Proving that a series converges I am given the series $\Sigma_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$, where $a_n \geq 0 \forall n$ , and $\Sigma_1^{\infty} a_n  $ converges. 
I was advised to expand $|\sqrt{a_n} - \frac{1}{n}|^2 $ . Doing this gives me expressions like $-\frac{2 \sqrt{a}}{n}+a+\frac{1}{n^2}$ , but I don't see how this is useful. 

Writing $(x-y)^2 = x^2 - 2xy + y^2$, we see that
$$ xy \leq \frac{1}{2} \left( x^2 + y^2 \right). $$
This implies that 
$$ \frac{\sqrt{a_n}}{n} \leq \frac{1}{2} \left( a_n + \frac{1}{n^2} \right). $$
Since both $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge, the series $\sum_{n=1}^{\infty} \frac{1}{2} \left( a_n + \frac{1}{n^2} \right)$ also converges and so by the comparison test, the series $\sum_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$ also converges.
