TITLE: Convergence of the series $\sum a_n$ implies the convergence of $\sum \frac{\sqrt a_n}{n}$, if $a_n>0$
QUESTION [8 upvotes]: 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.

REPLY [18 votes]: Cauchy-Schwarz inequality on partial sums give you
$$ \sum^N_{n=1}\frac{\sqrt{a_n}}{n}\leq \sqrt{\sum^N_{n=1}a_n}\sqrt{\sum^N_{n=1}\frac{1}{n^2}} $$