Let $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}|b_n|$ converging series. Proof that $\sum_{n=0}^{\infty}a_nb_n$ converges as well. What happens if $\sum_{n=0}^{\infty}b_n$ doesn't converge absolutely?
I would like to know if my proof for the statement holds and if my counter-example for second part works as well. Thank you in advance!
Proof. (First part)
First of all, as $\sum_{n=0}^{\infty}a_n$ converges, we deduce that $\lim_{n\to \infty}a_n=0$. Therefore, $\exists N \ \forall n\ge N$: $0<|a_n|<1$
Then, by the analogous reasoning we obtain that: $\exists N' \ \forall n\ge N': 0<|b_n|<1$
Let $n_0= \max\{N,N'\}$. Thus, $\forall n\ge n_0$ we have that:
$0<|a_n||b_n|=|a_nb_n|<|b_n|$. So, by comparaison, as $\sum_{n=0}^{\infty}|b_n|$ exists, we conclude that $\sum_{n=0}^{\infty}a_nb_n$ converges absolutely.
Counter-example (If $\sum_{n=0}^{\infty}b_n$ converges, but not absolutely)
Let $(a_n)=\frac{(-1)^n}{\sqrt[3]{(n+1)^2}}$ and $(b_n)=\frac{(-1)^n}{\sqrt[3]{n+1}}$. By alternating series criterion the series $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ converge ($\sum_{n=0}^{\infty}b_n$ doesn't converge absolutely). But,
$a_nb_n=\frac{1}{n+1}$ and so $\sum_{n=0}^{\infty}a_nb_n$ doesn't converge.