Convergence of the series $\sum a_n$ implies the convergence of $\sum \frac{\sqrt a_n}{n}$, if $a_n>0$ [duplicate]
Cauchy-Schwarz inequality on partial sums give you $$ \sum^N_{n=1}\frac{\sqrt{a_n}}{n}\leq \sqrt{\sum^N_{n=1}a_n}\sqrt{\sum^N_{n=1}\frac{1}{n^2}} $$