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Here is the problem: Given that $limlimits_{n to infty} a_n sum_{i=1}^n a_i^2=1$, Show $ limlimits_{n to infty} sqrt[3]{3n}a_n=1$.

Let ${S_n}$ be a (not strictly) increasing sequence by $S_n=sum_{i=1}^{n}a_i^2$, and $limlimits_{nrightarrowinfty}S_n=+infty$.

In fact, if $limlimits_{nrightarrowinfty}S_n$ is a finite value, then
$$limlimits_{nrightarrow infty}a_n^2=limlimits_{nrightarrow infty}(S_n-S_{n-1})=0$$
which implies
$$limlimits_{nrightarrow infty}a_nsum_{i=1}^na_i^2=limlimits_{nrightarrow infty}a_nS_n=0$$
it contradicts to the known conditions. So $limlimits_{nrightarrowinfty}S_n=+infty$ and $displaystylelimlimits_{nrightarrowinfty}frac{1}{S_n}=0$. Then

$$limlimits_{nrightarrow infty}a_n=limlimits_{nrightarrow infty}a_nS_ncdotfrac{1}{S_n}=1cdot 0=0$$

Hence we have

$$limlimits_{nrightarrowinfty}frac{1}{3na^3_n}=limlimits_{nrightarrowinfty}frac{S^3_n}{3n}=limlimits_{nrightarrowinfty}frac{S^3_n-S^3_{n-1}}{3}=1$$

where
begin{align*}
limlimits_{nrightarrowinfty}{(S^3_n-S^3_{n-1})}&=
limlimits_{nrightarrowinfty}(S_n-S_{n-1})(S^2_n+S_nS_{n-1}+S^2_{n-1})\&=limlimits_{nrightarrowinfty}a^2_n[S^2_n+S_n(S_n-a^2_n)+(S_n-a^2_n)^2]
\&=limlimits_{nrightarrowinfty}[3(a_nS_n)^2-3a^4_nS_n+a^6_n]\&=
limlimits_{nrightarrowinfty}[3(a_nsum_{i=1}^{n}a_i^2)^2-3a^3_n(a_nsum_{i=1}^na_i^2)+a^6_n]\&=3
end{align*}
so

$$limlimits_{nrightarrowinfty}{sqrt[3]{3n}a_n}=sqrt[3]{limlimits_{nrightarrowinfty}{3na^3_n}}=1$$