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I noticed that a lot of the time, people ask whether the following sum converges:

$$\sum_{n=1}^\infty\frac{\sin(n)}n$$

Though I've never stopped to ask what it equaled. According to this other post, the sum is given as

$$\sum_{n=1}^\infty\frac{\sin(n)}n=\frac{\pi-1}2$$

The solution involves realizing $\sin(n)=\Im e^{in}$ and the Taylor expansion for the natural logarithm.

While thats great and all, how can I prove this using only real numbers?

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    $\begingroup$ Find the (real) Fourier series of the function $x \mapsto \frac{\pi- x}{2}$ on $[0,2\pi]$. $\endgroup$ – Daniel Fischer Nov 26 '16 at 15:26
  • $\begingroup$ @DanielFischer If you could do that that'd be great $\endgroup$ – Simply Beautiful Art Nov 26 '16 at 15:30
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    $\begingroup$ How about this? $\endgroup$ – Daniel Fischer Nov 26 '16 at 15:33
  • $\begingroup$ @DanielFischer Oh, thanks for the nice find. $\endgroup$ – Simply Beautiful Art Nov 26 '16 at 15:37
  • $\begingroup$ I thought Fourier transformation involve Complex numbers. $\endgroup$ – Zaid Alyafeai Sep 4 '17 at 4:35
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As suggested in comments, lets use fourier series. =). From here we have the fourier series of $x$, valid in the range $[-\pi, \pi]$: $$ x = -2\sum_{n=1}^\infty\frac{(-1)^{n}}{n}\sin(nx) $$

If we insert: $x=\pi-1$, it will elliminate the $(-1)^n$ from the formula. $$ \sin(nx) = \sin(n\pi - n) = \sin(n\pi)\cos(n)-\cos(n\pi)\sin(n) = -(-1)^n\sin(n) $$

Then: $$ \pi-1 = 2\sum_{n=1}^\infty\frac{1}{n}\sin(n) \quad\implies\quad \sum_{n=1}^\infty\frac{\sin(n)}{n} = \frac{\pi-1}{2} $$

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