}{n^s}=sum_{n=1}^inftyfrac2{(2n)^s}=2^{1-s}sum_{n=1}^inftyfrac1{n^s}=2^{1-s}zeta(s)$$



                                                          $$zeta(s)-eta(s)=2^{1-s}zeta(s)implies (1-2^{1-s})zeta(s)=eta(s)$$




                                                          $$zeta(s)=frac1{1-2^{1-s}}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$




                                                          After a quick application of Euler's transform, we get a nice analytic continuation to the entire complex plane.



                                                          $${small E_1}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}=sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$



                                                          Finally giving us




                                                          $$zeta(s)=frac1{1-2^{1-s}}sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$




                                                          And at $s=-1$,



                                                          $$zeta(-1)=-frac1{12}$$






                                                          share|cite|improve this answer











                                                          $endgroup$
















                                                            8












                                                            8








                                                            8





                                                            $begingroup$

                                                            Here is a useful place for Euler's transform and the Dirichlet eta function:



                                                            $$zeta(s)=sum_{n=1}^inftyfrac1{n^s}$$



                                                            $$eta(s)=sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$



                                                            $$zeta(s)-eta(s)=sum_{n=1}^inftyfrac{1+(-1)^n}{n^s}=sum_{n=1}^inftyfrac2{(2n)^s}=2^{1-s}sum_{n=1}^inftyfrac1{n^s}=2^{1-s}zeta(s)$$



                                                            $$zeta(s)-eta(s)=2^{1-s}zeta(s)implies (1-2^{1-s})zeta(s)=eta(s)$$




                                                            $$zeta(s)=frac1{1-2^{1-s}}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$




                                                            After a quick application of Euler's transform, we get a nice analytic continuation to the entire complex plane.



                                                            $${small E_1}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}=sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$



                                                            Finally giving us




                                                            $$zeta(s)=frac1{1-2^{1-s}}sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$




                                                            And at $s=-1$,



                                                            $$zeta(-1)=-frac1{12}$$






                                                            share|cite|improve this answer











                                                            $endgroup$



                                                            Here is a useful place for Euler's transform and the Dirichlet eta function:



                                                            $$zeta(s)=sum_{n=1}^inftyfrac1{n^s}$$



                                                            $$eta(s)=sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$



                                                            $$zeta(s)-eta(s)=sum_{n=1}^inftyfrac{1+(-1)^n}{n^s}=sum_{n=1}^inftyfrac2{(2n)^s}=2^{1-s}sum_{n=1}^inftyfrac1{n^s}=2^{1-s}zeta(s)$$



                                                            $$zeta(s)-eta(s)=2^{1-s}zeta(s)implies (1-2^{1-s})zeta(s)=eta(s)$$




                                                            $$zeta(s)=frac1{1-2^{1-s}}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}$$




                                                            After a quick application of Euler's transform, we get a nice analytic continuation to the entire complex plane.



                                                            $${small E_1}sum_{n=1}^inftyfrac{(-1)^{n+1}}{n^s}=sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$



                                                            Finally giving us




                                                            $$zeta(s)=frac1{1-2^{1-s}}sum_{n=0}^inftyleft[frac1{2^{n+1}}sum_{k=0}^nbinom nkfrac{(-1)^k}{(k+1)^s}right]$$




                                                            And at $s=-1$,



                                                            $$zeta(-1)=-frac1{12}$$







                                                            share|cite|improve this answer














                                                            share|cite|improve this answer



                                                            share|cite|improve this answer








                                                            edited Jan 2 '18 at 14:01









                                                            John Bentin

                                                            11.2k22554




                                                            11.2k22554










                                                            answered Jan 31 '17 at 14:55









                                                            Simply Beautiful ArtSimply Beautiful Art

                                                            50.4k578181




                                                            50.4k578181























                                                                0












                                                                $begingroup$


                                                                Full disclosure: I saw a different question ask a similar question to the OP and it got closed before I could explain it. I'm mostly focusing on why people say the sum of the natural numbers is $-1/12$, and why that's wrong, in as simple a way as I can communicate and assuming as little knowledge as possible, save for perhaps a bit of basic calculus, since the OP of that question was only in high school.






