# Why $\sum^{\infty}_{n=1}\frac{\sin[n]}{n}=\frac{1}{2}(\pi-1)$? [duplicate]

From Stewart, we cannot find a calculus 2 easy way to prove this:

$$\sum^{\infty}_{n=1}\frac{\sin[n]}{n}=\frac{1}{2}(\pi-1)$$

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## marked as duplicate by Marvis, Austin Mohr, TMM, Cameron Buie, ArkamisNov 2 '12 at 21:02

What's $[n]$ suppose to be? perhaps the question has a typo? –  Jean-Sébastien Nov 2 '12 at 19:05
Kerry just means $\sin(n)$ –  user17762 Nov 2 '12 at 19:12
Yeah floor or $n$ is just $n$, but this may just be it because sum of $\sin(n)/n$ is $1/2(\pi-1)$ –  Jean-Sébastien Nov 2 '12 at 19:16
$\sum\limits^{\infty}_{n=1}\dfrac{\sin{n\theta}}{n}$ is the Fourier series for $f(\theta)=\dfrac{\pi-\theta}{2}, \;\; 0<\theta<2\pi,$ and converges uniformly on every closed interval $[\alpha, \beta], \;\; 0<\alpha<\beta<2\pi$ by Dirichlet's test.