How to prove $\sum\limits_{n=1}^{\infty}\frac{\sin n}n=\frac{\pi-1}{2}$

Question

One of my classmates challenged me to solve $\displaystyle\sum\limits_{n=1}^{\infty}\frac{\sin n}n=\;?$

With a simple c program I found that $\displaystyle\sum\limits_{n=1}^{1048576}\frac{\sin n}n\approx1.070796$. Later I found $\displaystyle1.070796\approx\frac{\pi-1}{2}$. My classmate told me I guessed right, but he ask me to prove it, and he gave me a hint that $\displaystyle e^{i\theta} = \cos\theta + i \sin\theta$, though I can't see the relationship between the question and the hint.

So how to prove $\displaystyle\sum\limits_{n=1}^{\infty}\frac{\sin n}n=\frac{\pi-1}{2}$?

possible duplicate of Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$ — Norbert, May 26 '13 at 06:17
@Norbert While the accepted answer to that question answers this one, aren't they different questions? This one asks what the series converges to. All that that one asks is for a proof of convergnece. — 2'5 9'2, May 26 '13 at 06:36
@Norbert The answer in your link confuses me. And what about my friend's hint that $\displaystyle e^{i\theta} = \cos\theta + i \sin\theta$? — johnchen902, May 26 '13 at 06:44
@johnchen902, if "your friend" really hinted you that $,e^{i\theta}=\cos\theta+i\sin\theta;$ and you did not know that before hand, it may be this exercise is not suited to your mathematics level. — DonAntonio, May 26 '13 at 08:42
@DonAntonio I know that beforehand. My friend is just giving me a clue to solve. — johnchen902, May 26 '13 at 08:49
What a crazy identity. Sin(n) will go all over the range of sine because sine is an ergodic function over the integers. Yet it sums to such a nice closed form. — Fomalhaut, Sep 29 '23 at 02:24

score 32 · Accepted Answer · edited Dec 09 '13 at 06:57

32

Make use of the following fact: For $\vert z \vert \leq 1$ and $z \neq 1$, we have $$\log(1-z) = - \sum_{k=1}^{\infty} \dfrac{z^k}k$$ Take $z=e^i$ and look at the imaginary part.

We hence have $$\log(1-e^i) = - \sum_{k=1}^{\infty} \dfrac{e^{ik}}k \implies \text{Imag}(\log(1-e^i)) = \text{Imag}\left(- \sum_{k=1}^{\infty} \dfrac{e^{ik}}k \right)$$ which gives us $$\sum_{k=1}^{\infty} \dfrac{\sin(k)}k = - \text{Imag}(\log(1-e^i))$$ $$\log(1-e^i) = \log \left(2\sin^2(1/2) - 2i \sin(1/2) \cos(1/2)\right) = \log(2\sin(1/2)e^{-i \pi/2}e^{i/2})$$ Hence, $$\text{Imag}(\log(1-e^i)) = \dfrac{1-\pi}2$$ which gives us $$\sum_{k=1}^{\infty} \dfrac{\sin(k)}k = \dfrac{\pi-1}2$$

edited Dec 09 '13 at 06:57

Martin Sleziak

53,687

answered May 26 '13 at 06:16

9

That, my friend, is an epic observation. :-) – P.K. May 26 '13 at 06:18
2

But my question is how we get that. Working it out is no simple task. (I mean the logarithm and all that, as it doesn't simplify nicely) – Simply Beautiful Art Nov 26 '16 at 14:40
1

What is $\text{Imag}$? My first thought, it stood for imagine? – Mr Pie Jan 29 '18 at 07:27
@user477343 I think it stands for "imaginary part." – BallBoy Mar 27 '18 at 16:29
@Y.Forman thanks for telling me :) – Mr Pie Mar 27 '18 at 21:01

How to prove $\sum\limits_{n=1}^{\infty}\frac{\sin n}n=\frac{\pi-1}{2}$

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