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

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.

## marked as duplicate by Arnaud D., GNUSupporter 8964民主女神 地下教會, Saad, jvdhooft, T. BongersMar 31 '18 at 13:42

• Cauchy-Schwarz or $ab\leq \frac{a^2+b^2}2$. – Gabriel Romon Mar 30 '18 at 12:18
• It would have been nice to see C. Dekel work out the answer on his/her own based on the @GabrielRomon comment. Or, if the later answers by Prathyush and Shubhashish were enthusiastic demonstrations of working out the details starting from Gabriel's hint, it would be nice to see those two acknowledge Gabriel for the help and inspiration. – Michael Mar 30 '18 at 14:26

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

• Don't want to be annoying in any way but I'd very much like to see you taking the limit: that is take a finite sum, apply your reasoning and thus, bound the partial sums. Hence, using positivity you'll be able to use monotonicity arguments. Otherwise - great solution worthy of an upvote!! – asdf Mar 30 '18 at 12:30
• @asdf no harm in being rigourous :-) I’ll get on it – Prathyush Poduval Mar 30 '18 at 12:32
• @PrathyushPoduval +1. Please edit your answer and explain what happened to $2$ in denominator. – Qurultay Mar 30 '18 at 12:34

By C-S inequality $$\sum_{i=1}^{n}\frac{\sqrt{a_i}}{i}\leq\sqrt{(\sum_{i=1}^{n}a_i)(\sum_{i=1}^{n}\frac{1}{{i^2}})}$$

$(\sum_{i=1}^{n}\frac{1}{{i^2}})$ converges to $\frac{\pi^2}{6}$ as $n\to\infty$

So, $lim_{n\to\infty}$ $\sum_{i=1}^{n}\frac{\sqrt{a_i}}{i}$ exists.