Show that $sumlimits_{n=1}^{infty}left(frac{a_{n+1}}{a_n}-1right)$ converges.
$begingroup$
$(a_n)$ is a sequence of strictly increasing positive numbers and has a lower as well as an upper bound. The question is, if the partial sum:
begin{align*}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)&=left(frac{a_{1+1}}{a_1}-1right)+left(frac{a_{2+1}}{a_2}-1right)+cdots +left(frac{a_{k+1}}{a_k}-1right)\&=frac{a_2}{a_1}+frac{a_3}{a_2}+cdots +frac{a_{k+1}}{a_k}-k.end{align*}
converges? Because $(a_n)$ is a sequence of strictly increasing positive numbers, we have $a_{n+1}>a_n$ for all $n$. Because $(a_n)$ is bounded, we have an upper bound, e.g. $a$, and a lower bound, e.g. $b$, such that $b leq a_n leq a$ for all $n$. This means that $b leq a_n < a_{n+1} leq a$...
The problem with that is: if $k rightarrow infty$, then the series diverges, because of the $-k$ at the end of the expression?! Is there some kind of trick or something to use here?!
Thanks for your help in advance.
Best Regards,
Ahmed Hossam
real-analysis calculus sequences-and-series real-numbers
$endgroup$
|
show 4 more comments
$begingroup$
$(a_n)$ is a sequence of strictly increasing positive numbers and has a lower as well as an upper bound. The question is, if the partial sum:
begin{align*}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)&=left(frac{a_{1+1}}{a_1}-1right)+left(frac{a_{2+1}}{a_2}-1right)+cdots +left(frac{a_{k+1}}{a_k}-1right)\&=frac{a_2}{a_1}+frac{a_3}{a_2}+cdots +frac{a_{k+1}}{a_k}-k.end{align*}
converges? Because $(a_n)$ is a sequence of strictly increasing positive numbers, we have $a_{n+1}>a_n$ for all $n$. Because $(a_n)$ is bounded, we have an upper bound, e.g. $a$, and a lower bound, e.g. $b$, such that $b leq a_n leq a$ for all $n$. This means that $b leq a_n < a_{n+1} leq a$...
The problem with that is: if $k rightarrow infty$, then the series diverges, because of the $-k$ at the end of the expression?! Is there some kind of trick or something to use here?!
Thanks for your help in advance.
Best Regards,
Ahmed Hossam
real-analysis calculus sequences-and-series real-numbers
$endgroup$
$begingroup$
Note that the sequence $(a_n)$ converges since it is increasing and bounded above. Hence, the fractions $frac{a_{n+1}}{a_n}$ tend to $1$ so that your summands $frac{a_{n+1}}{a_n}-1$ tend to $0$. Hence, the "$-k$" is not a problem. Of course summands tending to zero is just a necessary, not a sufficient condition for the series to converge.
$endgroup$
– Christoph
Dec 13 '18 at 12:08
$begingroup$
You mean $limlimits_{n rightarrow infty} left(frac{a_{n+1}}{a_n} - 1right) = limlimits_{n rightarrow infty} frac{a_{n+1}}{a_n} - limlimits_{n rightarrow infty} 1 = frac{limlimits_{n rightarrow infty} a_{n+1}}{limlimits_{n rightarrow infty} a_n} - 1 = frac{a}{a} - 1 = 1 - 1 = 0 $ ? Is this the same as $limlimits_{k rightarrow infty} sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:32
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:27
$begingroup$
The partial sums are increasing and bounded from above... $$sum_{n=1}^k left(frac{a_{n+1}}{a_n} -1 right) = sum_{n=1}^k frac{a_{n+1}-a_n}{a_n} le sum_{n=1}^k frac{a_{n+1}-a_n}{a_1} = frac{a_{k+1}-a_1}{a_1}$$
$endgroup$
– achille hui
Dec 13 '18 at 14:38
$begingroup$
The last equality is not clear...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 18:28
|
show 4 more comments
$begingroup$
$(a_n)$ is a sequence of strictly increasing positive numbers and has a lower as well as an upper bound. The question is, if the partial sum:
begin{align*}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)&=left(frac{a_{1+1}}{a_1}-1right)+left(frac{a_{2+1}}{a_2}-1right)+cdots +left(frac{a_{k+1}}{a_k}-1right)\&=frac{a_2}{a_1}+frac{a_3}{a_2}+cdots +frac{a_{k+1}}{a_k}-k.end{align*}
converges? Because $(a_n)$ is a sequence of strictly increasing positive numbers, we have $a_{n+1}>a_n$ for all $n$. Because $(a_n)$ is bounded, we have an upper bound, e.g. $a$, and a lower bound, e.g. $b$, such that $b leq a_n leq a$ for all $n$. This means that $b leq a_n < a_{n+1} leq a$...
