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

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.

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.
• Hi Jose, thank you! So, even though $\lim sin(a_k) = a_k$ as $k \to \infty$ there will never be a point where these two are equal? Mar 26, 2020 at 8:59
$$\sum b_k \sin (a_k)$$ is absolutely convergent because $$|\sin x | \leq |x|$$ and $$\sum a_kb_k$$ is absolutely convergent