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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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? $\endgroup$
    – oliver
    Mar 26, 2020 at 8:59
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    $\begingroup$ Yes, that is correct. $\endgroup$ Mar 26, 2020 at 9:00
  • $\begingroup$ Thank you very much! $\endgroup$
    – oliver
    Mar 26, 2020 at 9:01
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    $\begingroup$ I'm glad I could help. $\endgroup$ Mar 26, 2020 at 9:04
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$\sum b_k \sin (a_k)$ is absolutely convergent because $|\sin x | \leq |x|$ and $\sum a_kb_k$ is absolutely convergent

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  • $\begingroup$ Hi Kavi, thanks, that was actually more or less the idea of the official solution. $\endgroup$
    – oliver
    Mar 26, 2020 at 9:03

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