The problem with that is: if $k rightarrow infty$, then the series diverges, because of the $-k$ at the end of the expression?! Is there some kind of trick or something to use here?!
Thanks for your help in advance.
Best Regards,
Ahmed Hossam
real-analysis calculus sequences-and-series real-numbers
$endgroup$
$(a_n)$ is a sequence of strictly increasing positive numbers and has a lower as well as an upper bound. The question is, if the partial sum:
begin{align*}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)&=left(frac{a_{1+1}}{a_1}-1right)+left(frac{a_{2+1}}{a_2}-1right)+cdots +left(frac{a_{k+1}}{a_k}-1right)\&=frac{a_2}{a_1}+frac{a_3}{a_2}+cdots +frac{a_{k+1}}{a_k}-k.end{align*}
converges? Because $(a_n)$ is a sequence of strictly increasing positive numbers, we have $a_{n+1}>a_n$ for all $n$. Because $(a_n)$ is bounded, we have an upper bound, e.g. $a$, and a lower bound, e.g. $b$, such that $b leq a_n leq a$ for all $n$. This means that $b leq a_n < a_{n+1} leq a$...
The problem with that is: if $k rightarrow infty$, then the series diverges, because of the $-k$ at the end of the expression?! Is there some kind of trick or something to use here?!
Thanks for your help in advance.
Best Regards,
Ahmed Hossam
real-analysis calculus sequences-and-series real-numbers
real-analysis calculus sequences-and-series real-numbers
edited Dec 13 '18 at 12:24
Ahmed Hossam
asked Dec 13 '18 at 11:53
Ahmed HossamAhmed Hossam
126
126
$begingroup$
Note that the sequence $(a_n)$ converges since it is increasing and bounded above. Hence, the fractions $frac{a_{n+1}}{a_n}$ tend to $1$ so that your summands $frac{a_{n+1}}{a_n}-1$ tend to $0$. Hence, the "$-k$" is not a problem. Of course summands tending to zero is just a necessary, not a sufficient condition for the series to converge.
$endgroup$
– Christoph
Dec 13 '18 at 12:08
$begingroup$
You mean $limlimits_{n rightarrow infty} left(frac{a_{n+1}}{a_n} - 1right) = limlimits_{n rightarrow infty} frac{a_{n+1}}{a_n} - limlimits_{n rightarrow infty} 1 = frac{limlimits_{n rightarrow infty} a_{n+1}}{limlimits_{n rightarrow infty} a_n} - 1 = frac{a}{a} - 1 = 1 - 1 = 0 $ ? Is this the same as $limlimits_{k rightarrow infty} sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:32
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:27
$begingroup$
The partial sums are increasing and bounded from above... $$sum_{n=1}^k left(frac{a_{n+1}}{a_n} -1 right) = sum_{n=1}^k frac{a_{n+1}-a_n}{a_n} le sum_{n=1}^k frac{a_{n+1}-a_n}{a_1} = frac{a_{k+1}-a_1}{a_1}$$
$endgroup$
– achille hui
Dec 13 '18 at 14:38
$begingroup$
The last equality is not clear...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 18:28
|
show 4 more comments
$begingroup$
Note that the sequence $(a_n)$ converges since it is increasing and bounded above. Hence, the fractions $frac{a_{n+1}}{a_n}$ tend to $1$ so that your summands $frac{a_{n+1}}{a_n}-1$ tend to $0$. Hence, the "$-k$" is not a problem. Of course summands tending to zero is just a necessary, not a sufficient condition for the series to converge.
