# How to prove $\sum\limits_{n=1}^\infty\frac{\sin(n)}n=\frac{\pi-1}2$ using only real numbers.

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?

• Find the (real) Fourier series of the function $x \mapsto \frac{\pi- x}{2}$ on $[0,2\pi]$. – Daniel Fischer Nov 26 '16 at 15:26
• @DanielFischer If you could do that that'd be great – Simply Beautiful Art Nov 26 '16 at 15:30
• How about this? – Daniel Fischer Nov 26 '16 at 15:33
• @DanielFischer Oh, thanks for the nice find. – Simply Beautiful Art Nov 26 '16 at 15:37
• I thought Fourier transformation involve Complex numbers. – Zaid Alyafeai Sep 4 '17 at 4:35

## 1 Answer

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