Need to prove: $\sum_{n=1}^{\infty}{a_n}$ converges $\implies\sum_{n=1}^{\infty }\frac{\sqrt{a_n}}{n}$ converges

Question

I need to prove that for a positive sequence $a_n$:

$$\sum_{n=1}^{\infty}{a_n} \text{ converges} \implies \sum_{n=1}^{\infty }\frac{\sqrt{a_n}}{n} \text{ converges}$$

How can I prove it with only basic tools such the comparison test? I've been struggling with this one for pretty long time and I'd be glad for some help.

Answer 1

Let $S_N=\sum_{n=1}^{N }\frac{\sqrt{a_n}}{n}$, $A_N=\sum_{n=1}^{N } a_n$ and $b_n=\frac{\sqrt{a_n}}{n}$

By the AM-GM Inequality, for all $n$ we have $$0\le b_n = \frac{\sqrt{a_n}}{n} \le\frac{a_n+\frac{1}{n^2}}{2}= \frac{a_n}{2}+\frac{1}{2n^2}$$

Thus, $$0\le S_N\le\frac{A_N}{2}+\frac{1}{2}\sum_{n=1}^{N}\frac{1}{n^2}<\frac{A_N}{2}+\pi^2$$

Where I used the fact that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$

Since $\lim_{N\to\infty}A_N$ exist, by the comparison test, we have that $\lim_{N\to\infty}S_N$ also exists and thus we conclude the sum converges

Answer 2

By Cauchy-Schwarz inequality $$\sum_{i=1}^{n}\frac{\sqrt{a_i}}{i}\leq\sqrt{\left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}\frac{1}{{i^2}}\right)}$$

$\displaystyle\sum_{i=1}^{n}\frac{1}{{i^2}}$ converges to $\dfrac{\pi^2}{6}$ as $n\to\infty$. So, by comparison,$\displaystyle\sum_{i=1}^\infty\frac{\sqrt{a_i}}{i}$ converges.