$endgroup$
– Christoph
Dec 13 '18 at 12:08
$begingroup$
You mean $limlimits_{n rightarrow infty} left(frac{a_{n+1}}{a_n} - 1right) = limlimits_{n rightarrow infty} frac{a_{n+1}}{a_n} - limlimits_{n rightarrow infty} 1 = frac{limlimits_{n rightarrow infty} a_{n+1}}{limlimits_{n rightarrow infty} a_n} - 1 = frac{a}{a} - 1 = 1 - 1 = 0 $ ? Is this the same as $limlimits_{k rightarrow infty} sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:32
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:27
$begingroup$
The partial sums are increasing and bounded from above... $$sum_{n=1}^k left(frac{a_{n+1}}{a_n} -1 right) = sum_{n=1}^k frac{a_{n+1}-a_n}{a_n} le sum_{n=1}^k frac{a_{n+1}-a_n}{a_1} = frac{a_{k+1}-a_1}{a_1}$$
$endgroup$
– achille hui
Dec 13 '18 at 14:38
$begingroup$
The last equality is not clear...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 18:28
$begingroup$
Note that the sequence $(a_n)$ converges since it is increasing and bounded above. Hence, the fractions $frac{a_{n+1}}{a_n}$ tend to $1$ so that your summands $frac{a_{n+1}}{a_n}-1$ tend to $0$. Hence, the "$-k$" is not a problem. Of course summands tending to zero is just a necessary, not a sufficient condition for the series to converge.
$endgroup$
– Christoph
Dec 13 '18 at 12:08
$begingroup$
Note that the sequence $(a_n)$ converges since it is increasing and bounded above. Hence, the fractions $frac{a_{n+1}}{a_n}$ tend to $1$ so that your summands $frac{a_{n+1}}{a_n}-1$ tend to $0$. Hence, the "$-k$" is not a problem. Of course summands tending to zero is just a necessary, not a sufficient condition for the series to converge.
$endgroup$
– Christoph
Dec 13 '18 at 12:08
$begingroup$
You mean $limlimits_{n rightarrow infty} left(frac{a_{n+1}}{a_n} - 1right) = limlimits_{n rightarrow infty} frac{a_{n+1}}{a_n} - limlimits_{n rightarrow infty} 1 = frac{limlimits_{n rightarrow infty} a_{n+1}}{limlimits_{n rightarrow infty} a_n} - 1 = frac{a}{a} - 1 = 1 - 1 = 0 $ ? Is this the same as $limlimits_{k rightarrow infty} sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:32
$begingroup$
You mean $limlimits_{n rightarrow infty} left(frac{a_{n+1}}{a_n} - 1right) = limlimits_{n rightarrow infty} frac{a_{n+1}}{a_n} - limlimits_{n rightarrow infty} 1 = frac{limlimits_{n rightarrow infty} a_{n+1}}{limlimits_{n rightarrow infty} a_n} - 1 = frac{a}{a} - 1 = 1 - 1 = 0 $ ? Is this the same as $limlimits_{k rightarrow infty} sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:32
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:27
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:27
$begingroup$
The partial sums are increasing and bounded from above... $$sum_{n=1}^k left(frac{a_{n+1}}{a_n} -1 right) = sum_{n=1}^k frac{a_{n+1}-a_n}{a_n} le sum_{n=1}^k frac{a_{n+1}-a_n}{a_1} = frac{a_{k+1}-a_1}{a_1}$$
$endgroup$
– achille hui
Dec 13 '18 at 14:38
$begingroup$
The partial sums are increasing and bounded from above... $$sum_{n=1}^k left(frac{a_{n+1}}{a_n} -1 right) = sum_{n=1}^k frac{a_{n+1}-a_n}{a_n} le sum_{n=1}^k frac{a_{n+1}-a_n}{a_1} = frac{a_{k+1}-a_1}{a_1}$$
$endgroup$
– achille hui
Dec 13 '18 at 14:38
$begingroup$
The last equality is not clear...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 18:28
$begingroup$
The last equality is not clear...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 18:28
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
We are given
$(a_n)$ is a strictly increasing sequence- of positive terms
- bounded from above.
