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 $2\alpha >1 \implies \alpha >\frac12$.
A: hint:
The Cauchy-Schwarz inequality gives
$$\sum_{n=1}^\infty \frac{\sqrt{a_n}}{n}\leq \sqrt{\sum_{n=1}^\infty a_n}\,\sqrt{\sum_{n=1}^\infty \frac{1}{n^2}}<\infty.$$
A: Just to give an elementary proof (not relying on Cauchy-Schwartz), let $\alpha={1+p\over2}$. Then
$$\begin{align}
\left|\sqrt{a_n/n^{1+p}}-1/n^{1+p}\right|
&={\left|a_n/n^{1+p}-1/n^{2+2p}\right|\over\sqrt{a_n/n^{1+p}}+1/n^{1+p}}\\
&\le{\left|a_n/n^{1+p}-1/n^{2+2p}\right|\over1/n^{1+p}}\\
&=\left|a_n-1/n^{1+p}\right|\\
&\le a_n+1/n^{1+p}
\end{align}$$
and thus 
$$\begin{align}
\left|\sqrt{a_n/n^{1+p}}\right|
&=\left|\sqrt{a_n/n^{1+p}}-1/n^{1+p}+1/n^{1+p}\right|\\
&\le\left|\sqrt{a_n/n^{1+p}}-1/n^{1+p}\right|+1/n^{1+p}\\
&\le a_n+2/n^{1+p}
\end{align}$$
and so
$$\sum_{n=1}^\infty{\sqrt{a_n}\over n^\alpha}\le\sum_{n=1}^\infty a_n+2\sum_{n=1}^\infty{1\over n^{1+p}}\lt\infty$$
