Prove the convergence of $prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} $ and Find Its Limit
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Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.
I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?
limits convergence
edited Nov 22 at 14:35
gimusi
88.8k74394
asked Nov 22 at 14:06
weilam06
667
add a comment |
up vote
1
down vote
favorite
Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.
I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?
limits convergence
edited Nov 22 at 14:35
gimusi
88.8k74394
asked Nov 22 at 14:06
weilam06
667
There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19
What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21
$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.
I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?
limits convergence
edited Nov 22 at 14:35
gimusi
88.8k74394
asked Nov 22 at 14:06
weilam06
667
Suppose $p> 1$ and the sequence ${x_n}_{n=1}^{infty}$ has a general term of
$$x_n=prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} spacespacespacespacespacespacespacespacespacespacespacespacespacespacespacespace n=1,2,3, cdots$$
Show that the sequence ${x_n}_{n=1}^{infty}$ converges, and hence find
$$lim_{nrightarrowinfty}{x_n}$$
which is related to $p$ itself.
I have been attempted to find the convergence of the sequence using ratio test but failed. The general term has a form of alike the $p$-series. And also the question seems difficult to find its limit because the denominator is of $p^{th}$ power. How do I deal it?
limits convergence
limits convergence
edited Nov 22 at 14:35
gimusi
88.8k74394
asked Nov 22 at 14:06
weilam06
667
edited Nov 22 at 14:35
gimusi
88.8k74394
asked Nov 22 at 14:06
weilam06
667
edited Nov 22 at 14:35
gimusi
88.8k74394
edited Nov 22 at 14:35
gimusi
88.8k74394
edited Nov 22 at 14:35
gimusi
88.8k74394
88.8k74394
asked Nov 22 at 14:06
weilam06
667
asked Nov 22 at 14:06
weilam06
667
asked Nov 22 at 14:06
weilam06
667
667
There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19
What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21
$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23
add a comment |
There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19
What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21
$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23
There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19
There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19
What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21
What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21
$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23
$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
We have that
$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$
and
$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$
therefore the sequence converges for $pge 2$
- for $p=2 implies x_n to sqrt e$
- for $p>2 implies x_n to 1$
and diverges for $1<p<2$.
Refer also to the related
- Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$
- How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?
answered Nov 22 at 14:22
gimusi
88.8k74394
How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06
For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23
@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56
That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08
@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We have that
$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$
and
$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$
therefore the sequence converges for $pge 2$
- for $p=2 implies x_n to sqrt e$
- for $p>2 implies x_n to 1$
and diverges for $1<p<2$.
Refer also to the related
- Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$
- How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?
answered Nov 22 at 14:22
gimusi
88.8k74394
How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06
For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23
@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56
That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08
@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13
add a comment |
up vote
3
down vote
accepted
We have that
$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$
and
$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$
therefore the sequence converges for $pge 2$
- for $p=2 implies x_n to sqrt e$
- for $p>2 implies x_n to 1$
and diverges for $1<p<2$.
Refer also to the related
- Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$
- How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?
answered Nov 22 at 14:22
gimusi
88.8k74394
How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06
For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23
@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56
That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08
@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We have that
$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$
and
$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$
therefore the sequence converges for $pge 2$
- for $p=2 implies x_n to sqrt e$
- for $p>2 implies x_n to 1$
and diverges for $1<p<2$.
Refer also to the related
- Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$
- How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?
answered Nov 22 at 14:22
gimusi
88.8k74394
We have that
$$prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)}=e^{sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}}$$
and
$$sum^{n}_{k=1}{log left(1+frac{k}{n^p}right)}=sum^{n}_{k=1} left(frac{k}{n^p}+Oleft(frac{k^2}{n^{2p}}right)right)=$$$$=frac{n(n-1)}{2n^{p}}+Oleft(frac{n^3}{n^{2p}}right)=frac{1}{2n^{p-2}}+Oleft(frac{1}{n^{2p-3}}right)$$
therefore the sequence converges for $pge 2$
- for $p=2 implies x_n to sqrt e$
- for $p>2 implies x_n to 1$
and diverges for $1<p<2$.
Refer also to the related
- Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$
- How to evaluate $limlimits_{nto+infty} prodlimits_{k=1}^n (1+k/n^2)$?
answered Nov 22 at 14:22
gimusi
88.8k74394
edited Nov 23 at 0:17
answered Nov 22 at 14:22
gimusi
88.8k74394
answered Nov 22 at 14:22
gimusi
88.8k74394
answered Nov 22 at 14:22
gimusi
88.8k74394
88.8k74394
How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06
For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23
@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56
That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08
@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13
add a comment |
How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06
For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23
@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56
That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08
@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13
How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06
How do you attempt $sumlimits^{n}_{k=1}{log {(1+frac{k}{n^p})}^{n^p}}$~$frac{n^2}{2}$?
– weilam06
Nov 22 at 15:06
For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23
For n large $(1+k/n^p)^{n^p}to e^k$ and $sum k=n(n+1)/2$.
– gimusi
Nov 22 at 15:23
@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56
@weilam06 In my answer I've given the key fact to find a complete solution, I can add some more detail to make thinkg more clear and rigorous.
– gimusi
Nov 22 at 15:56
That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08
That's okay. I understand that $frac{1}{2n^{p-2}}$ is a $p-$series.
– weilam06
Nov 22 at 16:08
@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13
@weilam06 That'snice. I've just modified a little bit the key step in a more precise way! Bye
– gimusi
Nov 22 at 16:13
add a comment |
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There's no way it converges when $p=1$
– mathworker21
Nov 22 at 14:19
What if the case for $p>1$? I have edited the question.
– weilam06
Nov 22 at 14:21
$log(x_n) le sum_{k=1}^n log(1+frac{k}{n^p}) le sum_{k=1}^n frac{k}{n^p} = frac{1}{2}n^{2-p}$, which goes to $0$ if $p > 2$. Idk about $1 < p le 2$.
– mathworker21
Nov 22 at 14:23