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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.

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    $\begingroup$ You could use $pq\le{1\over2}(p^2+q^2)$ with $p=\sqrt a_n$ and $q=1/n$. $\endgroup$ – David Mitra Feb 19 '13 at 19:03
  • $\begingroup$ As Andre pointed out, this is called Cauchy Shwarz. Thanks! $\endgroup$ – Mark Feb 19 '13 at 19:13
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    $\begingroup$ This isn't Cauchy-Schwarz. $\endgroup$ – David Mitra Feb 19 '13 at 19:22
  • $\begingroup$ Oh oops. I didn't look closely enough. $\endgroup$ – Mark Feb 19 '13 at 19:31
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To someone accustomed to these things, this one could be an automatic no-thinking application of the Cauchy-Schwarz Inequality.

Remark: I like your way better. It confronts the actual series.

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  • $\begingroup$ Haha thanks. But it took a lot of thinking, whereas if I think I can learn to spot questions where Cauchy Schwarz can be applied, I will be able to do the problems faster. $\endgroup$ – Mark Feb 19 '13 at 19:07
  • $\begingroup$ Thinking, like working out, has benefits. And about "faster," it is unfortunate that the education systems reward quickness so much. $\endgroup$ – André Nicolas Feb 19 '13 at 19:12

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