Show the terms of a sequence $u_n$ is $0$.
$begingroup$
Given $u_n=sum_{k=1}^{infty} u_{n+k}^2$ and $sum_{n=1}^{infty}u_n$ converge. Show $u_k=0 forall k in mathbb{N}$.
Remark:This sequence is not increasing as $u_n-u_{n+1}=u_{n+1}^2geq0$.
It is bounded below by $0$.
If we can show $u_n=u_{n+1}$, we are done.
Help!
real-analysis
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
$endgroup$
add a comment |
$begingroup$
Given $u_n=sum_{k=1}^{infty} u_{n+k}^2$ and $sum_{n=1}^{infty}u_n$ converge. Show $u_k=0 forall k in mathbb{N}$.
Remark:This sequence is not increasing as $u_n-u_{n+1}=u_{n+1}^2geq0$.
It is bounded below by $0$.
If we can show $u_n=u_{n+1}$, we are done.
Help!
real-analysis
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
$endgroup$
$begingroup$
$u_1=u_2^2+u_3^2+....=0+0+0...=0$?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:02
$begingroup$
Then $u_1$ is not $u_2^2+...$.
$endgroup$
– Mindlack
Dec 13 '18 at 0:02
add a comment |
$begingroup$
Given $u_n=sum_{k=1}^{infty} u_{n+k}^2$ and $sum_{n=1}^{infty}u_n$ converge. Show $u_k=0 forall k in mathbb{N}$.
Remark:This sequence is not increasing as $u_n-u_{n+1}=u_{n+1}^2geq0$.
It is bounded below by $0$.
If we can show $u_n=u_{n+1}$, we are done.
Help!
real-analysis
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
$endgroup$
Given $u_n=sum_{k=1}^{infty} u_{n+k}^2$ and $sum_{n=1}^{infty}u_n$ converge. Show $u_k=0 forall k in mathbb{N}$.
Remark:This sequence is not increasing as $u_n-u_{n+1}=u_{n+1}^2geq0$.
It is bounded below by $0$.
If we can show $u_n=u_{n+1}$, we are done.
Help!
real-analysis
real-analysis
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
edited Dec 12 '18 at 23:39
mathpadawan
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
asked Dec 12 '18 at 23:31
mathpadawanmathpadawan
1,884422
1,884422
$begingroup$
$u_1=u_2^2+u_3^2+....=0+0+0...=0$?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:02
$begingroup$
Then $u_1$ is not $u_2^2+...$.
$endgroup$
– Mindlack
Dec 13 '18 at 0:02
add a comment |
$begingroup$
$u_1=u_2^2+u_3^2+....=0+0+0...=0$?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:02
$begingroup$
Then $u_1$ is not $u_2^2+...$.
$endgroup$
– Mindlack
Dec 13 '18 at 0:02
$begingroup$
$u_1=u_2^2+u_3^2+....=0+0+0...=0$?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:02
$begingroup$
$u_1=u_2^2+u_3^2+....=0+0+0...=0$?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:02
$begingroup$
Then $u_1$ is not $u_2^2+...$.
$endgroup$
– Mindlack
Dec 13 '18 at 0:02
$begingroup$
Then $u_1$ is not $u_2^2+...$.
$endgroup$
– Mindlack
Dec 13 '18 at 0:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $u_n=u_{n+1}+u_{n+1}^2$.
Thus, if for some $N$, $u_N=0$, we are done. Now we assume that $u$ is positive. Hence it is decreasing so converges to some $l geq 0$ such that $l=l+l^2$ thus $l=0$.
Now, let $a < 0$ be a real number, then $u_{n+1}^a-u_n^a=u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$.
Take $a=-1$, then $u_{n+1}^a-u_n^a rightarrow 1$, thus $u_n sim 1/n$ and $sum_n {u_n}=infty$, a contradiction.
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
$endgroup$
$begingroup$
Sorry, how do you get $u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$? What does ~ mean? Also how do you get the step after that, can you elaborate?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:18
1
$begingroup$
I am using Landau (“big Oh”) notations and the fact that $(1+x)^a-1/x rightarrow a$ (Actually, you can just do the computation explicitly with $a=-1$ to find that $u_{n+1}^a-u_n^a$ goes to $1$. The second point you raise is a consequence of Cesaro’s lemma.
$endgroup$
– Mindlack
Dec 13 '18 at 0:21
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%2f3037383%2fshow-the-terms-of-a-sequence-u-n-is-0%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
$begingroup$
Note that $u_n=u_{n+1}+u_{n+1}^2$.
Thus, if for some $N$, $u_N=0$, we are done. Now we assume that $u$ is positive. Hence it is decreasing so converges to some $l geq 0$ such that $l=l+l^2$ thus $l=0$.
Now, let $a < 0$ be a real number, then $u_{n+1}^a-u_n^a=u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$.
Take $a=-1$, then $u_{n+1}^a-u_n^a rightarrow 1$, thus $u_n sim 1/n$ and $sum_n {u_n}=infty$, a contradiction.
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
$endgroup$
$begingroup$
Sorry, how do you get $u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$? What does ~ mean? Also how do you get the step after that, can you elaborate?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:18
1
$begingroup$
I am using Landau (“big Oh”) notations and the fact that $(1+x)^a-1/x rightarrow a$ (Actually, you can just do the computation explicitly with $a=-1$ to find that $u_{n+1}^a-u_n^a$ goes to $1$. The second point you raise is a consequence of Cesaro’s lemma.
