$\frac1n\sum _{k=1}^na_k\to0$ if and only if $\frac1n\sum _{k=1}^na^2_k\to0$

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.

For the other direction, we can use Cauchy-Schwartz:

$$\frac{\sum_{k=1}^{n}a_k}{n} = \frac{\sum_{k=1}^{n}(a_k\cdot 1)}{n}\le \frac{(\sum_{k=1}^{n}a_k^2)^{1/2}(\sum_{k=1}^{n}1^2)^{1/2}}{n}$$ $$ = \left (\frac{ \sum_{k=1}^{n}a_k^2}{n}\right)^{1/2}\frac{ n^{1/2}}{n^{1/2}} \to 0$$