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.


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

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  • 1
    $\begingroup$ 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

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 $$ |S_n| \leq \left|e^i\frac{1-e^{in}}{1-e^i}\right| \leq \frac{2}{|1-e^i|} < \infty. $$


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 $$\sum_{k=1}^{n}\sin kx=\frac{\cos\left(\frac{1}2x\right)-\cos\left(n+\frac{1}2\right)x}{2\sin(x/2)},x\neq0,\pm\pi,\pm 2\pi,...$$


$\displaystyle \sum \frac{\sin n}{n}$ 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.
  • $\begingroup$ 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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