Let's say $M$ is an upper bound. (*1) and (*2) implies for all $n$,
$$a_{n+1} > a_n > 0 implies frac{a_{n+1}}{a_n} > 1 iff frac{a_{n+1}}{a_n} - 1 > 0$$
As a sequence indexed by $k$, the partial sums $s_k stackrel{def}{=}sumlimits_{n=1}^k left(frac{a_{n+1}}{a_n} - 1right)$ is a sum of positive terms, so it is strictly increasing. More precisely, we have
$$
s_{k+1} - s_{k} =
sumlimits_{n=1}^{k+1} left(frac{a_{n+1}}{a_n} - 1right)
-
sumlimits_{n=1}^{k} left(frac{a_{n+1}}{a_n} - 1right)
= frac{a_{k+2}}{a_{k+1}} - 1 > 0
implies s_{k+1} > s_{k}
$$
By (*1) and (*2) again, we have $a_{n+1} > a_n ge a_1 > 0$. This implies
$$s_k = sum_{n=1}^k left(frac{a_{n+1}}{a_n} - 1 right)
= sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_n}
le sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_1}
= frac{1}{a_1}sum_{n=1}^k (a_{n+1} - a_n)
$$
The last sum is a telescoping sum. Together with (*3), we obtain
$$s_k le frac{1}{a_1}(a_{k+1} - a_1) le frac{M-1}{a_1}$$
This means as a sequence, the partial sums is bounded from above too. Being increasing and bounded from above, the sequence $s_k$ converges. By definition, so does the series
$sumlimits_{n=1}^infty left(frac{a_{n+1}}{a_n} - 1right)$.
$endgroup$
add a comment |
$begingroup$
You can check that it is equivalent to determine whether the series of $logleft(frac{a_{n+1}}{a_n}right)$ converges, because $frac{a_{n+1}}{a_n} rightarrow 1$.
Edit: to be more clear:
$frac{a_{n+1}}{a_n}-1 sim logleft(frac{a_{n+1}}{a_n}right) > 0$ as $n$ goes to infinity.
Thus, the series $sum_n{frac{a_{n+1}}{a_n}-1}$ is convergent iff the series $sum_n{logleft(frac{a_{n+1}}{a_n}right)}$ is convergent.
$endgroup$
$begingroup$
You mean $limlimits_{krightarrow infty}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right) Leftrightarrow limlimits_{krightarrow infty}sumlimits_{n=1}^{k} log left(frac{a_{n+1}}{a_n}right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:38
$begingroup$
No. I mean that one series grows to $infty$ iff the other does.
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– Mindlack
Dec 13 '18 at 12:41
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Ok, which series is that one series and which is the other?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:43
$begingroup$
$frac{a_{k+1}}{a_k} rightarrow 1$ as $k rightarrow infty$, because the sequence $(a_n)$ has an upper bound. But this is only one term in $displaystylefrac{a_{2}}{a_1} + frac{a_{3}}{a_2} +cdots + frac{a_{k+1}}{a_k}-k$. All other fractions are $>1$, because $a_{n+1}>a_n$ ...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:17
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:28
|
show 4 more comments
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
We are given
$(a_n)$ is a strictly increasing sequence- of positive terms
- bounded from above.