                                                                The notion that the sum of the natural numbers, in any sense, is $-1/12$ is actually completely untrue and it results from a misconception some people have ... I actually saw a professional mathematician make this mistake in a lecture yesterday so I guess it penetrates even academia.



                                                                I'll try to give you the rough idea of what this, and what the misconception is.



                                                                We define the Riemann zeta function $zeta(s)$ by the summation



                                                                $$zeta(s) = sum_{n=1}^infty frac{1}{n^s} = 1 + frac{1}{2^s} + frac{1}{3^s} + frac{1}{4^s} + frac{1}{5^s} + ...$$



                                                                This $s$ can be any complex or real number, within a specific constant. If $s$ is real, then we must have $s>1$. If $s$ is complex, the real part of $s$, denoted $Re(s)$, must satisfy $Re(s)>1$. (Of course, both fall under this latter condition but we're going to focus on the former here since the complex inputs for the function aren't important to describe the phenomenon.)



                                                                Why do we impose this constraint? You might recall $p$-series from Calculus I or II in school: the convergence of this sum diverges if $sleq 1$ as a result. For example, if $s=1$:



                                                                $$zeta(1) "=" sum_{n=1}^infty frac{1}{n} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + ...$$



                                                                This is a well-known divergent sum, the harmonic series. As an example of a convergent sum, consider $s=2$:



                                                                $$zeta(2) = sum_{n=1}^infty frac{1}{n^2} = 1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + ... = frac{pi^2}{6}$$



                                                                This notion - the sum of the reciprocals of the perfect squares - is known as the Basel problem, and the result was first proved by Euler. 3Blue1Brown has a great video showing (somewhat intuitively) how the sum converges to this value if you're curious - https://www.youtube.com/watch?v=d-o3eB9sfls



                                                                So it should be relatively clear that the sum itself diverges when $s>1$. Okay, so how does this relate to the sum of the natural numbers, i.e. $1+2+3+4+...$?



                                                                Well, notice:



                                                                $$zeta(-1) "=" sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$






                                                                What you should be saying to yourself right now: "But wait! $zeta(s)$ isn't convergent for $s<1$! $-1$ is clearly less than $1$, what the hell are you doing?"






                                                                This is exactly where the misconceptions begin to start.



                                                                Flat out, the notion of $zeta(-1)$ is nonsensical right now. $zeta(s)$ only works for $s>1$ - it is a very bad idea to even consider messing with a clearly divergent sum like this.



                                                                However, out of our interests in the field of complex analysis, we can define the $zeta(s)$ in such a way that it can take in inputs $s leq 1$ (or $Re(s)leq 1$ for complex $s$). This is a process known as "analytic continuation." The notion is a bit above even my head, but the general notion is to create a definition for $zeta(s)$ (or any function) for $s leq 1$ such that the function is differentiable everywhere. The notion of differentiability also gives us a unique definition in this sense.



                                                                I'll skip over the details because I'm not qualified (it is discussed here), but what results is this definition for $zeta(s)$ for $s < 1$:



                                                                $$zeta(s) = 2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s)$$



                                                                Oof, what a messy definition. (Note: $s=1$ is still divergent for this!)



                                                                So this gives us the "grander" definition of the $zeta$ function for all $s$, negative or positive or complex, except $s=1$:



                                                                $$zeta(s) = )left{begin{matrix}
                                                                sum_{n=1}^infty n^{-s} & Re(s)>1 \
                                                                2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s) & Re(s)<1
                                                                end{matrix}right.$$



                                                                It can be shown, through this definition, $zeta(-1) = -1/12$.



                                                                However, if you plug $-1$ naively into the original zeta function - the summation over the natural numbers - and ignore that the function doesn't converge there, you get $1+2+3+4+5+...$. Taken alongside the fact that under the analytic continuation of the zeta function $zeta(-1) = -1/12$, this leads one to naively conclude



                                                                $$1+2+3+4+5+ ... = -frac{1}{12}$$



                                                                Now, obviously, this is untrue. As stated, $zeta(s)$ in its summation form does not converge for $s leq 1$, so it's just outright wrong to say it's equal in this way.



