Problem concerning limit My friend asked me the question while he is preparing the mathematical analysis exams.

Let $\left\{a_{n}\right\}$ be a sequence satisfying
  $\lim\limits_{n\to\infty}\left(a_{n}\sum\limits_{k=1}^{n}a_{k}^{2}\right)=1$. Prove that $\lim\limits_{n\to\infty}\left(\,\sqrt[3]{3n\,}\,\ a_n\,\right) =1$.

Here is my attempt:  let $S_n= \sum_{k=1}^{n} a_k^2$, then we get 
$$\lim_{n\to\infty} (S_n-S_{n-1}) S_n^2=1$$
What we need is $$\lim_{n\to\infty} \sqrt[3]{9n^2} (S_n-S_{n-1})=1$$
After trying Stolz theorem, I still cannot get the term $\sqrt[3]{9n^2}$. I wonder how to get this result? Any hints or solutions are welcomed, thanks!
 A: Since $S_n - S_{n-1} = a_n^2 \sim S_n^{-2}$, $S_n$ should be expected to behave like solutions to the differential equation $y' = y^{-2}$, like $y = \sqrt[3]3 x^{1/3}$. In particular, $S_n^3$ should grow like a linear function.
$S_n^3 - S_{n-1}^3 = (S_n - S_{n-1})^3 - 3(S_n - S_{n-1})^2S_n + 3(S_n - S_{n-1})S_n^2$. These three terms are equivalent to $S_n^{-6}, -3S_n^{-3}, +3$ respectively.
If $S_n$ is convergent, then $S_n - S_{n-1}$ converges to $0$, while $S_n^{-2}$ converges to something other than $0$, so they can't be equivalent. Hence $S_n$ is not convergent but diverges to $+ \infty$, and $S_n^3 - S_{n-1}^3$ converges to $3$.
The Stolz-Cesaro theorem then says that $S_n^3 \sim 3n$, hence $S_n \sim \sqrt[3]{3n}$. Since $a_n \sim S_n^{-1}$, we have $a_n \sim (3n)^{-1/3}$, which is what is wanted.
A: Proof:
$S_n=\sum_{k=1}^nx^k, \: (x_nS_n)$ tend to 1. The sequence $(S_n)$ is strictly increasing if $(S_n)$ converge to $L$, then $x_n\to \frac{1}{L}>0$ and we will have $S_n=\sum_{k=1}^nx_k^2\to +\infty$ contradiction. We have
$$\lim_{n\to+\infty} S_n\to +\infty$$ and $$\lim_{n\to +\infty}x_n=0$$
We also have $x_nS_{n-1}\to 1$ when $n\to +\infty$ 
and 
$S_n^3-S_{n-1}^3=x_n^2(S_n^2+S_nS_{n-1}+S_{n-1}^2)\to 3$ when $n\to +\infty$
With Stolz-Cesaro:
$$\lim_{n\to +\infty} \frac{S_n}{n}=\lim_{n\to +\infty} \frac{S_n^3-S_{n-1}^3}{n-(n-1)}=3$$
We deduce that
$$\frac{\sqrt[3]{3n}}{S_n}\to1$$
$$\sqrt[3]{3n}x_n=\frac{\sqrt[3]{3n}}{S_n}x_nS_n\to1$$
