Show the terms of a sequence $u_n$ is $0$.












3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $u_1=u_2^2+u_3^2+....=0+0+0...=0$?
    $endgroup$
    – mathpadawan
    Dec 13 '18 at 0:02










  • $begingroup$
    Then $u_1$ is not $u_2^2+...$.
    $endgroup$
    – Mindlack
    Dec 13 '18 at 0:02
















3












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    $u_1=u_2^2+u_3^2+....=0+0+0...=0$?
    $endgroup$
    – mathpadawan
    Dec 13 '18 at 0:02










  • $begingroup$
    Then $u_1$ is not $u_2^2+...$.
    $endgroup$
    – Mindlack
    Dec 13 '18 at 0:02














3












3








3


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 23:39







mathpadawan

















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


















  • $begingroup$
    $u_1=u_2^2+u_3^2+....=0+0+0...=0$?
    $endgroup$
    – 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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
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1 Answer
1






active

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active

oldest

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active

oldest

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1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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