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$$\text{If} \ S=\displaystyle\sum_{n=1}^{\infty}\dfrac{\sin (n)}{n}, \ \text{then what is} \ 2S+1$$

I know that $\sum \frac{\sin(n)}{n}$ converges. But now what do I do?

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  • $\begingroup$ That depends a bit on just how you know that the series converges. $\endgroup$ – Gerry Myerson Nov 14 '15 at 4:18
  • $\begingroup$ @GerryMyerson What do you mean. not quite understanding $\endgroup$ – rashid Nov 14 '15 at 4:20
  • $\begingroup$ You write that you know that the series converges. How do you know that the series converges? $\endgroup$ – Gerry Myerson Nov 14 '15 at 4:21
  • $\begingroup$ Hint: what is the value of the sum from $-1$ to $-\infty$? $\endgroup$ – Steven Stadnicki Nov 14 '15 at 4:32
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I'm not going to prove this here, but we know that (ask another question if you don't): $$\sum_{k=1}^{\infty} \frac{x^k }{k} = -\log(1-x)$$ For any $x$ s.t. $|x|\leq 1, x\neq 1$: $$S= \text{Im} \sum_{n=1}^{\infty} \frac{e^{i n}}{n} = -\text{Im}(\log(1-e^{i} ))$$ Now with some trig: $$1-e^i = 1-\cos 1 -i \sin 1 = 2\sin \frac{1}{2}\left(\sin \frac{1}{2} - i \cos \frac{1}{2} \right) = 2\exp\left(\frac{(1-\pi)i}{2}\right) \sin \frac{1}{2}$$ Therefore, $$S= \text{Im}(\log(1-e^{i} )) = \text{Im}\left(\frac{(\pi-1)i}{2} - \log \left(2\sin \frac{1}{2}\right) \right) = \frac{\pi-1}{2} $$ $$\therefore 2S+1 =\pi$$

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