Convergence of $\sum_{n=0}^{\infty}a_n/n$ if $\sum_{n=0}^{\infty}a_n^2$ exists

Question

Let $\sum_{n=0}^{\infty}a_n^2$ a convergent series. Prove that $\sum_{n=0}^{\infty}a_n/n$ converges or give a counter-example.

I know that there is already a post concerning this statement with a lot of answers, but i did't find a similar approach to mine in that post and an answer to my question.

After a long searching for a counter-example (failed), i started to prove the statement. First, i tried to prove it by comparaison but didn't succeed. So, i tried to pass by Cauchy definition. I'm not sure at all if it holds what i did. If someone could give a feedback and point to my errors, i would really appreciate it.

Proof

As $\sum_{n=0}^{\infty}a_n^2$ exists, then $\lim_{n\to \infty}a_n^2=0$. Thus, $\forall \epsilon>0 \ \exists N \ \forall n\ge N:$ $|a_n|<\sqrt{\epsilon}$

Let's proof that $\sum_{n=0}^{\infty}\dfrac{a_n}{n}$ is Cauchy. By definition of Cauchy we have to satisfy the following: $\forall \epsilon>0 \ \exists N' \ \forall n\ge N'$:

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|<\epsilon$

But,

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le \sum_{k=N'}^{n}\dfrac{|a_k|}{k}\le \sqrt{\epsilon}\sum_{k=N'}^{n}\dfrac{1}{k}$

Let $(x_n)=\frac{1}{n}$. As $\lim_{n\to\infty}x_n=0$ we have by definition that: $\forall \epsilon>0 \ \exists N'' \ \forall n\ge N'':$ $|x_n|<\sqrt{\epsilon}$. Therefore,

$\sqrt{\epsilon}\sum_{k=N'}^{n}\dfrac{1}{k}<\sqrt{\epsilon}\sqrt{\epsilon}=\epsilon$.

So, if we choose $N'=\max\{N,N''\}$, we satisfy Cauchy definition.

Edit: As $x_n$ is a converging sequence: $\exists M>0$: $|x_n|\le M \ \forall n \in \mathbf{N}$. We could then take $|a_n|<\frac{\epsilon}{M(n-N)}$.

Thus,

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le \frac{\epsilon}{M(n-N)}\sum_{k=N'}^{n}\dfrac{1}{k}\le \frac{\epsilon}{M(n-N)}\cdot M(n-N')$

If we take $N'=N$, we have then:

$\Big|\sum_{k=N'}^{n}\dfrac{a_k}{k}\Big|\le\epsilon$

Does it hold this way?

Answer 1

From $x_n<\sqrt \epsilon$, you unfortunately only get $$\sqrt \epsilon\sum_{k=N'}^n\frac1n=\sqrt \epsilon\sum_{k=N'}^nx_n<\sqrt \epsilon\sum_{k=N'}^n\sqrt \epsilon =(n-N+1)\epsilon\not<\epsilon. $$

Answer 2

By Cauchy-Schwarz inequality, we have that $\sum_{n=1}^{N}|a_{n}|\cdot|\frac{1}{n}|\leq\left\{ \sum_{n=1}^{N}a_{n}^{2}\right\} ^{\frac{1}{2}}\left\{ \sum_{n=1}^{N}\frac{1}{n^{2}}\right\} ^{\frac{1}{2}}\leq\left\{ \sum_{n=1}^{\infty}a_{n}^{2}\right\} ^{\frac{1}{2}}\left\{ \sum_{n=1}^{\infty}\frac{1}{n^{2}}\right\} ^{\frac{1}{2}}<\infty$. Therefore, $\left(\sum_{n=1}^{N}|a_{n}|\cdot|\frac{1}{n}|\right)_{N}$ is a bounded increasing sequence and hence $\lim_{N\rightarrow\infty}\sum_{n=1}^{N}|a_{n}|\cdot|\frac{1}{n}|$ exists. That is, the infinite series $\sum_{n=1}^{\infty}\frac{a_{n}}{n}$ converges absolutely.