$\frac1n\sum _{k=1}^na_k\to0$ if and only if $\frac1n\sum _{k=1}^na^2_k\to0$ 
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?
 A: Your direction is fine. To prove the other direction first pick $\epsilon >0$ and split up the sum as follows, letting $\chi$ denote an indicator function,
$$
\frac{1}{n}\sum_{k=1}^n a_k = \frac{1}{n}\sum_{k=1}^n a_k\,\chi_{a_k < \epsilon} + \frac{1}{n}\sum_{k=1}^n a_k\,\chi_{a_k \geq \epsilon}.
$$
Now $a_k \geq \epsilon$ implies $a_k \leq a_k^2 / \epsilon$. So we have
$$
\frac{1}{n}\sum_{k=1}^n a_k \leq \frac{1}{n}\sum_{k=1}^n \epsilon\,\chi_{a_k < \epsilon} + \frac{1}{n}\sum_{k=1}^n (a_k^2 / \epsilon)\,\chi_{a_k \geq \epsilon}.
$$
This gives us 
$$
\frac{1}{n}\sum_{k=1}^n a_k \leq \epsilon + \frac{1}{\epsilon} (\frac{1}{n}\sum_{k=1}^n a_k^2).
$$
Now supposing $\frac{1}{n}\sum_{k=1}^n a_k^2$ tends to zero, take the limit $n \rightarrow \infty$ to obtain
$$
\limsup_{n \rightarrow \infty}\frac{1}{n}\sum_{k=1}^n a_k \leq \epsilon.
$$
As $\epsilon > 0$ was arbitrarily chosen, you have that $\frac{1}{n}\sum_{k=1}^n a_k$ tends to zero.
A: For the other direction, we can use Cauchy-Schwartz:
$$\frac{\sum_{k=1}^{n}a_k}{n} = \frac{\sum_{k=1}^{n}(a_k\cdot 1)}{n}\le \frac{(\sum_{k=1}^{n}a_k^2)^{1/2}(\sum_{k=1}^{n}1^2)^{1/2}}{n}$$ $$   = \left (\frac{  \sum_{k=1}^{n}a_k^2}{n}\right)^{1/2}\frac{  n^{1/2}}{n^{1/2}}  \to 0$$
