Does $\sum_{n=1}^\infty a_n^2 \text{ converges },\ a_n \geq 0 \implies \sum_{n=1}^\infty \frac{a_n}{n} \text{ converges}$?

Note that,

$$\frac{a_n}{n} \leq \frac{a_n^2+\frac{1}{n^2}}{2}$$ since $a^2+b^2 \geq 2ab$

So the series converges absolutely.

You can apply Cauchy-Schwarz directly: $$ \sum_n\frac{a_n}n\leq\left(\sum_na_n^2\right)^{1/2}\left(\sum_n\frac1{n^2}\right)^{1/2}. $$

Lets split the sum $\sum_{n=1}^\infty a_n^2$ into two sums (we can since it converges absolutely) , one is the sum of all $a_n^2$ where $a_n\leq \frac1n$ and the other of all $a_n^2$ such that $a_n>\frac1n$, now for the first sum we have $\frac{a_n}n\leq \frac1{n^2}$ and for second we have $a_n^2=a_n\cdot a_n>a_n\cdot\frac1n$ so we have that the corresponding two sums of $\frac{a_n} n$ also converge so we can combine them and rearrange (because both converge absolutely) them into $\sum_{n=1}^\infty\frac{a_n}n$