# Convergence of the series $\sum a_n$ implies the convergence of $\sum \frac{\sqrt a_n}{n}$, if $a_n>0$ [duplicate]

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 BanerjeeApr 27 '13 at 15:40

• 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. – Ben Apr 27 '13 at 15:09
• @Ben If we could construct such sequence $b_n$, this may prove this result. – srijan Apr 27 '13 at 15:10
• 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. – David Mitra Apr 27 '13 at 15:11
• Since, it is a positive sequence, it is enough to find an upper bound of partial sums. – i707107 Apr 27 '13 at 15:11
• Yes. But the two series involved are $\sum a_n$ and $\sum {1\over n^2}$. Both are convergent. – David Mitra Apr 27 '13 at 15:14

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

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