Absolute convergence of $\sum_{n=0}^{\infty}a_nb_n$ if $\sum_{n=0}^{\infty}b_n^2$ and $\sum_{n=0}^{\infty}a_n^2$ converge Let $(b_n)$ and $(a_n)$ sequences of real numbers. If $\sum_{n=0}^{\infty}b_n^2$ and $\sum_{n=0}^{\infty}a_n^2$ converge, prove that $\sum_{n=0}^{\infty}a_nb_n$ converges absolutely.
I would like to have a feedback on my proof and know if everything is correct, please.
Proof.
As $\sum_{n=0}^{\infty}b_n^2$ converges, $\lim_{n\to\infty}b_n^2=0$. Therefore, $b_n^2$ is bounded and we have that: $\exists M>0$: $|b_n|<\sqrt{M}\ \forall n\in\mathbf{N}$ .
By analogous reasoning we obtain that $\forall \epsilon>0\ \exists N \ \forall n\ge N'$: $|a_n|<\frac{\epsilon}{\sqrt{M}(n-N)}$.
Let's show that $\sum_{n=0}^{\infty}|a_nb_n|$ is Cauchy. By definition we have to show the following: $\forall \epsilon>0 \ \exists N' \ \forall n\ge N'$:
$|\sum_{k=N'}^{n}|a_kb_k||<\epsilon$
But,
$|\sum_{k=N'}^{n}|a_kb_k||=\sum_{k=N'}^{n}|a_kb_k|=\sum_{k=N''}^{n}|a_k||b_k|\le \sqrt{M}\sum_{k=N'}^{n}|a_n|\le \sqrt{M}\cdot \frac{\epsilon}{\sqrt{M}(n-N)}\cdot (n-N')$
If we set $N'=N$, we obtain that
$|\sum_{k=N'}^{n}|a_kb_k||\le \epsilon$. We conclude then that $\sum_{n=0}^{\infty}a_nb_n$ converges absolutely.

Your proof cannot be correct, as in the end you only used that (i) $(b_n)_n$ was bounded, and (ii) $\lim_{n\to \infty} a_n =0$. But those those assumptions alone cannot imply convergence (let alone absolute convergence) of $\sum_n a_n b_n$, as otherwise we would get convergence of $\sum_n \frac{1}{n}\cdot 1$.
Now, why is it false? Well, you write

By analogous reasoning we obtain that $\forall \epsilon>0\ \exists N \ \forall n\ge N$: $|a_n|<\frac{\epsilon}{\sqrt{M}(n-N)}$.

This step, for instance, is not true. It may holds for fixed $N,M$ (with a fixed $n$ on the RHS), but how do you prove it holds for all $n\geq N$? Here is a counterexample:
$$
a_n = \frac{1}{n^{2/3}}
$$
Note that $\sum_n a_n^2 < \infty$, but the statement you wrote does not hold.
