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Does $\sum\frac{\sin n}{n}$ converge?

I have tried the comparison test, root test and ratio test but still can't prove it is convergent or divergent.

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marked as duplicate by Seirios, user1729, Paul, user17762, Joe Mar 28 '13 at 0:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The linked "possible duplicate" addresses convergence of a series with slightly larger terms (harder problem), and it has a comment that claims a closed form for the series discussed here (assuming limits $n=1$ to $\infty$ were intended). But I couldn't find an exact dup (yet). $\endgroup$ – hardmath Mar 27 '13 at 12:42
  • $\begingroup$ The nice Answer to this Question shows how to derive the exact limit $(\pi - 1)/2$ for this series. $\endgroup$ – hardmath Mar 31 '13 at 0:55
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You can use Dirichlet's test: the sequence $\frac{1}{n}$ is decreasingly converging to $0$, so you have to prove that $$ S_n=\sum_{k=1}^n \sin k $$ is bounded.

Here is a quick way to prove it: using $S_n = \Im(\sum_{k=1}^n e^{ik})$ and the inequality $|\Im(z)| \leq |z|$, we have $$ |S_n| \leq \left|e^i\frac{1-e^{in}}{1-e^i}\right| \leq \frac{2}{|1-e^i|} < \infty. $$

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Hint: I found this hint from my old notes. I hope it helps you.

$$\sum_{n=1}^{\infty}\frac{\sin nx}{n}$$ is uniformly convergent in any interval which doesn't include $$0,\pm\pi,\pm 2\pi,...$$

To see this use the Dirichlet test and this fact that $$\sum_{k=1}^{n}\sin kx=\frac{\cos\left(\frac{1}2x\right)-\cos\left(n+\frac{1}2\right)x}{2\sin(x/2)},x\neq0,\pm\pi,\pm 2\pi,...$$

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$\displaystyle \sum \frac{\sin n}{n}$ converges

  1. $\displaystyle \sum_{n = 1}^M \sin n$ is bounded always (why?)
  2. $\dfrac{1}{n} \to 0$ and is decreasing as $n \to \infty$
  3. We can conclude by Dirichlet's Test for convergence, which is sort of a generalization of the 'alternating series test,' that our series converges.
  4. This convergence is not absolute. In particular, $\max\{|\sin n|, |\sin (n + 1)|\}> \delta > 0$ for some $\delta$ (not depending on $n$), and so combining every two adjacent terms of the absolute series will diverge by comparison with the harmonic series.
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  • $\begingroup$ you forgot the absolute values in (4) $\endgroup$ – mau Mar 27 '13 at 10:20
  • $\begingroup$ @mau: yes, I did. Thank you. $\endgroup$ – davidlowryduda Mar 27 '13 at 10:40
  • $\begingroup$ In (4) do you mean $\max \{|\sin n|,|\sin(n+1)|\} \gt \delta \gt 0$ for some $\delta$ not depending on $n$? $\endgroup$ – hardmath Mar 27 '13 at 12:36
  • $\begingroup$ @hardmath, yes, that's right $\endgroup$ – davidlowryduda Mar 27 '13 at 18:31

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