TITLE: If $\lim_\limits{n\to\infty}(a_n\sum_{i=1}^{n}a_i^2)=1$, then $\lim_\limits{n\to\infty}\sqrt[3]{(3n)}a_n=1$
QUESTION [9 upvotes]: If $\lim\limits_{n\to\infty}(a_n\sum\limits_{i=1}^{n}a_i^2)=1$, prove $\lim\limits_{n\to\infty}\sqrt[3]{(3n)}a_n=1$.
Since $a_n^2\geqslant0$, if$\sum\limits_{i=1}^{n}a_i^2\to M,a_n\to 0, $then $\lim\limits_{n\to\infty}(a_n\sum\limits_{i=1}^{n}a_i^2)=0$, it's contradictory to $\displaystyle\lim\limits_{n\to\infty}(a_n\sum\limits_{i=1}^{n}a_i^2)=1.$
So I conclude $\sum\limits_{i=1}^{n}a_i^2\to+\infty$, then I get 
$a_n\sim\frac{1}{\sum\limits_{i=1}^{n}a_i^2}\to 0$
$\lim\limits_{n\to\infty}\sqrt[3]{(3n)}a_n=\lim\limits_{n\to\infty}\frac{\sqrt[3]{(3n)}}{\sum\limits_{i=1}^{n}a_i^2}=\lim\limits_{n\to\infty}\frac{\sqrt[3]{3n}-\sqrt[3]{3(n-1)}}{a_n^2}$(according to $Stolz~theorem$)
However, I have no idea how to continue it, I sincerely appreciate your help.

REPLY [5 votes]: You are almost there. Consider $s_n=\sum\limits_{i=1}^{n} a_i^2$.
We have $$\lim s_{n+1}/s_n=1+\lim(s_{n+1}-s_n)/s_n=1+\lim a_{n+1}^2/s_n=1.$$
Also, 
$$\lim s_{n+1}^3-s_n^3=\lim (s_{n+1}^2+s_{n+1}s_n+s_n^2)a_{n+1}^2=\lim (1+s_n/s_{n+1}+(s_n/s_{n+1})^2)(s_{n+1}a_{n+1})^2=3.$$
Thus, $\lim s_n^3/3n=1$. As $\lim_n a_ns_n=1$, $\lim_n\frac{1}{\sqrt[3]{3n}a_n}=\lim_n \frac{s_n}{\sqrt[3]{3n}}\frac{1}{a_ns_n}=1$. The result follows immediately.