Let $a_n$ be a sequence of real numbers such that $\lim(a_n\sum_{k = 1}^n a_k^2) = 1$ Let $a_n$ be a sequence of real numbers such that $\lim(a_n\sum_{k = 1}^n a_k^2) = 1$ . Prove that $\lim((3n)^{\frac{1}{3}}a_n)=1$
I'm more concerned with how I can derive the prove of this question

Let $S_n=a_1^2+\cdots+a_n^2$. The sequence $(S_n)_{n\ge1}$ is nondecreasing. If it is convergent then $\lim_{n\to\infty}a_n=0$ and this contradicts the hypotgesis $\lim_{n\to\infty}a_nS_n=1$. Thus, 
$\lim_{n\to\infty}S_n=+\infty$ and $\lim_{n\to\infty}a_n=
\lim_{n\to\infty}\frac{a_nS_n}{S_n}=0$. Now
$$S_{n}^3-S_{n-1}^3=S_{n}^3-(S_{n}-a_n^2)^3=3(a_nS_n)^2-3a_n^3(a_nS_n)+a_n^6$$
Hence
$$\lim_{n\to\infty}(S_{n}^3-S_{n-1}^3)=3$$
So, by Cesàro's lemma we get
$$\lim_{n\to\infty}\frac{S_{n}^3}{n}=3$$
Or equivalently
$$\lim_{n\to\infty}\frac{\sqrt[3]{3n}}{S_{n}}=1$$
Finally
$$\lim_{n\to\infty}\sqrt[3]{3n}a_n=\lim_{n\to\infty}(a_nS_n)\frac{\sqrt[3]{3n}}{S_{n}}=1$$
Done.