                                                                That this has almost become prolific as a sort of meme and spread to people not knowledgeable enough to no any better doesn't help matters.






                                                                A digression on "convergence," Ramanujan, and partial sums:






                                                                Interestingly, Ramanujan came up with a similar result without making use of the zeta function, by manipulating the infinite sum $1+2+3+4+5+...$, letting it equal some constant $c$, and then manipulations yielded $c = -frac{1}{12}$. I've seen people tempted to defend the result $zeta(-1) = -frac{1}{12}$ since they're basically the same thing. So before that notion comes up, I'd also like to note that while it depends on your notion of "convergence" (perhaps there's a context in which the sum can be said to converge?), most people use the definition of convergence as the limit of the partial sums.



                                                                In which case, let us consider the partial sums of $1+2+3+4+5+...$. We let $P(n)$ denote the sum of the first $n$ terms in the sum. Clearly,




                                                                • $P(1) = 1$

                                                                • $P(2) = 3$

                                                                • $P(3) = 6$

                                                                • $P(4) = 10$

                                                                • $P(5) = 15$

                                                                • $P(6) = 21$

                                                                • $P(7) = 28$

                                                                • ...

                                                                • $P(n) = frac{n(n+1)}{2}$


                                                                Clearly,



                                                                $$lim_{n to infty} P(n) = lim_{n to infty} frac{n(n+1)}{2} = infty$$



                                                                i.e. the sum diverges.



                                                                This is not the only way one can define convergence: for example, Cesaro convergence looks at the average of the partial sums and their limit. This is a context in which



                                                                $$sum_{n=1}^infty (-1)^n = (-1) + 1 +(-1) + 1 +(-1) + 1 +... = frac{1}{2}$$



                                                                It should be clear that the averages for the summation $1+2+3+4+5+...$ also diverge - and certainly don't approach $-1/12$!



                                                                So in that sense we can't utilize Ramanujan's summation since the notion of manipulating the summation as he did (which involved multiplying the sum by scalars and adding/subtracting such things from the original sum) is generally forbidden for divergent sums.






                                                                Thus, the end conclusion and what needs to be said:






                                                                $$zeta(-1) = -frac{1}{12} ; ; ; ; ; text{BUT} ; ; ; ; ; zeta(-1) neq sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$



                                                                And therefore,



                                                                $$-frac{1}{12} neq 1+2+3+4+5+...$$






                                                                share|cite|improve this answer









                                                                $endgroup$


















                                                                  0












                                                                  $begingroup$


                                                                  Full disclosure: I saw a different question ask a similar question to the OP and it got closed before I could explain it. I'm mostly focusing on why people say the sum of the natural numbers is $-1/12$, and why that's wrong, in as simple a way as I can communicate and assuming as little knowledge as possible, save for perhaps a bit of basic calculus, since the OP of that question was only in high school.






                                                                  The notion that the sum of the natural numbers, in any sense, is $-1/12$ is actually completely untrue and it results from a misconception some people have ... I actually saw a professional mathematician make this mistake in a lecture yesterday so I guess it penetrates even academia.



                                                                  I'll try to give you the rough idea of what this, and what the misconception is.



                                                                  We define the Riemann zeta function $zeta(s)$ by the summation



                                                                  $$zeta(s) = sum_{n=1}^infty frac{1}{n^s} = 1 + frac{1}{2^s} + frac{1}{3^s} + frac{1}{4^s} + frac{1}{5^s} + ...$$



                                                                  This $s$ can be any complex or real number, within a specific constant. If $s$ is real, then we must have $s>1$. If $s$ is complex, the real part of $s$, denoted $Re(s)$, must satisfy $Re(s)>1$. (Of course, both fall under this latter condition but we're going to focus on the former here since the complex inputs for the function aren't important to describe the phenomenon.)



