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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.

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    $\begingroup$ Cauchy-Schwarz or $ab\leq \frac{a^2+b^2}2$. $\endgroup$ Mar 30, 2018 at 12:18
  • $\begingroup$ 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. $\endgroup$
    – Michael
    Mar 30, 2018 at 14:26

2 Answers 2

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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

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    $\begingroup$ 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!! $\endgroup$
    – asdf
    Mar 30, 2018 at 12:30
  • $\begingroup$ @asdf no harm in being rigourous :-) I’ll get on it $\endgroup$ Mar 30, 2018 at 12:32
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    $\begingroup$ @PrathyushPoduval +1. Please edit your answer and explain what happened to $2$ in denominator. $\endgroup$
    – Qurultay
    Mar 30, 2018 at 12:34
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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.

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