Proving the convergence of the following complex series

Hi I was working on a problem from the book written by Polya and sgezo:

Suppose that the numbers $z_1,z_2,...z_n,...$ are all in the right half plane $\text{Re }z_n\ge 0$ and that the series $z_1+z_2+....+z_n+....$ and $z_1^2+z_2^2+....+z_n^2+...$ converge. Then $|z_1|^2+|z_2|^2+...+|z_n|^2+...$ converges too. Give a counter example when the $z_n$ are not restricted on the right half plane.

I was thinking like since $$\sum_{n=1}^\infty z_n^2\,\,\text{converges}\Rightarrow \sum_{n=1}^\infty (x_n+iy_n)^2=\sum_{n=1}^\infty (x_n^2-y_n^2+i2x_ny_n)\Rightarrow \sum_{n=1}^\infty x_n^2-y_n^2 \text{ and }\sum_{n=1}^\infty 2x_ny_n \text{converge}$$

I need to show $\sum_{n=1}^\infty x_n^2+y_n^2=\sum_{n=1}^\infty |z_n|^2$ converges.

I'm stuck here, Any help would be highly appreciated.

• Quick remainder: $i^2 = -1$ ;) – Hermès Nov 5 '16 at 23:29
• @Hermès, Thanks man, How silly of me. Then how should I do it.. Thanks again – mint Nov 5 '16 at 23:34
• Here's my two cents: you know that $\sum z_n$ converges so $\sum x_n$ converges aswell, so for $n$ large enough, $x_n\leq 1$ so $x_n^2\leq x_n$ thus $\sum x_n^2$ converges. Finally you have $\sum x_n^2 +y_n^2\leq \sum y_n^2 -x_n^2 + 2\sum x_n^2$. – Andrei.B Nov 6 '16 at 0:11
• For the counter-example, try $z_n=\dfrac{i^n}{(n+1)^{1/4}}$. – Andrei.B Nov 6 '16 at 21:13