                                                                  Why do we impose this constraint? You might recall $p$-series from Calculus I or II in school: the convergence of this sum diverges if $sleq 1$ as a result. For example, if $s=1$:



                                                                  $$zeta(1) "=" sum_{n=1}^infty frac{1}{n} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + ...$$



                                                                  This is a well-known divergent sum, the harmonic series. As an example of a convergent sum, consider $s=2$:



                                                                  $$zeta(2) = sum_{n=1}^infty frac{1}{n^2} = 1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + ... = frac{pi^2}{6}$$



                                                                  This notion - the sum of the reciprocals of the perfect squares - is known as the Basel problem, and the result was first proved by Euler. 3Blue1Brown has a great video showing (somewhat intuitively) how the sum converges to this value if you're curious - https://www.youtube.com/watch?v=d-o3eB9sfls



                                                                  So it should be relatively clear that the sum itself diverges when $s>1$. Okay, so how does this relate to the sum of the natural numbers, i.e. $1+2+3+4+...$?



                                                                  Well, notice:



                                                                  $$zeta(-1) "=" sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$






                                                                  What you should be saying to yourself right now: "But wait! $zeta(s)$ isn't convergent for $s<1$! $-1$ is clearly less than $1$, what the hell are you doing?"






                                                                  This is exactly where the misconceptions begin to start.



                                                                  Flat out, the notion of $zeta(-1)$ is nonsensical right now. $zeta(s)$ only works for $s>1$ - it is a very bad idea to even consider messing with a clearly divergent sum like this.



                                                                  However, out of our interests in the field of complex analysis, we can define the $zeta(s)$ in such a way that it can take in inputs $s leq 1$ (or $Re(s)leq 1$ for complex $s$). This is a process known as "analytic continuation." The notion is a bit above even my head, but the general notion is to create a definition for $zeta(s)$ (or any function) for $s leq 1$ such that the function is differentiable everywhere. The notion of differentiability also gives us a unique definition in this sense.



                                                                  I'll skip over the details because I'm not qualified (it is discussed here), but what results is this definition for $zeta(s)$ for $s < 1$:



                                                                  $$zeta(s) = 2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s)$$



                                                                  Oof, what a messy definition. (Note: $s=1$ is still divergent for this!)



                                                                  So this gives us the "grander" definition of the $zeta$ function for all $s$, negative or positive or complex, except $s=1$:



                                                                  $$zeta(s) = )left{begin{matrix}
                                                                  sum_{n=1}^infty n^{-s} & Re(s)>1 \
                                                                  2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s) & Re(s)<1
                                                                  end{matrix}right.$$



                                                                  It can be shown, through this definition, $zeta(-1) = -1/12$.



                                                                  However, if you plug $-1$ naively into the original zeta function - the summation over the natural numbers - and ignore that the function doesn't converge there, you get $1+2+3+4+5+...$. Taken alongside the fact that under the analytic continuation of the zeta function $zeta(-1) = -1/12$, this leads one to naively conclude



                                                                  $$1+2+3+4+5+ ... = -frac{1}{12}$$



                                                                  Now, obviously, this is untrue. As stated, $zeta(s)$ in its summation form does not converge for $s leq 1$, so it's just outright wrong to say it's equal in this way.



                                                                  That this has almost become prolific as a sort of meme and spread to people not knowledgeable enough to no any better doesn't help matters.






                                                                  A digression on "convergence," Ramanujan, and partial sums:






                                                                  Interestingly, Ramanujan came up with a similar result without making use of the zeta function, by manipulating the infinite sum $1+2+3+4+5+...$, letting it equal some constant $c$, and then manipulations yielded $c = -frac{1}{12}$. I've seen people tempted to defend the result $zeta(-1) = -frac{1}{12}$ since they're basically the same thing. So before that notion comes up, I'd also like to note that while it depends on your notion of "convergence" (perhaps there's a context in which the sum can be said to converge?), most people use the definition of convergence as the limit of the partial sums.



