Given $\Sigma a_n \to \alpha$ show that $\Sigma \frac{\sqrt{a_n}}{n} \to \beta$

Question

I am trying to prove that if $\Sigma a_n$ is convergent, then $\Sigma {\sqrt {a_n} \over n}$ is also convergent.

I tried to use the comparison test but $\sqrt{a_n} >a_n $ so I couldn't go that rout.

Next I tried to prove that the second sequence is Cauchy given the first one is, as in

$$|\sum_{i=m}^{n} a_i| \lt \epsilon \Rightarrow |\sum_{i=m}^{n} {\sqrt {a_i} \over i}| \lt \epsilon'$$

but I am not sure how I can get there ...

Answer 1

You can estimate partial sums using Cauchy-Shwartz. Indeed, $$\left(\sum_{k=1}^n \frac{\sqrt{a_k}}{k}\right)^2\le \sum_{k=1}^n{a_k}\sum_{k=1}^n\frac{1}{k^2}.$$