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Problem: Suppose a sequence $\{x_n\}_{n\in\mathbb{N}}$ satisfies that $$\lim_{n\to\infty}\bigg(x_n\cdot\sum_{k=0}^nx_k^2\bigg)=1.$$ Prove that $\lim_{n\to\infty}\sqrt[3]{3n}x_n=1$.

I do not have much idea about how to construct precisely this limit since it is hard to imagine how the "$3$" under the root symbol comes. Can anyone give me some hints? Many thanks!

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Don't worry about the "3". Any way, $\displaystyle{\large% \lim_{n \to \infty}\left(n\,x_{n}\right) = {1 \over 3}}$. –  Felix Marin Sep 22 '13 at 5:59
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@FelixMartin: Yeah, it's just a trick to distract you. A sequence $(a_n)$ converges if and only if $(\sqrt[3]{a_n})$ does and $$\sqrt[3]{\lim_{n\to\infty} a_n}=\lim_{n\to\infty} \sqrt[3]{a_n},$$ since the function $x\mapsto\sqrt[3]{x}$ is continuous. –  triple_sec Sep 22 '13 at 6:18
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I believe there is a typo. If $3nx_n\to1$ then $\sum x_n^2$ converges, but then $x_n\sum_{k=0}^n x_k^2\to0$. –  user8268 Sep 22 '13 at 6:29
    
In fact, if $x_n$ behaves as $n^{-\alpha}$ then $x^n\sum_k^n x_k^2$ is roughly $n^{1-3\alpha}$. We should thus have $\sqrt[3]{n}x_n\to cst$ (whatever the constant is). –  user8268 Sep 22 '13 at 6:35
    
Would you please make more explanation about how to get this approximation, i.e., why $x_n\sum_{k=1}^nx_k^2$ is roughly $n^{1-3a}$? Thanks! –  user65018 Sep 22 '13 at 15:04

1 Answer 1

Let $y_n:=\sum_{k=0}^{n}x_k^2$, i.e. $x_ny_n\to1$, i.e. $(y_n-y_{n-1})^{1/2}y_n\to1$, i.e. $y_n^3-y_{n-1}y_n^2\to1$. From this and from $y_n\to\infty$ (a consequence of $x_n\to0$) we know $y_{n-1}/y_n\to1$. But now $$y_n^3-y_{n-1}^3=(y_n-y_{n-1})y_n^2(1+y_{n-1}/y_n+(y_{n-1}/y_n)^2)\to3$$ so that $y_n^3/n\to 3$, and thus (given $x_ny_n\to 1$) $\sqrt[3]{3n}x_n\to1$.

edit why $y_n^3/n\to3$: If for some $\epsilon>0$ and for infinitely many $n$'s $y_n^3>(3+\epsilon)n$ then for infinitely many $n$'s $y_n^3-y_{n-1}^3>3+\epsilon/2$ (say). Likewise for $y_n^3<(3-\epsilon)n$.

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Would you please make a little explanation about how we arrived at $y_n^3/n\to 3$? Thanks a lot! –  user65018 Sep 23 '13 at 0:56
    
From $y_n^3-y_{n-1}^3 \to 3$, summing from $n$ to $m$ such that $n/m \to 0$, $y_m^3-y_n^3 \to 3(m-n)$. The $n$ terms are small, so this implies $y_m^3 \to 3m$ or $y_m^3/m \to 3$. –  marty cohen Sep 23 '13 at 2:08

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