Absolute convergence of $\sum_{n=1}^{\infty}z_n^2$

Question

There are two complex series given ($z \in \mathbb{C}$.
We know that: $$\sum_{n=1}^{\infty}z_n^2$$ converges absolutely.
We are to show that $$\sum_{n=1}^{\infty} \frac{z_n}{n}$$ also converges absolutely.
I tried to prove it with no success unfortunately. How should the proof look like?

Answer 1

Hint: Cauchy-Schwarz.

In more detail (hidden, place your mouse over to show it):

$$ \sum_{n=1}^N \left\lvert \frac{z_n}{n}\right\rvert \leq \sqrt{\sum_{n=1}^N \lvert z_n\rvert^2}\sqrt{\sum_{n=1}^N \frac{1}{n^2}} $$

Final comment: instead of Cauchy-Schwarz, you could use the AM-GM inequality as well.

Answer 2

This follows from Cauchy-Schwarz, applied to $\sum |z_n|^2$ and $\sum 1/n^2$.