I need help to solve following problem from Rudin's Mathematical analysis book:

Convergence of the series $\sum a_n$ implies the convergence of $\sum \dfrac{\sqrt {a_n}}{n}$, if $a_n>0$

I tried to construct a suitable convergence sequence $b_n$ such that $\sum b_n$ converges and $a_n \leq b_n$ but, I am not able to find such sequence $b_n$ .

Thanks for the help and sugestions.


marked as duplicate by user940, David Mitra, Micah, Julian Kuelshammer, Ishan Banerjee Apr 27 '13 at 15:40

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  • $\begingroup$ I don't think finding such a $b_n$ will help. This would just show you what is true by hypothesis: that $a_n$ converges. $\endgroup$ – Ben Apr 27 '13 at 15:09
  • $\begingroup$ @Ben If we could construct such sequence $b_n$, this may prove this result. $\endgroup$ – srijan Apr 27 '13 at 15:10
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    $\begingroup$ Use the inequality $pq\le{1\over 2}(p^2+q^2)$ with $p=\sqrt {a_n}$ and $q=1/n$. Then use the Comparision test. $\endgroup$ – David Mitra Apr 27 '13 at 15:11
  • $\begingroup$ Since, it is a positive sequence, it is enough to find an upper bound of partial sums. $\endgroup$ – i707107 Apr 27 '13 at 15:11
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    $\begingroup$ Yes. But the two series involved are $\sum a_n$ and $\sum {1\over n^2}$. Both are convergent. $\endgroup$ – David Mitra Apr 27 '13 at 15:14

Cauchy-Schwarz inequality on partial sums give you $$ \sum^N_{n=1}\frac{\sqrt{a_n}}{n}\leq \sqrt{\sum^N_{n=1}a_n}\sqrt{\sum^N_{n=1}\frac{1}{n^2}} $$

  • $\begingroup$ @smiley 06 Its quite an interesting answer. Haven't though of this? Thanks $\endgroup$ – srijan Apr 27 '13 at 15:15

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