# If $\sum (a_n)^2$ converges then $\sum \frac{a_n}{n}$ converges (one solution)

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

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

• This is a shining example of the use of the Cauchy--Schwarz inequality. – Pedro Tamaroff Jan 27 at 20:36
• 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. – User8128 Jan 27 at 20:37
• $|a_n|\le a_n^2\iff a_n\le -1 \ \vee \ a_n=0 \ \vee \ a_n\ge 1$ – Ixion Jan 27 at 20:39
• 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 $\sum \frac{1}{n^2} < \infty$... – Calvin Khor Jan 27 at 21:08
• @CalvinKhor I do not see why the convergence of $1/n^{3/2}$ can be more eligible for the answer than that of $1/n^2$. – 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 $$\displaystyle\sum_{n=1}^\infty\frac1{n^2}$$.

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

• 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. – Juan Daniel Valdivia Fuentes Jan 27 at 20:47