# Prove that if $\sum |a_n|^2$ converges then $\sum \frac{a_n}{n}$ converges [duplicate]

Let $\{a_n\}\in \mathbb{C}$. Prove that if $\sum |a_n|^2$ converges than $\sum \frac{a_n}{n}$ converges.

Note that this problem is taken from the first chapter on series of a calculus book, so it should be solvable with very basic tools (e.g. the only convergence test introduced was the comparison test and nothing about absolute convergence is assumed).

I have tried a few thing but nothing worked. Could you give me a hint on how to approach the problem?

## marked as duplicate by Steven Stadnicki, Watson, Community♦Sep 1 '16 at 22:27

• (Note that the other question talks about $\mathbb{R}$, but there's nothing there that doesn't really apply here. Also, this has to be one of the most-duplicated analysis questions on this site - I count at least a half-dozen different instances...) – Steven Stadnicki Sep 1 '16 at 17:53
The other two answers, Robert Z's and anonymus', use Cauchy-Schwarz to establish (absolute) convergence of the series with general term $\frac{a_n}{n}$. As an alternative, one can use the AM-GM inequality:$^{(\ast)}$ indeed, $$\frac{\lvert a_n\rvert }{n} = \sqrt{\frac{\lvert a_n\rvert^2 }{n^2}} \leq \frac{\lvert a_n\rvert^2+\frac{1}{n^2}}{2}$$ for all $n\geq 1$. It only remains to conclude by theorems of comparison.
${}^{(\ast)}$ "Inequality of arithmetic and geometric means."
Use Cauchy-Schwarz inequality and the fact that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is convergent: $$\left(\sum_{n=1}^N \frac{|a_n|}{n}\right)^2\leq \sum_{n=1}^N |a_n|^2 \cdot \sum_{n=1}^N \frac{1}{n^2}\leq \sum_{n=1}^{\infty} |a_n|^2 \cdot \sum_{n=1}^{\infty} \frac{1}{n^2}<+\infty$$ Hence $\sum_{n=1}^{\infty} \frac{a_n}{n}$ is absolutely convergent and therefore also convergent.
With Cauchy-Schwarz inequality and the fact that $\sum_n 1/n^2 < \infty$, the result follows.