$b_n$ bounded, $\sum a_n$ converges absolutely, then $\sum a_nb_n$ also

Solution 1:

Recall that if $\sum a_n$ is finite then $c\cdot\sum a_n = \sum ca_n$. This is true even if the series is not absolutely convergent.

The reason is simple, $\displaystyle\sum a_n = \lim_{k\to\infty}\sum_{n<k} a_n$, and when multiplying by a constant it can jump into the limit.

Now your reasoning is true. If $b_n\le c$ then $\sum a_nb_n\le \sum ca_n\le c\sum a_n$, and the latter converges.

For the second question, in the first one the assumption was the series is absolutely convergent. Take a sequence which is not of this form.

For example $a_n = \dfrac{(-1)^n}{n}$, and $b_n=(-1)^n$. Now what is $\sum a_nb_n$?