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

Question

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.

Answer 1

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 $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$.

Answer 2

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?