Evaluation of this infinite series












0












$begingroup$


Given the following infinite series:



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



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






real-analysis







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04
















0












$begingroup$


Given the following infinite series:



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



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






real-analysis







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04














0












0








0





$begingroup$


Given the following infinite series:



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



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






real-analysis







share|cite|improve this question









$endgroup$




Given the following infinite series:



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



Find the summation of this series?



My approach:



The above series expands to:



Sum = $(frac{1}{1^p}-frac{1}{2^p}) +(frac{1}{2^p}-frac{1}{3^p}) + (frac{1}{3^p}-frac{1}{4^p})+ (frac{1}{4^p}-frac{1}{5^p}).... (frac{1}{(infty)^p}-frac{1}{(infty+2)^p})$



Since, from second term onward, all neighboring terms cancel each other out all the 'middle' terms will cancel each other out. Thus it will evaluate to:



Sum = $(frac{1}{1^p} - frac{1}{(infty+2)^p})$



As, for any finite real/complex value of: $p$ $(infty+2)^p = infty$. Thus,



Sum = $(frac{1}{1^p} - frac{1}{infty})$.



As, $frac{1}{infty} = 0$. Thus:



Sum = $frac{1}{1^p}$. (for any finite value of $p$).



Query: Is the above argument correct and can we can replace the series with $frac{1}{1^p}$ in any equation or not?






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 19 '18 at 12:33









TheoryQuest1TheoryQuest1

1498




1498








  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04














  • 2




    $begingroup$
    You're throwing around $infty$ like it's an actual number...
    $endgroup$
    – Andrew Li
    Dec 19 '18 at 12:35










  • $begingroup$
    No it is not sound. Fortunately there is a very simple way to make it rigorous.
    $endgroup$
    – Did
    Dec 19 '18 at 12:38










  • $begingroup$
    @Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:59






  • 2




    $begingroup$
    It is. $ $ $ $ $ $
    $endgroup$
    – Did
    Dec 19 '18 at 13:04








2




2




$begingroup$
You're throwing around $infty$ like it's an actual number...
$endgroup$
– Andrew Li
Dec 19 '18 at 12:35




$begingroup$
You're throwing around $infty$ like it's an actual number...
$endgroup$
– Andrew Li
Dec 19 '18 at 12:35












$begingroup$
No it is not sound. Fortunately there is a very simple way to make it rigorous.
$endgroup$
– Did
Dec 19 '18 at 12:38




$begingroup$
No it is not sound. Fortunately there is a very simple way to make it rigorous.
$endgroup$
– Did
Dec 19 '18 at 12:38












$begingroup$
@Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:59




$begingroup$
@Did I was wondering if the method below (in the answer) is what you had in mind when you said that there is a simple way to make it rigorous?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:59




2




2




$begingroup$
It is. $ $ $ $ $ $
$endgroup$
– Did
Dec 19 '18 at 13:04




$begingroup$
It is. $ $ $ $ $ $
$endgroup$
– Did
Dec 19 '18 at 13:04










1 Answer
1






active

oldest

votes


















4












$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56
















4












$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56














4












4








4





$begingroup$

Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$






share|cite|improve this answer









$endgroup$



Yes you can, although the symbol $infty$ can be handled in a different way. For example, call



$$
S_N = sum_{n=1}^Nfrac{1}{n^p} - frac{1}{(n + 1)^p}
$$



You showed that



$$
S_N = 1 - frac{1}{(N + 1)^p}
$$



The sum you are interested in is



$$
S = lim_{N to infty} S_N = 1
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 12:37









caveraccaverac

14.8k31130




14.8k31130












  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56


















  • $begingroup$
    Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:52












  • $begingroup$
    (2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
    $endgroup$
    – TheoryQuest1
    Dec 19 '18 at 12:53








  • 1




    $begingroup$
    @TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
    $endgroup$
    – caverac
    Dec 19 '18 at 12:56
















$begingroup$
Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:52






$begingroup$
Thank you. As I understand, when we say $S = lim_{N to infty} S_N = 1$ instead of $frac{1}{1^p}$ we essentially imply that as $lim_{N to infty} then frac{1}{(N + 1)^p} to 0$. Thus, it is never exactly 0 but goes asymptotically to 0. Now, here is what I am struggling with. (1) If I remember correctly, $.99999...$ with infinite digits is considered equal to 1. Why can't we have the same argument for this case?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:52














$begingroup$
(2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:53






$begingroup$
(2) In any equation, do we replace the series, with $S = lim_{N to infty} S_N = 1$ as the sum instead of 1 (as I thought) ?
$endgroup$
– TheoryQuest1
Dec 19 '18 at 12:53






1




1




$begingroup$
@TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
$endgroup$
– caverac
Dec 19 '18 at 12:56




$begingroup$
@TheoryQuest1 You can replace the series as 1, as it converges to 1. You can see this other post, maybe it will help you
$endgroup$
– caverac
Dec 19 '18 at 12:56


















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