$endgroup$
– Mindlack
Dec 13 '18 at 0:21
add a comment |
$begingroup$
Note that $u_n=u_{n+1}+u_{n+1}^2$.
Thus, if for some $N$, $u_N=0$, we are done. Now we assume that $u$ is positive. Hence it is decreasing so converges to some $l geq 0$ such that $l=l+l^2$ thus $l=0$.
Now, let $a < 0$ be a real number, then $u_{n+1}^a-u_n^a=u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$.
Take $a=-1$, then $u_{n+1}^a-u_n^a rightarrow 1$, thus $u_n sim 1/n$ and $sum_n {u_n}=infty$, a contradiction.
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
$endgroup$
$begingroup$
Sorry, how do you get $u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$? What does ~ mean? Also how do you get the step after that, can you elaborate?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:18
1
$begingroup$
I am using Landau (“big Oh”) notations and the fact that $(1+x)^a-1/x rightarrow a$ (Actually, you can just do the computation explicitly with $a=-1$ to find that $u_{n+1}^a-u_n^a$ goes to $1$. The second point you raise is a consequence of Cesaro’s lemma.
$endgroup$
– Mindlack
Dec 13 '18 at 0:21
add a comment |
$begingroup$
Note that $u_n=u_{n+1}+u_{n+1}^2$.
Thus, if for some $N$, $u_N=0$, we are done. Now we assume that $u$ is positive. Hence it is decreasing so converges to some $l geq 0$ such that $l=l+l^2$ thus $l=0$.
Now, let $a < 0$ be a real number, then $u_{n+1}^a-u_n^a=u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$.
Take $a=-1$, then $u_{n+1}^a-u_n^a rightarrow 1$, thus $u_n sim 1/n$ and $sum_n {u_n}=infty$, a contradiction.
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
$endgroup$
Note that $u_n=u_{n+1}+u_{n+1}^2$.
Thus, if for some $N$, $u_N=0$, we are done. Now we assume that $u$ is positive. Hence it is decreasing so converges to some $l geq 0$ such that $l=l+l^2$ thus $l=0$.
Now, let $a < 0$ be a real number, then $u_{n+1}^a-u_n^a=u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$.
Take $a=-1$, then $u_{n+1}^a-u_n^a rightarrow 1$, thus $u_n sim 1/n$ and $sum_n {u_n}=infty$, a contradiction.
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
answered Dec 13 '18 at 0:02
MindlackMindlack
4,900211
4,900211
$begingroup$
Sorry, how do you get $u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$? What does ~ mean? Also how do you get the step after that, can you elaborate?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:18
1
$begingroup$
I am using Landau (“big Oh”) notations and the fact that $(1+x)^a-1/x rightarrow a$ (Actually, you can just do the computation explicitly with $a=-1$ to find that $u_{n+1}^a-u_n^a$ goes to $1$. The second point you raise is a consequence of Cesaro’s lemma.
$endgroup$
– Mindlack
Dec 13 '18 at 0:21
add a comment |
$begingroup$
Sorry, how do you get $u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$? What does ~ mean? Also how do you get the step after that, can you elaborate?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:18
1
$begingroup$
I am using Landau (“big Oh”) notations and the fact that $(1+x)^a-1/x rightarrow a$ (Actually, you can just do the computation explicitly with $a=-1$ to find that $u_{n+1}^a-u_n^a$ goes to $1$. The second point you raise is a consequence of Cesaro’s lemma.
$endgroup$
– Mindlack
Dec 13 '18 at 0:21
$begingroup$
Sorry, how do you get $u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$? What does ~ mean? Also how do you get the step after that, can you elaborate?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:18
$begingroup$
Sorry, how do you get $u_{n+1}^a(1-(1+u_{n+1})^a)sim -au_{n+1}^{a+1}$? What does ~ mean? Also how do you get the step after that, can you elaborate?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:18
1
1
$begingroup$
I am using Landau (“big Oh”) notations and the fact that $(1+x)^a-1/x rightarrow a$ (Actually, you can just do the computation explicitly with $a=-1$ to find that $u_{n+1}^a-u_n^a$ goes to $1$. The second point you raise is a consequence of Cesaro’s lemma.
$endgroup$
– Mindlack
Dec 13 '18 at 0:21
$begingroup$
I am using Landau (“big Oh”) notations and the fact that $(1+x)^a-1/x rightarrow a$ (Actually, you can just do the computation explicitly with $a=-1$ to find that $u_{n+1}^a-u_n^a$ goes to $1$. The second point you raise is a consequence of Cesaro’s lemma.
$endgroup$
– Mindlack
Dec 13 '18 at 0:21
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.
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%2f3037383%2fshow-the-terms-of-a-sequence-u-n-is-0%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
$begingroup$
$u_1=u_2^2+u_3^2+....=0+0+0...=0$?
$endgroup$
– mathpadawan
Dec 13 '18 at 0:02
$begingroup$
Then $u_1$ is not $u_2^2+...$.
$endgroup$
– Mindlack
Dec 13 '18 at 0:02