                                                                  In which case, let us consider the partial sums of $1+2+3+4+5+...$. We let $P(n)$ denote the sum of the first $n$ terms in the sum. Clearly,




                                                                  • $P(1) = 1$

                                                                  • $P(2) = 3$

                                                                  • $P(3) = 6$

                                                                  • $P(4) = 10$

                                                                  • $P(5) = 15$

                                                                  • $P(6) = 21$

                                                                  • $P(7) = 28$

                                                                  • ...

                                                                  • $P(n) = frac{n(n+1)}{2}$


                                                                  Clearly,



                                                                  $$lim_{n to infty} P(n) = lim_{n to infty} frac{n(n+1)}{2} = infty$$



                                                                  i.e. the sum diverges.



                                                                  This is not the only way one can define convergence: for example, Cesaro convergence looks at the average of the partial sums and their limit. This is a context in which



                                                                  $$sum_{n=1}^infty (-1)^n = (-1) + 1 +(-1) + 1 +(-1) + 1 +... = frac{1}{2}$$



                                                                  It should be clear that the averages for the summation $1+2+3+4+5+...$ also diverge - and certainly don't approach $-1/12$!



                                                                  So in that sense we can't utilize Ramanujan's summation since the notion of manipulating the summation as he did (which involved multiplying the sum by scalars and adding/subtracting such things from the original sum) is generally forbidden for divergent sums.






                                                                  Thus, the end conclusion and what needs to be said:






                                                                  $$zeta(-1) = -frac{1}{12} ; ; ; ; ; text{BUT} ; ; ; ; ; zeta(-1) neq sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$



                                                                  And therefore,



                                                                  $$-frac{1}{12} neq 1+2+3+4+5+...$$






                                                                  share|cite|improve this answer









                                                                  $endgroup$
















                                                                    0












                                                                    0








                                                                    0





                                                                    $begingroup$


                                                                    Full disclosure: I saw a different question ask a similar question to the OP and it got closed before I could explain it. I'm mostly focusing on why people say the sum of the natural numbers is $-1/12$, and why that's wrong, in as simple a way as I can communicate and assuming as little knowledge as possible, save for perhaps a bit of basic calculus, since the OP of that question was only in high school.






                                                                    The notion that the sum of the natural numbers, in any sense, is $-1/12$ is actually completely untrue and it results from a misconception some people have ... I actually saw a professional mathematician make this mistake in a lecture yesterday so I guess it penetrates even academia.



                                                                    I'll try to give you the rough idea of what this, and what the misconception is.



                                                                    We define the Riemann zeta function $zeta(s)$ by the summation



                                                                    $$zeta(s) = sum_{n=1}^infty frac{1}{n^s} = 1 + frac{1}{2^s} + frac{1}{3^s} + frac{1}{4^s} + frac{1}{5^s} + ...$$



                                                                    This $s$ can be any complex or real number, within a specific constant. If $s$ is real, then we must have $s>1$. If $s$ is complex, the real part of $s$, denoted $Re(s)$, must satisfy $Re(s)>1$. (Of course, both fall under this latter condition but we're going to focus on the former here since the complex inputs for the function aren't important to describe the phenomenon.)



                                                                    Why do we impose this constraint? You might recall $p$-series from Calculus I or II in school: the convergence of this sum diverges if $sleq 1$ as a result. For example, if $s=1$:



                                                                    $$zeta(1) "=" sum_{n=1}^infty frac{1}{n} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + ...$$



                                                                    This is a well-known divergent sum, the harmonic series. As an example of a convergent sum, consider $s=2$:



                                                                    $$zeta(2) = sum_{n=1}^infty frac{1}{n^2} = 1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + ... = frac{pi^2}{6}$$



                                                                    This notion - the sum of the reciprocals of the perfect squares - is known as the Basel problem, and the result was first proved by Euler. 3Blue1Brown has a great video showing (somewhat intuitively) how the sum converges to this value if you're curious - https://www.youtube.com/watch?v=d-o3eB9sfls



                                                                    So it should be relatively clear that the sum itself diverges when $s>1$. Okay, so how does this relate to the sum of the natural numbers, i.e. $1+2+3+4+...$?



