Given that the convergence of the series $\sum a_n$ implies convergence of $\sum \frac{\sqrt{a_n}}{n}$. I saw a lot answers and hints saying this can be proved using Cauchy–Schwarz inequality, but none with details, I'm going to try to fill in the details and see If I understand the idea fully.
We know that a series of nonnegative numbers is convergent if and only it is bounded. Let M_1>0 be the bound for $\sum a_n$ and $M_2$ be the bound for $\sum\frac{1}{n^2}$. I think the idea is to look at the sum as a dot product of 2 vectors from $(a_0,a_1,\dots,a_N),(1,\frac{1}{2},\frac{1}{3},\dots, \frac{1}{N})\in \mathbb{R}^N$ and hence we apply the C-S inequality $$(a_0,a_1,\dots,a_N)\cdot(1,\frac{1}{2},\frac{1}{3},\dots, \frac{1}{N})=\sum^N\frac{a_n}{n}\leq \sqrt{\sum^Na_n^2}\sqrt{\sum^N\frac{1}{n^2}}$$ If we take N to infinity, then the right hand side of the inequality is bounded by $M_1^2M_2<\infty$. Hence $\sum\frac{a_n}{n}$ is bounded and so it is convergent.
But not sure if I'm missing any details. Thanks!