Let's say $M$ is an upper bound. (*1) and (*2) implies for all $n$,
$$a_{n+1} > a_n > 0 implies frac{a_{n+1}}{a_n} > 1 iff frac{a_{n+1}}{a_n} - 1 > 0$$
As a sequence indexed by $k$, the partial sums $s_k stackrel{def}{=}sumlimits_{n=1}^k left(frac{a_{n+1}}{a_n} - 1right)$ is a sum of positive terms, so it is strictly increasing. More precisely, we have
$$
s_{k+1} - s_{k} =
sumlimits_{n=1}^{k+1} left(frac{a_{n+1}}{a_n} - 1right)
-
sumlimits_{n=1}^{k} left(frac{a_{n+1}}{a_n} - 1right)
= frac{a_{k+2}}{a_{k+1}} - 1 > 0
implies s_{k+1} > s_{k}
$$
By (*1) and (*2) again, we have $a_{n+1} > a_n ge a_1 > 0$. This implies
$$s_k = sum_{n=1}^k left(frac{a_{n+1}}{a_n} - 1 right)
= sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_n}
le sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_1}
= frac{1}{a_1}sum_{n=1}^k (a_{n+1} - a_n)
$$
The last sum is a telescoping sum. Together with (*3), we obtain
$$s_k le frac{1}{a_1}(a_{k+1} - a_1) le frac{M-1}{a_1}$$
This means as a sequence, the partial sums is bounded from above too. Being increasing and bounded from above, the sequence $s_k$ converges. By definition, so does the series
$sumlimits_{n=1}^infty left(frac{a_{n+1}}{a_n} - 1right)$.
$endgroup$
add a comment |
$begingroup$
We are given
$(a_n)$ is a strictly increasing sequence- of positive terms
- bounded from above.
Let's say $M$ is an upper bound. (*1) and (*2) implies for all $n$,
$$a_{n+1} > a_n > 0 implies frac{a_{n+1}}{a_n} > 1 iff frac{a_{n+1}}{a_n} - 1 > 0$$
As a sequence indexed by $k$, the partial sums $s_k stackrel{def}{=}sumlimits_{n=1}^k left(frac{a_{n+1}}{a_n} - 1right)$ is a sum of positive terms, so it is strictly increasing. More precisely, we have
$$
s_{k+1} - s_{k} =
sumlimits_{n=1}^{k+1} left(frac{a_{n+1}}{a_n} - 1right)
-
sumlimits_{n=1}^{k} left(frac{a_{n+1}}{a_n} - 1right)
= frac{a_{k+2}}{a_{k+1}} - 1 > 0
implies s_{k+1} > s_{k}
$$
By (*1) and (*2) again, we have $a_{n+1} > a_n ge a_1 > 0$. This implies
$$s_k = sum_{n=1}^k left(frac{a_{n+1}}{a_n} - 1 right)
= sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_n}
le sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_1}
= frac{1}{a_1}sum_{n=1}^k (a_{n+1} - a_n)
$$
The last sum is a telescoping sum. Together with (*3), we obtain
$$s_k le frac{1}{a_1}(a_{k+1} - a_1) le frac{M-1}{a_1}$$
This means as a sequence, the partial sums is bounded from above too. Being increasing and bounded from above, the sequence $s_k$ converges. By definition, so does the series
$sumlimits_{n=1}^infty left(frac{a_{n+1}}{a_n} - 1right)$.
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add a comment |
$begingroup$
We are given
$(a_n)$ is a strictly increasing sequence- of positive terms
- bounded from above.
Let's say $M$ is an upper bound. (*1) and (*2) implies for all $n$,
$$a_{n+1} > a_n > 0 implies frac{a_{n+1}}{a_n} > 1 iff frac{a_{n+1}}{a_n} - 1 > 0$$
As a sequence indexed by $k$, the partial sums $s_k stackrel{def}{=}sumlimits_{n=1}^k left(frac{a_{n+1}}{a_n} - 1right)$ is a sum of positive terms, so it is strictly increasing. More precisely, we have
$$
s_{k+1} - s_{k} =
sumlimits_{n=1}^{k+1} left(frac{a_{n+1}}{a_n} - 1right)
-
sumlimits_{n=1}^{k} left(frac{a_{n+1}}{a_n} - 1right)
= frac{a_{k+2}}{a_{k+1}} - 1 > 0
implies s_{k+1} > s_{k}
$$
By (*1) and (*2) again, we have $a_{n+1} > a_n ge a_1 > 0$. This implies
$$s_k = sum_{n=1}^k left(frac{a_{n+1}}{a_n} - 1 right)
= sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_n}
le sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_1}
= frac{1}{a_1}sum_{n=1}^k (a_{n+1} - a_n)
$$
The last sum is a telescoping sum. Together with (*3), we obtain
$$s_k le frac{1}{a_1}(a_{k+1} - a_1) le frac{M-1}{a_1}$$
This means as a sequence, the partial sums is bounded from above too. Being increasing and bounded from above, the sequence $s_k$ converges. By definition, so does the series
$sumlimits_{n=1}^infty left(frac{a_{n+1}}{a_n} - 1right)$.