                                                                    Well, notice:



                                                                    $$zeta(-1) "=" sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$






                                                                    What you should be saying to yourself right now: "But wait! $zeta(s)$ isn't convergent for $s<1$! $-1$ is clearly less than $1$, what the hell are you doing?"






                                                                    This is exactly where the misconceptions begin to start.



                                                                    Flat out, the notion of $zeta(-1)$ is nonsensical right now. $zeta(s)$ only works for $s>1$ - it is a very bad idea to even consider messing with a clearly divergent sum like this.



                                                                    However, out of our interests in the field of complex analysis, we can define the $zeta(s)$ in such a way that it can take in inputs $s leq 1$ (or $Re(s)leq 1$ for complex $s$). This is a process known as "analytic continuation." The notion is a bit above even my head, but the general notion is to create a definition for $zeta(s)$ (or any function) for $s leq 1$ such that the function is differentiable everywhere. The notion of differentiability also gives us a unique definition in this sense.



                                                                    I'll skip over the details because I'm not qualified (it is discussed here), but what results is this definition for $zeta(s)$ for $s < 1$:



                                                                    $$zeta(s) = 2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s)$$



                                                                    Oof, what a messy definition. (Note: $s=1$ is still divergent for this!)



                                                                    So this gives us the "grander" definition of the $zeta$ function for all $s$, negative or positive or complex, except $s=1$:



                                                                    $$zeta(s) = )left{begin{matrix}
                                                                    sum_{n=1}^infty n^{-s} & Re(s)>1 \
                                                                    2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s) & Re(s)<1
                                                                    end{matrix}right.$$



                                                                    It can be shown, through this definition, $zeta(-1) = -1/12$.



                                                                    However, if you plug $-1$ naively into the original zeta function - the summation over the natural numbers - and ignore that the function doesn't converge there, you get $1+2+3+4+5+...$. Taken alongside the fact that under the analytic continuation of the zeta function $zeta(-1) = -1/12$, this leads one to naively conclude



                                                                    $$1+2+3+4+5+ ... = -frac{1}{12}$$



                                                                    Now, obviously, this is untrue. As stated, $zeta(s)$ in its summation form does not converge for $s leq 1$, so it's just outright wrong to say it's equal in this way.



                                                                    That this has almost become prolific as a sort of meme and spread to people not knowledgeable enough to no any better doesn't help matters.






                                                                    A digression on "convergence," Ramanujan, and partial sums:






                                                                    Interestingly, Ramanujan came up with a similar result without making use of the zeta function, by manipulating the infinite sum $1+2+3+4+5+...$, letting it equal some constant $c$, and then manipulations yielded $c = -frac{1}{12}$. I've seen people tempted to defend the result $zeta(-1) = -frac{1}{12}$ since they're basically the same thing. So before that notion comes up, I'd also like to note that while it depends on your notion of "convergence" (perhaps there's a context in which the sum can be said to converge?), most people use the definition of convergence as the limit of the partial sums.



                                                                    In which case, let us consider the partial sums of $1+2+3+4+5+...$. We let $P(n)$ denote the sum of the first $n$ terms in the sum. Clearly,




                                                                    • $P(1) = 1$

                                                                    • $P(2) = 3$

                                                                    • $P(3) = 6$

                                                                    • $P(4) = 10$

                                                                    • $P(5) = 15$

                                                                    • $P(6) = 21$

                                                                    • $P(7) = 28$

                                                                    • ...

                                                                    • $P(n) = frac{n(n+1)}{2}$


                                                                    Clearly,



                                                                    $$lim_{n to infty} P(n) = lim_{n to infty} frac{n(n+1)}{2} = infty$$



                                                                    i.e. the sum diverges.



                                                                    This is not the only way one can define convergence: for example, Cesaro convergence looks at the average of the partial sums and their limit. This is a context in which



                                                                    $$sum_{n=1}^infty (-1)^n = (-1) + 1 +(-1) + 1 +(-1) + 1 +... = frac{1}{2}$$



                                                                    It should be clear that the averages for the summation $1+2+3+4+5+...$ also diverge - and certainly don't approach $-1/12$!



