Prove that $\sum ^\infty _{n=1} \sqrt{a_n}\frac{1}{n^s}$ converges for $s>\frac{1}{2}$

Question

Assume that $a_n \ge 0$ for $n=0,1.....$ and $,\sum a_n< \infty$ prove that $$\sum ^\infty _{n=1} \sqrt{a_n}\frac{1}{n^s}$$ converges for $s>\frac{1}{2}$

my idea

$|\sum ^\infty _{n=1} \sqrt{a_n}\frac{1}{n^s}|^2 \leq (\sum _{n=1}^\infty \sqrt{a_n})^2(\sum ^\infty _{n=1} \frac{1}{n^s})^2$

I am trying solve this way but not get the answer..please help me

Answer 1

Cauchy-Schwarz: \begin{align*} \sum\sqrt{a_{n}}\cdot\dfrac{1}{n^{s}}\leq\left(\sum a_{n}\right)^{1/2}\left(\sum\dfrac{1}{n^{2s}}\right)^{1/2}<\infty, \end{align*} since $2s>1$.