Prove the convergence of $prodlimits^{n}_{k=1}{left(1+frac{k}{n^p}right)} $ and Find Its Limit
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
1
asked Nov 22 at 14:06
weilam06
1109
add a comment |
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
1
asked Nov 22 at 14:06
weilam06
1109
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 |
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
1
asked Nov 22 at 14:06
weilam06
1109
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
1
asked Nov 22 at 14:06
weilam06
1109
edited Nov 22 at 14:35
gimusi
1
asked Nov 22 at 14:06
weilam06
1109
edited Nov 22 at 14:35
gimusi
1
edited Nov 22 at 14:35
gimusi
1
edited Nov 22 at 14:35
gimusi
1
1
asked Nov 22 at 14:06
weilam06
1109
asked Nov 22 at 14:06
weilam06
1109
asked Nov 22 at 14:06
weilam06
1109
1109
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
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
1
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009177%2fprove-the-convergence-of-prod-limitsn-k-1-left1-fracknp-right%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
1
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 |
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
1
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 |
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
1
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
1
edited Nov 23 at 0:17
answered Nov 22 at 14:22
gimusi
1
answered Nov 22 at 14:22
gimusi
1
answered Nov 22 at 14:22
gimusi
1
1
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009177%2fprove-the-convergence-of-prod-limitsn-k-1-left1-fracknp-right%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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