this is a Homework I did a few days ago, my solution differs from the official solution, but the conclusion is correct. Yet I'm not sure if this is just coincidence as my solution is very simple. I am grateful if you could have a look.
The question:
$\sum_{k=0}^\infty a_k$ converges absolutely and $\sum_{k=0}^\infty b_k$ converges Does this imply that $\sum_{n=0}^\infty b_ksin(a_k)$ converges?
So I thought that because $\sum_{n=0}^\infty a_k$ converges absolutely we have that $\lim{n\to \infty}$ of $a_k= 0$ . $$\lim_{x\to 0} \frac {\sin(x)}{x} = 1$$ Therefore I thought: $$\lim_{k\to \infty} \frac {\sin(a_k)}{a_k} = 1$$
So there is some $N$ after which $$\sin(a_k) \approx a_k$$
And $\sum_{n=0}^\infty b_ka_k$ converges. So I break apart the Series to $$S_N = \sum_{k=0}^N b_k\sin(a_k)$$ as:
$$\sum_{n=0}^\infty b_k\sin(a_k) = S_N + \sum_{n=N+1}^\infty b_ka_k$$
I think it must be wrong now. But I can't see why? I apologize about the formatting, I'm still not very good at it.