# Prove $\sum_{n=1}^\infty |z_n|^2$ converges.

I had this practice problem but I am stuck after I tried things below. Could anyone give me a hint on how to proceed?

$\sum_{n=1}^\infty z_n$ and $\sum_{n=1}^\infty z_n^2$ both converge with $\Re(z_n) \geq 0$ for each $z_n \in \mathbb{C}$. Show $\sum_{n=1}^\infty |z_n|^2$ also converges.

Things I tried so far:

Let $z_n = x_n + i y_n$, where $x_n = \Re(z_n) \geq 0, y_n \in \mathbb{R}$, $n \in \mathbb{N}$.

$\sum_{n=1}^\infty z_n$ convergent implies that infinite series formed by $\Re(z_n)$ and $\Im(z_n)$ are convergent series respectively, i.e., $\sum_{n=1}^\infty x_n$ and $\sum_{n=1}^\infty y_n$ both converge.

By the same token, apply the above argument to convergent series $\sum_{n=1}^\infty z_n^2$, we have $\sum_{n=1}^\infty \Re(z_n^2) = \sum_{n=1}^\infty x_n^2 - y_n^2$ converges and $\sum_{n=1}^\infty \Im(z_n^2) = \sum_{n=1}^\infty 2 x_n y_n$ converges.

When it comes to the absolute convergence of $\sum_{n=1}^\infty z_n^2$, how can I relate $\sum_{n=1}^\infty |z_n|^2 = \sum_{n=1}^\infty x_n^2 + y_n^2$ to what we have so far?

And how does $x_n \geq 0$ come into play? I sense Cauchy-Schwarz might help but not sure of exactly how.

$\sum x_n$ converges and $x_n \ge 0$ implies that $x_n^2$ converges: That is true since $x_n <1$ for large $n$ and so $$0 \le x_n^2 \le x_n$$

for large $n$.

So both $\sum x_n^2 - y_n^2$ and $\sum x_n^2$ converges, then so is $\sum -y_n^2$. Hence

$$\sum x_n^2 + y_n^2$$ converges.

(Note that when $\sum x_n$ is only conditionally convergent, $\sum x_n^2$ might not be convergent. Can you think of any example?)

• I found "When $x_n$ changes sign, $\sum x_n^2$ might not be convergent (any example?) " to be a little confusing right there in the middle. Perhaps it would be better as a note at the end? – zhw. Sep 30 '17 at 18:35
• @zhw. is it better now (thanks!)? – user99914 Sep 30 '17 at 18:40
• Yes, I just made a small edit. – zhw. Oct 1 '17 at 19:18
• As for the example of $\sum x_n$ converges while $\sum x_n^2$ does not, mentioned by @JohnMa consider alternating series $\sum (-1)^n \frac{1}{\sqrt{n}}$ where $x_n = (-1)^n \frac{1}{\sqrt{n}}$ has sign changes. – swoopin_swallow Oct 3 '17 at 18:08