# If $\sum\limits_{n=1}^\infty a_n$ is convergent with $a_n$ positive, then $\sum\limits_{n=1}^\infty \frac {{(a_n)}^{1/2}}{n}$ is also convergent

Question: $$\sum_{n=1}^\infty a_n$$ is convergent with $$a_n$$ positive, prove the series $$\sum_{n=1}^\infty \frac {{(a_n)}^{1/2}}{n}$$ is convergent.

The hint given is: $$x^2+y^2\geq2xy$$.

But I cannot realize anything from the hint, so asking here for help. I know it has something to do with comparison theorem. Anyone can help ? thanks!

If there is another method without using this hint, please feel free to tell me also! thanks! And also please provide the general method for solving this kind of question if there is any. Thanks so much!

## 2 Answers

Take the hint with $x = a_n^{1/2}$ and $y = 1/n$.

• Thx! But I would like to ask is there any general method for solving this kind of question? given a series converges, then prove another more or less similar series converges. – UnusualSkill Apr 22 '15 at 15:37
• in other words, without given the hint, how can I approach this ques? – UnusualSkill Apr 22 '15 at 15:39

You can either use $\sqrt{ab}\leq\frac{a+b}{2}$ to state: $$\sum_{n\geq 1}\sqrt{\frac{a_n}{n^2}}\leq \frac{1}{2}\left(\sum_{n\geq 1}a_n+\sum_{n\geq 1}\frac{1}{n^2}\right)\tag{1}$$ or the Cauchy-Schwarz inequality that gives: $$\sum_{n\geq 1}\sqrt{\frac{a_n}{n^2}}\leq \sqrt{\frac{\pi^2}{6}\sum_{n\geq 1}a_n}\tag{2}$$ since: $$\sum_{n\geq 1}\frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}.\tag{3}$$