# A simpler proof that this series converges

I am given that $a_n \ge 0$ and $\sum a_n$ converges.
I need to show that $\sum \frac{\sqrt{a_n}}{n}$ converges also.

After a long time, I came up with a solution by re-ordering the series into:
$\sum_{a_n < \frac{1}{n^2}} \frac{\sqrt{a_n}}{n} + \sum_{a_n \ge \frac{1}{n^2}} \frac{\sqrt{a_n}}{n}$ which is bounded by $\sum\frac{1}{n^2} + \sum{a_n}$

I was wondering if there is a more basic solution because I don't want to be using ad-hoc methods on an exam.

• You could use $pq\le{1\over2}(p^2+q^2)$ with $p=\sqrt a_n$ and $q=1/n$. – David Mitra Feb 19 '13 at 19:03
• As Andre pointed out, this is called Cauchy Shwarz. Thanks! – Mark Feb 19 '13 at 19:13
• This isn't Cauchy-Schwarz. – David Mitra Feb 19 '13 at 19:22
• Oh oops. I didn't look closely enough. – Mark Feb 19 '13 at 19:31