For $\sum a_n$ convergent and also $\forall n \in \mathbb{N}$: $a_n\geq0$, show that: $$ \sum \frac{\sqrt{a_n}}{n^\alpha}$$ converges for $\alpha> \frac{1}{2}$.
My idea was to use the Direct comparison test, I'm pretty sure that's the way to prove it, and to find a bigger series that also converges, but I haven't been able to find one appropriate series.
It would be very helpful if you could give me a hint on how this bigger series. Thank you very much
By Cauchy-Schwarz inequality, we have that $$\sum_{n=1}^N \frac{\sqrt{a_n}}{n^\alpha}=\sum_{n=1}^N \sqrt{a_n}\cdot\frac{1}{n^\alpha}\leq \left(\sum_{n=1}^N a_n\right)^{1/2}\cdot \left(\sum_{n=1}^N \frac{1}{n^{2\alpha}}\right)^{1/2}.$$ Can you take it from here?