0
$\begingroup$

I am trying to show that if the series $\sum_\limits{n=1}^{\infty}a_n^2$ converges, then the series $\sum_\limits{n=1}^{\infty}{a_n \over n}$ converges absolutely. How might I approach this?

$\endgroup$

marked as duplicate by Mark McClure, Willie Wong, user91500, zhoraster, Rohan Dec 20 '16 at 6:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

Hint:

$$xy \le \frac12(x^2 + y^2)$$

Put $x = \frac1n$ and $y=|a_n|$.

$\endgroup$
2
$\begingroup$

$$\sum_{n\geq 1}\frac{|a_n|}{n}\stackrel{\text{CS}}{\leq}\sqrt{\sum_{n\geq 1}a_n^2 \sum_{n\geq 1}\frac{1}{n^2}} = \frac{\pi}{\sqrt{6}}\left(\sum_{n\geq 1}a_n^2\right)^{1/2}.$$

$\endgroup$
1
$\begingroup$

see that $\frac{|a_n|}{n}$ $\leq$ $\frac{a_n^2}{2}$ +$\frac{1}{2n^2}$ by A.M G.M inequality. so

$\sum_{n=1}^{\infty}\frac{|a_n|}{n}$ $\leq$ $\frac{1}{2}$$\sum_{n=1}^{\infty} a_n^2$ +$\frac{1}{2}$$\sum_{n=1}^{\infty}\frac{1}{n^2}$

$\sum a_n^2$ and $\sum \frac{1}{n^2}$ are convergent. so by comparison test $\sum\frac{|a_n|}{n}$ is convergent. so $\sum \frac{a_n}{n}$ is convergent absolutely

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.