                                                                    So in that sense we can't utilize Ramanujan's summation since the notion of manipulating the summation as he did (which involved multiplying the sum by scalars and adding/subtracting such things from the original sum) is generally forbidden for divergent sums.






                                                                    Thus, the end conclusion and what needs to be said:






                                                                    $$zeta(-1) = -frac{1}{12} ; ; ; ; ; text{BUT} ; ; ; ; ; zeta(-1) neq sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$



                                                                    And therefore,



                                                                    $$-frac{1}{12} neq 1+2+3+4+5+...$$






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                                                                    $endgroup$




                                                                    Full disclosure: I saw a different question ask a similar question to the OP and it got closed before I could explain it. I'm mostly focusing on why people say the sum of the natural numbers is $-1/12$, and why that's wrong, in as simple a way as I can communicate and assuming as little knowledge as possible, save for perhaps a bit of basic calculus, since the OP of that question was only in high school.






                                                                    The notion that the sum of the natural numbers, in any sense, is $-1/12$ is actually completely untrue and it results from a misconception some people have ... I actually saw a professional mathematician make this mistake in a lecture yesterday so I guess it penetrates even academia.



                                                                    I'll try to give you the rough idea of what this, and what the misconception is.



                                                                    We define the Riemann zeta function $zeta(s)$ by the summation



                                                                    $$zeta(s) = sum_{n=1}^infty frac{1}{n^s} = 1 + frac{1}{2^s} + frac{1}{3^s} + frac{1}{4^s} + frac{1}{5^s} + ...$$



                                                                    This $s$ can be any complex or real number, within a specific constant. If $s$ is real, then we must have $s>1$. If $s$ is complex, the real part of $s$, denoted $Re(s)$, must satisfy $Re(s)>1$. (Of course, both fall under this latter condition but we're going to focus on the former here since the complex inputs for the function aren't important to describe the phenomenon.)



                                                                    Why do we impose this constraint? You might recall $p$-series from Calculus I or II in school: the convergence of this sum diverges if $sleq 1$ as a result. For example, if $s=1$:



                                                                    $$zeta(1) "=" sum_{n=1}^infty frac{1}{n} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + ...$$



                                                                    This is a well-known divergent sum, the harmonic series. As an example of a convergent sum, consider $s=2$:



                                                                    $$zeta(2) = sum_{n=1}^infty frac{1}{n^2} = 1 + frac{1}{2^2} + frac{1}{3^2} + frac{1}{4^2} + frac{1}{5^2} + ... = frac{pi^2}{6}$$



                                                                    This notion - the sum of the reciprocals of the perfect squares - is known as the Basel problem, and the result was first proved by Euler. 3Blue1Brown has a great video showing (somewhat intuitively) how the sum converges to this value if you're curious - https://www.youtube.com/watch?v=d-o3eB9sfls



                                                                    So it should be relatively clear that the sum itself diverges when $s>1$. Okay, so how does this relate to the sum of the natural numbers, i.e. $1+2+3+4+...$?



                                                                    Well, notice:



                                                                    $$zeta(-1) "=" sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$






                                                                    What you should be saying to yourself right now: "But wait! $zeta(s)$ isn't convergent for $s<1$! $-1$ is clearly less than $1$, what the hell are you doing?"






                                                                    This is exactly where the misconceptions begin to start.



                                                                    Flat out, the notion of $zeta(-1)$ is nonsensical right now. $zeta(s)$ only works for $s>1$ - it is a very bad idea to even consider messing with a clearly divergent sum like this.



                                                                    However, out of our interests in the field of complex analysis, we can define the $zeta(s)$ in such a way that it can take in inputs $s leq 1$ (or $Re(s)leq 1$ for complex $s$). This is a process known as "analytic continuation." The notion is a bit above even my head, but the general notion is to create a definition for $zeta(s)$ (or any function) for $s leq 1$ such that the function is differentiable everywhere. The notion of differentiability also gives us a unique definition in this sense.



