Show that, if $\sum (a_n)^2$ converges, then $\sum \frac{a_n}{n}$ also converges (Sug: use Cauchy-Schwarz inequality).

I figured out that $0 \leq \left( \sum \frac{a_n}{n} \right)^2 \leq \left( \sum a_n^2 \right) \left( \sum \frac{1}{n^2} \right)$. Then I concluded $\left( \sum \frac{a_n}{n} \right)^2$ converges, but how can I assure then that $\left( \sum \frac{a_n}{n} \right)$ converges? I tried to set up a limit such that:

$$\lim\limits_{n \to +\infty} (s_n)^2 = a$$

Then manipulate it to prove $\lim\limits_{n \to +\infty} (s_n)$ converges, but I couldn't do it. Thanks in advance.