real analysis - $b_n$ bounded, $sum a_n$ converges absolutely, then $sum a_nb_n$ also

a) Prove that if $\sum a_n$ converges absolutely and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges absolutely.

I wanted to use the comparison test to show it's true, but I think I got something mixed up. Here's what I have so far.

$b_n$ is bounded $\Rightarrow \exists c > 0: |b_k| < c, \forall k \in \mathbb N$

$\sum a_nb_n < \sum a_nc < c\sum a_n$. I'm very tempted to use the comparison test here and say, as $\sum a_n$ converges absolutetly, then so does $c\sum a_n$, but for that I needed the inverted relation, right? I would need $\sum a_n > c\sum a_n$, which is not true. However isn't it obvious that something absolutely convergent multiplied by a constant is also absolutetly convergent? Is there a mathematical way to write this? Thanks a lot in advance.

b) Refute with a counter example: if $\sum a_n$ converges and $b_n$ is a bounded sequence, then also $\sum a_nb_n$ converges.

Is this even possible? I mean, it has to be, but it doesn't make sense to me. If something converges, it means it's bounded, right? And I thought, well, since bounded + bounded = bounded, then bounded*bounded would also get me something bounded again.

Anyway, my idea would be to use for $a_n$ an alternating series that ist only convergent, for example $(-1)^n \frac 1 {n^2}$. But something tells me I'm trying to prove $a_n b_n$ is not absolutely convergent.

Thanks a lot in advance guys!

Answer

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

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 $\sum a_nb_n$?