$endgroup$
We are given
$(a_n)$ is a strictly increasing sequence- of positive terms
- bounded from above.
Let's say $M$ is an upper bound. (*1) and (*2) implies for all $n$,
$$a_{n+1} > a_n > 0 implies frac{a_{n+1}}{a_n} > 1 iff frac{a_{n+1}}{a_n} - 1 > 0$$
As a sequence indexed by $k$, the partial sums $s_k stackrel{def}{=}sumlimits_{n=1}^k left(frac{a_{n+1}}{a_n} - 1right)$ is a sum of positive terms, so it is strictly increasing. More precisely, we have
$$
s_{k+1} - s_{k} =
sumlimits_{n=1}^{k+1} left(frac{a_{n+1}}{a_n} - 1right)
-
sumlimits_{n=1}^{k} left(frac{a_{n+1}}{a_n} - 1right)
= frac{a_{k+2}}{a_{k+1}} - 1 > 0
implies s_{k+1} > s_{k}
$$
By (*1) and (*2) again, we have $a_{n+1} > a_n ge a_1 > 0$. This implies
$$s_k = sum_{n=1}^k left(frac{a_{n+1}}{a_n} - 1 right)
= sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_n}
le sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_1}
= frac{1}{a_1}sum_{n=1}^k (a_{n+1} - a_n)
$$
The last sum is a telescoping sum. Together with (*3), we obtain
$$s_k le frac{1}{a_1}(a_{k+1} - a_1) le frac{M-1}{a_1}$$
This means as a sequence, the partial sums is bounded from above too. Being increasing and bounded from above, the sequence $s_k$ converges. By definition, so does the series
$sumlimits_{n=1}^infty left(frac{a_{n+1}}{a_n} - 1right)$.
answered Dec 14 '18 at 3:58
achille huiachille hui
96.1k5132260
96.1k5132260
add a comment |
add a comment |
$begingroup$
You can check that it is equivalent to determine whether the series of $logleft(frac{a_{n+1}}{a_n}right)$ converges, because $frac{a_{n+1}}{a_n} rightarrow 1$.
Edit: to be more clear:
$frac{a_{n+1}}{a_n}-1 sim logleft(frac{a_{n+1}}{a_n}right) > 0$ as $n$ goes to infinity.
Thus, the series $sum_n{frac{a_{n+1}}{a_n}-1}$ is convergent iff the series $sum_n{logleft(frac{a_{n+1}}{a_n}right)}$ is convergent.
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You mean $limlimits_{krightarrow infty}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right) Leftrightarrow limlimits_{krightarrow infty}sumlimits_{n=1}^{k} log left(frac{a_{n+1}}{a_n}right)$?
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– Ahmed Hossam
Dec 13 '18 at 12:38
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No. I mean that one series grows to $infty$ iff the other does.
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– Mindlack
Dec 13 '18 at 12:41
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Ok, which series is that one series and which is the other?
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– Ahmed Hossam
Dec 13 '18 at 12:43
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$frac{a_{k+1}}{a_k} rightarrow 1$ as $k rightarrow infty$, because the sequence $(a_n)$ has an upper bound. But this is only one term in $displaystylefrac{a_{2}}{a_1} + frac{a_{3}}{a_2} +cdots + frac{a_{k+1}}{a_k}-k$. All other fractions are $>1$, because $a_{n+1}>a_n$ ...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:17
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$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:28
|
show 4 more comments
$begingroup$
You can check that it is equivalent to determine whether the series of $logleft(frac{a_{n+1}}{a_n}right)$ converges, because $frac{a_{n+1}}{a_n} rightarrow 1$.
