calculus - Prove that $lim_{n to infty}3nx_n^3 = 1$.

If $\{x_n\}$ is a given sequence satisfying
$$\lim_{n \to \infty}x_n\sum_{k=1}^{n}x_k^2=1,$$
prove that
$$\lim_{n \to \infty}3nx_n^3 = 1.$$
I have tried to use Stolz–Cesàro theorem but failed. Can anyone give a hint? Thanks.

Answer

Question: let $a_{n}$ be a sequence of real numbers such that $\lim_{n\to\infty}a_{n}\sum_{k=1}^{n}a^2_{k}=1$, show that: $$\lim_{n\to\infty}3na^3_{n}=1$$

solution: Set $S_{n}=\sum_{k=1}^{n}a^2_{k}$, then the condition $a_{n}S_{n}\to 1$ implies that
$S_{n}\to \infty$ and $a_{n}\to 0$ as $n\to\infty$,Hence we also have that $a_{n}S_{n-1}\to 1$ as $n\to\infty$,Therefore
$$S^3_{n}-S^3_{n-1}=a^2_{n}(S^2_{n}+S_{n}S_{n-1}+S^2_{n-1})\to 3,n\to \infty$$
Thus by the Stolz-Cesaro lemma,$\dfrac{S^3_{n}}{n}\to 3,n\to \infty$,since
$a_{n}S_{n}\to 1$,so
$$\lim_{n\to\infty}3na^3_{n}=1$$