Suppose $\sum{a_n}$ converges. How do I prove that $\sum{\frac{\sqrt{a_n}}{n}}$ converges [duplicate]

Suppose $\sum{a_n}$ converges. How do I prove that $\sum{\frac{\sqrt{a_n}}{n}}$ converges.

So I know that just because $\sum{a_n}$ converges, do not mean I can say anything about the converges of its square root. So I know that I can prove that if $\sum{\sqrt{a_n}}$ converges then $\sum{\frac{\sqrt{a_n}}{n}}$ converges by Abel's Test, but I do not know where to start for the case where $\sum{\frac{\sqrt{a_n}}{n}}$ diverges

marked as duplicate by Lord Shark the Unknown, Jyrki Lahtonen, Taroccoesbrocco, Brahadeesh, Claude LeiboviciAug 13 '18 at 10:05

• So are we to assume that $\forall{n}:a_n\geq0$? Otherwise $\exists{n}:\sqrt{a_n}\not\in\mathbb{R}$. – barak manos Dec 2 '14 at 13:22
From Cauchy-Schwarz inequality $$\left(\sum_{n=1}^N\frac{\sqrt{a_n}}{n}\right)^2\le\sum_{n=1}^N(\sqrt{a_n})^2 \cdot \sum_{n=1}^N\frac{1}{n^2}\le \sum_{n=1}^\infty a_n \cdot \sum_{n=1}^\infty\frac{1}{n^2}$$
Therefore $$\sum_{n=1}^\infty\frac{\sqrt{a_n}}{n}\le \sqrt{\sum_{n=1}^\infty a_n \cdot \sum_{n=1}^\infty\frac{1}{n^2} } <\infty$$
Using the AM-GM inequality, $\;\;\displaystyle\frac{\sqrt{a_n}}{n}\le\frac{1}{2}\left(a_n+\frac{1}{n^2}\right)$, so the result follows from the Comparison Test.