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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?

All advice is welcome^_^
Thanks.

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  • $\begingroup$ @Landscape : Yes. I missed it. I'll update now. $\endgroup$
    – Analysis
    Jun 13, 2013 at 3:08
  • $\begingroup$ @Maesumi : But then, $\sum1$ does not converge which is not fulfilled the condition. $\endgroup$
    – Analysis
    Jun 13, 2013 at 3:49

2 Answers 2

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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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  • $\begingroup$ @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. $\endgroup$
    – user63181
    Jun 13, 2013 at 3:31
  • $\begingroup$ 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. $\endgroup$
    – 23rd
    Jun 13, 2013 at 3:39
  • $\begingroup$ @Landscape A hint shouldn't contains more detail so I'd be happy if you undelete it. $\endgroup$
    – user63181
    Jun 13, 2013 at 3:46
  • $\begingroup$ I have undeleted my answer. Thank you. $\endgroup$
    – 23rd
    Jun 13, 2013 at 3:53
  • $\begingroup$ Nice work, as usual! $\endgroup$
    – amWhy
    May 31, 2014 at 11:24
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Hint: $$\left(\frac{a_n}{n^p}\right)^{\frac{1}{2}}\le\frac{1}{2}\left(a_n+\frac{1}{n^p}\right).$$

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  • $\begingroup$ Wow. Really simple! How do you think this inequality... Thanks! $\endgroup$
    – Analysis
    Jun 13, 2013 at 3:11
  • $\begingroup$ @Analysis: You are welcome. Since Sami Ben Romdhane posted a full answer a little earlier than me, let me delete my answer later. $\endgroup$
    – 23rd
    Jun 13, 2013 at 3:14
  • $\begingroup$ @Analysis: I am sorry for being capricious. After some discussion with Sami Ben Romdhane, I undeleted my answer. $\endgroup$
    – 23rd
    Jun 13, 2013 at 3:57

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