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I'm trying to determine whether the series $\sum_{n=1}^{\infty} \sin(a_n)$ converges and if it converge absolutely if we know that $\sum_{n=1}^{\infty} a_n$ converges absolutely.

You can rephrase the question asking if $\sum_{n=1}^{\infty} \sin(\frac1{n^2})$ converges.

Edit: I thought this was basically like the general case, but was pointed out it was not.

I found this: Convergence/Divergence of $\sum_{n=1}^{\infty} \sin(1/n)$ But it's for 1/n.

I'm pretty sure that it doesn't converge absolutely because it's a periodic function. As for regular convergence, I'm not really sure how to check.

Note: we can't use integrals because we haven't covered that.

Any advice would be appreciated.

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GinKin
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    Can you already use $\lvert \sin x\rvert \leqslant \lvert x\rvert$? – Daniel Fischer Dec 09 '13 at 17:14
  • I think so yes. – GinKin Dec 09 '13 at 17:15
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    Goes down fast enough, so converges. Absolutely, but the terms are positive, so that says nothing more. And actually it will show no periodicity, all angles will be between $0$ and $1$, – André Nicolas Dec 09 '13 at 17:15
  • @GinKIN is it collected from NET model queston? – Supriyo Dec 09 '13 at 17:25
  • Nope, no idea what that is. – GinKin Dec 09 '13 at 17:33
  • When you say, "you can rephrase the question..." don't you mean, "we can consider the special case of ...?" The general case certainly doesn't follow from the case $a_n=\frac1{n^2}$. – Thomas Andrews Dec 09 '13 at 18:04
  • Also, "because it's a periodic function..." What is a periodic function, $\sin(x)$? Why would that matter? – Thomas Andrews Dec 09 '13 at 18:05
  • @DanielFischer do you know what it's called, I would like to see the proof, beacuse in the linked question you can do this: $\sin(\frac1{n})\le\frac1{n^2}\le\frac1{n}$. $\frac1{n^2}$ we know converge, but $\sin(\frac1{n})$ diverge. – GinKin Dec 09 '13 at 18:54
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    $\sin \left(\frac1n\right) \leqslant \frac{1}{n^2}$ is wrong for $n \geqslant 2$. You have $x - \frac{x^3}{6} \leqslant \sin x \leqslant x$ for $0 \leqslant x \leqslant \pi$. – Daniel Fischer Dec 09 '13 at 18:59
  • @ThomasAndrews I assumed that basically any converging series behaves like $\frac1{n^2}$. Can you elaborate on the general case ? (I'll edit the question for the general case now) About the periodic function, what I meant is that $\sin x$ doesn't go to a certain value but rotate between 1 and -1, from my intuition I think that if we take the absolute value of $sin$, it will rotate but twice over the positive side, so the terms will "stack" and won't cancel each other and it will diverge. – GinKin Dec 09 '13 at 19:06
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    Regarding periodicity: You are only looking at the values of $\sin x$ in a range near $0$, it doesn't matter at all that the function is periodic. If this is true or false, it depends only on the behavior of $\sin x$ near $0$. @GinKin – Thomas Andrews Dec 09 '13 at 19:17

1 Answers1

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Yes, the series is converges absolutely

if $\sum\limits_{n=1}^{\infty} a_n$ converges absolutely, then $\sum\limits_{n=1}^{\infty} \sin a_n$ also converges absolutely

$$0\le |\sin a_n| \le |a_n|$$

ziang chen
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