$\lim_{n\to\infty}\sqrt[3]{3n}\sigma_n=1$

Question

We put : $S_n=\displaystyle\sum_{k=1}^{k=n}\sigma^2_k$ Where $(\sigma_n)_{n\geq 1}$ is a real sequence, its boundaries are positive.
_Assume that $(\lim_{n\to\infty}\sigma_n S_n=1)$
Prove that :$\lim_{n\to\infty}\sqrt[3]{3n}\sigma_n=1$

I tried to prove this, but I did not reach any result.

If possible, find any method or idea that would help us in the solution, thank you in advance

Answer 1

First results:

$\lim S_n=+\infty$ (otherwise: $\{S_n\}$ converges, $\lim \sigma_n^2=0$ , $\lim \sigma_n^2 S_n=0 $ , contradiction)

$\lim \sigma_n = 0 $ (because $\ \lim \sigma_nS_n=1\ $ and $\ \lim S_n=+\infty$)

$S_n \sim S_{n-1}$ (because $\ S_{n-1} = S_n-\sigma_n$)

The idea:

$S_n^3-S_{n-1}^3 = (S_n-S_{n-1})(S_n^2+S_nS_{n-1}+S_{n-1}^2) $

$(S_n^3-S_{n-1}^3) \sim 3\sigma_n^2 S_n^2$

$\lim (S_n^3-S_{n-1}^3) = 3$

By Cesaro: $\ \lim \dfrac{S_n^3-S_0^3}{n}=3 $

The conclusion:

$S_n\sim \root{3}\of{3n}\ $ , therefore $\ \sigma_n \sim \dfrac{1}{\root{3}\of3{n}} $