TITLE: $\frac1n\sum _{k=1}^na_k\to0$ if and only if $\frac1n\sum _{k=1}^na^2_k\to0$
QUESTION [6 upvotes]: If $(a_n)$ is a sequence in $(0,1)$, show that $\frac1n\sum _{k=1}^na_k\to0$ if and only if $\frac1n\sum _{k=1}^na^2_k\to0$

My try:
$\implies$: Since $a_k\in (0,1)$, we have $0\le\frac1n\sum _{k=1}^na^2_k\le \frac1n\sum _{k=1}^na_k$, and if RHS goes to zero, then by squeeze theorem, $\frac1n\sum _{k=1}^na^2_k\to0$
How to show other direction?

REPLY [4 votes]: For the other direction, we can use Cauchy-Schwartz:
$$\frac{\sum_{k=1}^{n}a_k}{n} = \frac{\sum_{k=1}^{n}(a_k\cdot 1)}{n}\le \frac{(\sum_{k=1}^{n}a_k^2)^{1/2}(\sum_{k=1}^{n}1^2)^{1/2}}{n}$$ $$   = \left (\frac{  \sum_{k=1}^{n}a_k^2}{n}\right)^{1/2}\frac{  n^{1/2}}{n^{1/2}}  \to 0$$