Solution Verification: $\sum_{n = 1}^{\infty} a_n \lt \infty \implies \sum_{n = 1}^{\infty} \dfrac{\sqrt{a_n}}{n} \lt \infty$ if $a_n \ge 0$

Question

I would appreciate it if someone can verify my answer to the following and maybe point me towards alternate solutions. Thanks in advance. I'm not entirely convinced with the use of the Comparison Test here.

Prove the convergence of $ \sum_{n = 1}^{\infty} a_n$ implies the convergence of $\sum_{n = 1}^{\infty} \dfrac{\sqrt{a_n}}{n}$ if $a_n \ge 0$ for each $n \in \Bbb N$.

Solution:

If $a_n \ge \frac 1 n$ for infinitely many $n \in \Bbb N$ it follows that $\sum a_n$ diverges leading to a contradiction. Then there is $m \in \Bbb N$ such that $n \ge m \implies a_n \lt \frac 1 n$. Therefore,

$$ n \ge m \implies \dfrac{\sqrt{a_n}}{n} \lt \dfrac{\sqrt{\frac{1}n{}}}{n} = \dfrac{1}{n^{\frac 3 2}} $$

Since $\sum \dfrac{1}{n^{\frac 3 2}} $ converges so does $\sum \dfrac{\sqrt{a_n}}{n}$ by the Comparison Test.

Answer 1

Hint: You can use Cauchy-Schwarz inequality.