Edit: to be more clear:
$frac{a_{n+1}}{a_n}-1 sim logleft(frac{a_{n+1}}{a_n}right) > 0$ as $n$ goes to infinity.
Thus, the series $sum_n{frac{a_{n+1}}{a_n}-1}$ is convergent iff the series $sum_n{logleft(frac{a_{n+1}}{a_n}right)}$ is convergent.
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You mean $limlimits_{krightarrow infty}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right) Leftrightarrow limlimits_{krightarrow infty}sumlimits_{n=1}^{k} log left(frac{a_{n+1}}{a_n}right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:38
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No. I mean that one series grows to $infty$ iff the other does.
$endgroup$
– Mindlack
Dec 13 '18 at 12:41
$begingroup$
Ok, which series is that one series and which is the other?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:43
$begingroup$
$frac{a_{k+1}}{a_k} rightarrow 1$ as $k rightarrow infty$, because the sequence $(a_n)$ has an upper bound. But this is only one term in $displaystylefrac{a_{2}}{a_1} + frac{a_{3}}{a_2} +cdots + frac{a_{k+1}}{a_k}-k$. All other fractions are $>1$, because $a_{n+1}>a_n$ ...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:17
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:28
|
show 4 more comments
$begingroup$
You can check that it is equivalent to determine whether the series of $logleft(frac{a_{n+1}}{a_n}right)$ converges, because $frac{a_{n+1}}{a_n} rightarrow 1$.
Edit: to be more clear:
$frac{a_{n+1}}{a_n}-1 sim logleft(frac{a_{n+1}}{a_n}right) > 0$ as $n$ goes to infinity.
Thus, the series $sum_n{frac{a_{n+1}}{a_n}-1}$ is convergent iff the series $sum_n{logleft(frac{a_{n+1}}{a_n}right)}$ is convergent.
$endgroup$
You can check that it is equivalent to determine whether the series of $logleft(frac{a_{n+1}}{a_n}right)$ converges, because $frac{a_{n+1}}{a_n} rightarrow 1$.
Edit: to be more clear:
$frac{a_{n+1}}{a_n}-1 sim logleft(frac{a_{n+1}}{a_n}right) > 0$ as $n$ goes to infinity.
Thus, the series $sum_n{frac{a_{n+1}}{a_n}-1}$ is convergent iff the series $sum_n{logleft(frac{a_{n+1}}{a_n}right)}$ is convergent.
edited Dec 13 '18 at 14:27
answered Dec 13 '18 at 11:59
MindlackMindlack
4,780210
4,780210
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You mean $limlimits_{krightarrow infty}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right) Leftrightarrow limlimits_{krightarrow infty}sumlimits_{n=1}^{k} log left(frac{a_{n+1}}{a_n}right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:38
$begingroup$
No. I mean that one series grows to $infty$ iff the other does.
$endgroup$
– Mindlack
Dec 13 '18 at 12:41
$begingroup$
Ok, which series is that one series and which is the other?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:43
$begingroup$
$frac{a_{k+1}}{a_k} rightarrow 1$ as $k rightarrow infty$, because the sequence $(a_n)$ has an upper bound. But this is only one term in $displaystylefrac{a_{2}}{a_1} + frac{a_{3}}{a_2} +cdots + frac{a_{k+1}}{a_k}-k$. All other fractions are $>1$, because $a_{n+1}>a_n$ ...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:17
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:28
|
show 4 more comments
$begingroup$
You mean $limlimits_{krightarrow infty}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right) Leftrightarrow limlimits_{krightarrow infty}sumlimits_{n=1}^{k} log left(frac{a_{n+1}}{a_n}right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:38
$begingroup$
No. I mean that one series grows to $infty$ iff the other does.
