Q:

How to prove convergence of $\sum\limits_{n=1}^\infty \frac{\sqrt a_n}{n^\alpha}$

Assume that $a_n >0$, $n\in\mathbb{N}$, and that $\sum\limits_{n=1}^\infty a_n $ converges.
Show that for $\alpha>1/2$ the series
$$\sum\limits_{n=1}^\infty \frac{\sqrt a_n}{n^\alpha}$$ converges as well
I think we can use Abel's convergence test but I'm not sure if that is the way to go

A:

By Cauchy–Schwarz inequality we have
$$\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n^{\alpha}}\leq \sqrt{\sum_{n=1}^\infty a_n}\,\sqrt{\sum_{n=1}^\infty \frac{1}{n^{2\alpha}}}$$
and 
$$\sum_{n=1}^\infty \frac{1}{n^{2\alpha}}$$
converges if and only if $2\alpha >1 \implies \alpha >\frac12$.

