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Suppose a sequence of complex number $(z_n)_{n=1}^\infty$. Show that if the series ( of positive numbers) $\sum_{n=1}^\infty {|z_n|^2}$ converges, then the series (of positive numbers) $\sum_{n=1}^\infty \frac{|z_n|}{n}$ also converges.

My attempt: Let $z_n=x_n+iy_n$ where $x_n,y_n \in \mathbb{R}$ for all $n \geq 1$. Notice that $|z_n|^2=x_n^2 +y_n^2$ and also $$\frac{|z_n|}{n}=\frac{\sqrt{x_n^2+y_n^2}}{n} \leq \frac{x_n^2 +y_n^2}{n}\leq x_n^2 +y_n^2$$ for $n \geq 1$. By comparison test, the series $\sum_{n=1}^\infty \frac{|z_n|}{n}$ converges. Is my proof correct?

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    $\begingroup$ No. Why should we be able to say that $\sqrt{x_n^2+y_n^2}\le x_n^2+y_n^2$? As it turns out, that will almost never be the case, since $x_n^2+y_n^2<1$ for all but finitely-many $n$. $\endgroup$ – Cameron Buie Aug 22 '13 at 16:04
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    $\begingroup$ Why bother with complex series, when all you need is a real-valued series (you only look at $|z_k|$ anyways)? Then we have for $(a_n)_{n\in\mathbb{N}} \subset \mathbb{R}^+$: $$\sum_{n\in\mathbb{N}} a_n^2 < \infty \Rightarrow \sum_{n\in\mathbb{N}} \frac{a_n}{n} < \infty$$ $\endgroup$ – AlexR Aug 22 '13 at 16:11
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    $\begingroup$ Try using Cauchy-Schwartz inequality for sequences $|z_n|$ and $1/n$. $\endgroup$ – user8268 Aug 22 '13 at 16:11
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(to update my comment to an answer)

By Cauch-Schwarz inequality $(\sum |z_n|/n)^2\leq (\sum |z_n|^2)(\sum 1/n^2)$. (use it for sums form 1 to $N$ and take $N$ to $\infty$, or use it for infinite sums, if you know it also in this form)

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