My solution is detailed, I would like to know if it is correct or not.

As the series converges and each $(a_n)^2$ is positive, then exists $K> 0$ such that : $(a_1)^2+(a_2)^2+ \cdots + (a_n)^2 < K, \forall n \in \mathbb{N}$.

Given $n \in \mathbb{N}$ : $ \lvert \dfrac{a_n}{n} \rvert = \dfrac{\lvert a_n \rvert}{n} \leq \dfrac{(a_n)^2}{n} \leq (a_n)^2$ so :

$\dfrac{\lvert a_1 \rvert}{1} + \dfrac{\lvert a_2 \rvert}{2} + \cdots \dfrac{\lvert a_n \rvert}{n} \leq (a_1)^2+(a_2)^2 + \cdots (a_n)^2<K$

Finally the series $\sum \dfrac{\lvert a_n \vert}{n}$ converge, consequently $\sum \dfrac{a_n}{n} $ converge.

  • 1
    $\begingroup$ This is a shining example of the use of the Cauchy--Schwarz inequality. $\endgroup$ – Pedro Tamaroff Jan 27 at 20:36
  • $\begingroup$ I don’t see how you can say $\lvert a_n\rvert / n \le a_n^2/n$, so I’m not sure this proof works. What you want here is the Cauchy-Schwarz inequality. $\endgroup$ – User8128 Jan 27 at 20:37
  • $\begingroup$ $|a_n|\le a_n^2\iff a_n\le -1 \ \vee \ a_n=0 \ \vee \ a_n\ge 1$ $\endgroup$ – Ixion Jan 27 at 20:39
  • $\begingroup$ If $\sum a_n^2$ converges, then so does $\sum |a_n|^3$ by comparison test, and now by Holder's inequality, $$ \sum \frac{|a_n| }{n} \le \left (\sum |a_n|^3 \right )^{1/3} \left( \sum\frac 1{n^{3/2}} \right)^{2/3} <\infty $$ without using $\sum \frac{1}{n^2} < \infty$... $\endgroup$ – Calvin Khor Jan 27 at 21:08
  • $\begingroup$ @CalvinKhor I do not see why the convergence of $1/n^{3/2} $ can be more eligible for the answer than that of $1/n^2$. $\endgroup$ – user Jan 27 at 21:25

No, that is not correct. You have no reason to assume that$$(\forall n\in\mathbb{N}):\frac{\lvert a_n\rvert}n\leqslant{a_n}^2.$$

The statement that you want to prove is a consequence of the Cauchy-Schwarz inequality and of the convergence of the series $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$.

  • $\begingroup$ Yes, you are right. Is true but if $\lvert a_n \rvert > 1$, thanks! :) $\endgroup$ – Juan Daniel Valdivia Fuentes Jan 27 at 20:40
  • $\begingroup$ @JuanDanielValdiviaFuentesJ If $|a_n|>1$ neither of the series can converge. $\endgroup$ – user Jan 27 at 21:31

Another way can be by using the AM-GM inequality: For positive $x,y$ we have $\frac{x^2+y^2}{2} \geq xy $. Now put $x=|a_k|$ and $y=\dfrac{1}{k}$. Then, we have

$$ a_k^2 + \frac{1}{k^2} \geq 2\frac{|a_k|}{k} $$

Adding up, we see that

$$ \sum_{k=1}^n a_k^2 + \sum_{k=1}^n \frac{1}{k^2} \geq 2 \sum \frac{ |a_k| }{k} $$

can you finish it?

  • $\begingroup$ Yes, the series $\sum (a_n)^2 $ and $\sum \dfrac{1}{n^2}$ converge. So the serie $\sum \dfrac{\lvert a_n \rvert}{n}$ converge. But I can not use the fact that the series $\sum \dfrac{1}{n^2}$, converges because I had been told to find a way without using it. $\endgroup$ – Juan Daniel Valdivia Fuentes Jan 27 at 20:47

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