Does $\frac1n\sum\limits_{k=1}^na_k^2\to\rho$ with $0\le\rho<1$ imply$\prod\limits_{k=1}^na_k\to0$? Let $\{a_n\}$ be a sequence of real numbers such that $\lim_{n\to \infty}\frac{\sum_{j=1}^n a_j^2}{n}=\rho$, and $0\le\rho<1$. The goal is to check whether the following is true $$\lim_{n\to \infty}\prod_{j=1}^n a_j=0$$ 
Indeed, the infinite product diverges to $0$ if the sequence $\{a_n\}$ is a sequence of non-negative real numbers. This can be seen simply applying AM-GM inequality. However, I am not sure how to proceed for the general real sequence case. Any help/suggestion is appreciated. Thanks in advance.
EDIT: I could solve the problem as indicated in the comments and as per @Winther's advice I posted my answer. However, it will be really nice to see other approaches that can solve the problem. I, for one, am not finding any other, so I am waiting for a different answer to be posted. 

Let $m> n$ be such that $l=m-n$ satisfies  $$\left|\frac{\sum_{j=n}^{m-1} a_j^2}{l}-\rho\right|<\epsilon$$ for all $m>n$, where $\epsilon>0$ is chosen such that $\rho+\epsilon<1$. Then, an application of AM-GM inequality gives $$\prod_{j=n}^{m-1} a_j^2\le \left(\frac{\sum_{j=m}^n a_j^2}{l}\right)^l\le (\rho+\epsilon)^l$$ Since this is true for all $l\ge 0$, it follows that $$\lim_{m\to \infty}\prod_{j=n}^{m-1} a_j^2=0$$ which shows the divergence to $0$.
