$\sum _{1}^{\infty}a_n^2/n$ converges then $(1/N)\sum_{1}^{N}a_n$ converges to zero.

Fix any $N_0$ and write: \begin{align*} \frac{1}{N} \sum\limits_{n = 1}^N |a_n| &= \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sum\limits_{n = N_0 + 1}^N |a_n| }{N} \\ &= \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sum\limits_{n = N_0 + 1}^N \frac{|a_n|}{\sqrt{n}} \cdot \sqrt{n} }{N} \\ &\le \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \frac{\sqrt{\left(\sum\limits_{n = N_0 + 1}^N \frac{a_n^2}{n} \right) \left(\sum\limits_{n = N_0 + 1}^N n \right) }}{N} \\ &\le \frac{\sum\limits_{n = 1}^{N_0} |a_n|}{N} + \sqrt{\left(\sum\limits_{n = N_0 + 1}^N \frac{a_n^2}{n} \right)}. \end{align*} Here on the third line I use CS inequality. To obtain the forth line I simply bound $\sum\limits_{n = N_0 + 1}^N n$ from above by $N^2$.

Now by taking $N\to\infty$ we get $$\limsup\limits_{N\to\infty} \frac{1}{N} \sum\limits_{n = 1}^N |a_n| \le \sqrt{\left(\sum\limits_{n = N_0 + 1}^\infty \frac{a_n^2}{n} \right)}. $$ As $N_0$ can chosen arbitrarily, the right hand side can be made arbitrarily small --- here we use the fact that $\sum\limits_{n = 1}^\infty \frac{a_n^2}{n} $ is convergent.