If $\sum_{n=1}^{\infty} a_n^{2}$ converges, then so does $\sum_{n=1}^{\infty} \frac {a_n}{n}$

What you did is correct; in fact you can show that if $\{a_n\}$ and $\{b_n\}$ are two sequences of real numbers and $\sum_{n\geq 0}a_n^2$ and $\sum_{n\geq 0}b_n^2$ are convergent then the series $\sum_{n=0}^{+\infty}|a_nb_n|$ is convergent, noting that $0\leq |a_nb_n|\leq \max(a_n^2,b_n^2)\leq a_n^2+b_n^2$.

Your particular case is $b_n=\frac 1n$.

Another approach is to note that for any positive integer $N,$ we have $\sum_{n=1}^{N} \frac{|a_n|}{n} \leq \sqrt{ \sum_{n= 1}^{N} a_{n}^{2}} \sqrt{ \sum_{n=1}^{N} \frac{1}{n^2}}$, and this is in turn less than $\frac{\pi}{\sqrt{6}}\sqrt{ \sum_{n= 1}^{N} a_{n}^{2} }.$ The first inequality follows by the Cauchy-Schwarz inequality, and the second follows by Euler's formula $\frac{\pi^2}{6} = \sum_{n=1}^{\infty} \frac{1}{n^2}.$

You are right.. you can also simplify things further using that $|ab| \leq {a^2 + b^2\over 2}$ for any $a$ and $b$, so that $|{a_n \over n}| \leq {a_n^2 \over 2} + {1 \over 2n^2}$ and thus your series converges absolutely as you are saying.