TITLE: Does $\sum_{n=1}^\infty a_n^2 \text{ converges },\ a_n \geq 0 \implies \sum_{n=1}^\infty \frac{a_n}{n} \text{ converges}$?
QUESTION [6 upvotes]: Let $\sum_{n=1}^\infty a_n^2$ a converging series with $a_n \geq 0$. Decide whether $\sum_{n=1}^\infty \frac{a_n}{n}$ is converging.
Note: you may consider the fact that $\sum_{n=1}^\infty \frac{1}{n^2}$ is converging.

I'm having problems with the above task. Given the hint in the note, I believe the latter series is indeed converging and am trying to prove it. Unfortunately, I fail to find a promising approach to do so.
Obviously, $\sum_{n=1}^\infty a_n^2 \neq \left(\sum_{n=1}^\infty a_n\right) \cdot \left(\sum_{n=1}^\infty a_n \right)$, which would be a rookie mistake. Similarly, $\sqrt{\sum_{n=1}^\infty a_n^2} \neq \sum_{n=1}^\infty a_n$. Therefore, the square seems difficult to get rid of. I thought of arguing using the Cauchy product, however, I'm not sure whether it could be applied "in reverse" to show that if the series based on $a_n^2$ is converging, then the one based on $a_n$ is, too.
What approach would you take to solve this problem?

REPLY [3 votes]: Lets split the sum $\sum_{n=1}^\infty a_n^2$ into two sums (we can since it converges absolutely) , one is the sum of all $a_n^2$ where $a_n\leq \frac1n$ and the other of all $a_n^2$ such that $a_n>\frac1n$, now for the first sum we have $\frac{a_n}n\leq \frac1{n^2}$ and for second we have $a_n^2=a_n\cdot a_n>a_n\cdot\frac1n$ so we have that the corresponding two sums of $\frac{a_n} n$ also converge so we can combine them and rearrange (because both converge absolutely) them into $\sum_{n=1}^\infty\frac{a_n}n$