$\sum_{n=0}^\infty a_k$ converges absolutely and $\sum_{n=0}^\infty b_k$ converges Does this imply that $\sum_{n=0}^\infty b_k\sin(a_k)$ converges?

Question

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.

Answer 1

It is wrong because it assumes that $\sum_{k=N+1}^\infty b_ka_k=\sum_{k=N+1}^\infty b_k\sin(a_k)$, which is not true.

Answer 2

$\sum b_k \sin (a_k)$ is absolutely convergent because $|\sin x | \leq |x|$ and $\sum a_kb_k$ is absolutely convergent