# Convergence of a series $\sum_{n=1}^\infty\left(\frac{a_n}{n^p}\right)^\frac{1}{2}$

I've got a question about the convergence of a series during studying analysis.

If I know that a series of positive real numbers $$\sum_{n=1}^\infty a_n$$ converge, why does $$\sum_{n=1}^\infty\left(\frac{a_n}{n^p}\right)^\frac{1}{2}$$ also converge for $p>1$?

Although I know about many convergence tests, I don't know how to apply those tests for this case. Since this problem is the form of "series A converge → series B converge", I've been thinking that it must be verified by using some "comparison" tests. Is this thinking correct?

Thanks.

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@Landscape : Yes. I missed it. I'll update now. –  Analysis Jun 13 '13 at 3:08
@Maesumi : But then, $\sum1$ does not converge which is not fulfilled the condition. –  Analysis Jun 13 '13 at 3:49

Since $$ab\leq a^2+b^2$$ we have $$\sum_{n=1}^\infty\left(\frac{a_n}{n^p}\right)^\frac{1}{2}\leq \sum_{n=1}^\infty a_n+\sum_{n=1}^\infty\frac{1}{n^p}$$

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@landscape Why you deleted your answer? As I have not deleted my answer (and I do not do) you must not remove yours, that's what you said. –  Sami Ben Romdhane Jun 13 '13 at 3:31
Dear Sami Ben Romdhane, I deleted mine because I thought it was a little embarrassing that my answer posted later than yours, but shared the same idea with yours and contained less details than yours. If you don't mind, I will undelete it. –  23rd Jun 13 '13 at 3:39
@Landscape A hint shouldn't contains more detail so I'd be happy if you undelete it. –  Sami Ben Romdhane Jun 13 '13 at 3:46
I have undeleted my answer. Thank you. –  23rd Jun 13 '13 at 3:53
Hint: $$\left(\frac{a_n}{n^p}\right)^{\frac{1}{2}}\le\frac{1}{2}\left(a_n+\frac{1}{n^p}\right).$$