                                                                    I'll skip over the details because I'm not qualified (it is discussed here), but what results is this definition for $zeta(s)$ for $s < 1$:



                                                                    $$zeta(s) = 2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s)$$



                                                                    Oof, what a messy definition. (Note: $s=1$ is still divergent for this!)



                                                                    So this gives us the "grander" definition of the $zeta$ function for all $s$, negative or positive or complex, except $s=1$:



                                                                    $$zeta(s) = )left{begin{matrix}
                                                                    sum_{n=1}^infty n^{-s} & Re(s)>1 \
                                                                    2^s pi^{s-1} sin left(dfrac{pi s}2right) Gamma(1-s) zeta(1-s) & Re(s)<1
                                                                    end{matrix}right.$$



                                                                    It can be shown, through this definition, $zeta(-1) = -1/12$.



                                                                    However, if you plug $-1$ naively into the original zeta function - the summation over the natural numbers - and ignore that the function doesn't converge there, you get $1+2+3+4+5+...$. Taken alongside the fact that under the analytic continuation of the zeta function $zeta(-1) = -1/12$, this leads one to naively conclude



                                                                    $$1+2+3+4+5+ ... = -frac{1}{12}$$



                                                                    Now, obviously, this is untrue. As stated, $zeta(s)$ in its summation form does not converge for $s leq 1$, so it's just outright wrong to say it's equal in this way.



                                                                    That this has almost become prolific as a sort of meme and spread to people not knowledgeable enough to no any better doesn't help matters.






                                                                    A digression on "convergence," Ramanujan, and partial sums:






                                                                    Interestingly, Ramanujan came up with a similar result without making use of the zeta function, by manipulating the infinite sum $1+2+3+4+5+...$, letting it equal some constant $c$, and then manipulations yielded $c = -frac{1}{12}$. I've seen people tempted to defend the result $zeta(-1) = -frac{1}{12}$ since they're basically the same thing. So before that notion comes up, I'd also like to note that while it depends on your notion of "convergence" (perhaps there's a context in which the sum can be said to converge?), most people use the definition of convergence as the limit of the partial sums.



                                                                    In which case, let us consider the partial sums of $1+2+3+4+5+...$. We let $P(n)$ denote the sum of the first $n$ terms in the sum. Clearly,




                                                                    • $P(1) = 1$

                                                                    • $P(2) = 3$

                                                                    • $P(3) = 6$

                                                                    • $P(4) = 10$

                                                                    • $P(5) = 15$

                                                                    • $P(6) = 21$

                                                                    • $P(7) = 28$

                                                                    • ...

                                                                    • $P(n) = frac{n(n+1)}{2}$


                                                                    Clearly,



                                                                    $$lim_{n to infty} P(n) = lim_{n to infty} frac{n(n+1)}{2} = infty$$



                                                                    i.e. the sum diverges.



                                                                    This is not the only way one can define convergence: for example, Cesaro convergence looks at the average of the partial sums and their limit. This is a context in which



                                                                    $$sum_{n=1}^infty (-1)^n = (-1) + 1 +(-1) + 1 +(-1) + 1 +... = frac{1}{2}$$



                                                                    It should be clear that the averages for the summation $1+2+3+4+5+...$ also diverge - and certainly don't approach $-1/12$!



                                                                    So in that sense we can't utilize Ramanujan's summation since the notion of manipulating the summation as he did (which involved multiplying the sum by scalars and adding/subtracting such things from the original sum) is generally forbidden for divergent sums.






                                                                    Thus, the end conclusion and what needs to be said:






                                                                    $$zeta(-1) = -frac{1}{12} ; ; ; ; ; text{BUT} ; ; ; ; ; zeta(-1) neq sum_{n=1}^infty frac{1}{n^{-1}} = sum_{n=1}^infty n = 1+2+3+4+5+...$$



                                                                    And therefore,



                                                                    $$-frac{1}{12} neq 1+2+3+4+5+...$$







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                                                                    answered Nov 29 '18 at 1:52









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                                                                        protected by Bruno Joyal Dec 6 '13 at 22:43



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