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

Question

Possible Duplicate:
Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$

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

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

What's $[n]$ suppose to be? perhaps the question has a typo? — Jean-Sébastien, Nov 02 '12 at 19:05
@BabakSorouh: But floor of a natural would be quite redundant. — Hagen von Eitzen, Nov 02 '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 02 '12 at 19:16
Also look here http://math.stackexchange.com/questions/161960/sum-inequality-sum-k-1n-frac-sin-kk-le-pi-1/183471#183471 — , Nov 02 '12 at 19:25
This is not a duplicate. I need a proof acceptable at calculus 2 student's level. — Kerry, Nov 02 '12 at 22:05

score 0 · Answer 1 · answered Nov 02 '12 at 19:58

0

$\sum\limits^{\infty}_{n=1}\dfrac{\sin{n\theta}}{n}$ 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.

answered Nov 02 '12 at 19:58

M. Strochyk

8,397

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

1 Answers1

Linked