$endgroup$
– Mindlack
Dec 13 '18 at 12:41
$begingroup$
Ok, which series is that one series and which is the other?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:43
$begingroup$
$frac{a_{k+1}}{a_k} rightarrow 1$ as $k rightarrow infty$, because the sequence $(a_n)$ has an upper bound. But this is only one term in $displaystylefrac{a_{2}}{a_1} + frac{a_{3}}{a_2} +cdots + frac{a_{k+1}}{a_k}-k$. All other fractions are $>1$, because $a_{n+1}>a_n$ ...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:17
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:28
$begingroup$
You mean $limlimits_{krightarrow infty}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right) Leftrightarrow limlimits_{krightarrow infty}sumlimits_{n=1}^{k} log left(frac{a_{n+1}}{a_n}right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:38
$begingroup$
You mean $limlimits_{krightarrow infty}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right) Leftrightarrow limlimits_{krightarrow infty}sumlimits_{n=1}^{k} log left(frac{a_{n+1}}{a_n}right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:38
$begingroup$
No. I mean that one series grows to $infty$ iff the other does.
$endgroup$
– Mindlack
Dec 13 '18 at 12:41
$begingroup$
No. I mean that one series grows to $infty$ iff the other does.
$endgroup$
– Mindlack
Dec 13 '18 at 12:41
$begingroup$
Ok, which series is that one series and which is the other?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:43
$begingroup$
Ok, which series is that one series and which is the other?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:43
$begingroup$
$frac{a_{k+1}}{a_k} rightarrow 1$ as $k rightarrow infty$, because the sequence $(a_n)$ has an upper bound. But this is only one term in $displaystylefrac{a_{2}}{a_1} + frac{a_{3}}{a_2} +cdots + frac{a_{k+1}}{a_k}-k$. All other fractions are $>1$, because $a_{n+1}>a_n$ ...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:17
$begingroup$
$frac{a_{k+1}}{a_k} rightarrow 1$ as $k rightarrow infty$, because the sequence $(a_n)$ has an upper bound. But this is only one term in $displaystylefrac{a_{2}}{a_1} + frac{a_{3}}{a_2} +cdots + frac{a_{k+1}}{a_k}-k$. All other fractions are $>1$, because $a_{n+1}>a_n$ ...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:17
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:28
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:28
|
show 4 more comments
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$begingroup$
Note that the sequence $(a_n)$ converges since it is increasing and bounded above. Hence, the fractions $frac{a_{n+1}}{a_n}$ tend to $1$ so that your summands $frac{a_{n+1}}{a_n}-1$ tend to $0$. Hence, the "$-k$" is not a problem. Of course summands tending to zero is just a necessary, not a sufficient condition for the series to converge.
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– Christoph
Dec 13 '18 at 12:08
$begingroup$
You mean $limlimits_{n rightarrow infty} left(frac{a_{n+1}}{a_n} - 1right) = limlimits_{n rightarrow infty} frac{a_{n+1}}{a_n} - limlimits_{n rightarrow infty} 1 = frac{limlimits_{n rightarrow infty} a_{n+1}}{limlimits_{n rightarrow infty} a_n} - 1 = frac{a}{a} - 1 = 1 - 1 = 0 $ ? Is this the same as $limlimits_{k rightarrow infty} sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)$?
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 12:32
$begingroup$
$$sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)=underbrace{left(overbrace{frac{a_{1+1}}{a_1}}^{>1}-1right)}_{>0}+underbrace{left(overbrace{frac{a_{2+1}}{a_2}}^{>1}- 1right)}_{>0}+cdots +underbrace{left(overbrace{frac{a_{k+1}}{a_k}}^{rightarrow 1} -1 right)}_{rightarrow 0}$$ if the terms in the sum converge to $0$, does this mean that, the partial sum converges to something???
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 14:27
$begingroup$
The partial sums are increasing and bounded from above... $$sum_{n=1}^k left(frac{a_{n+1}}{a_n} -1 right) = sum_{n=1}^k frac{a_{n+1}-a_n}{a_n} le sum_{n=1}^k frac{a_{n+1}-a_n}{a_1} = frac{a_{k+1}-a_1}{a_1}$$
$endgroup$
– achille hui
Dec 13 '18 at 14:38
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The last equality is not clear...
$endgroup$
– Ahmed Hossam
Dec 13 '18 at 18:28