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

Here is an exercise, on analysis which i am stuck.

• How do I prove that if $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$, then the sequence $\{F_{n}(x)\}$ is boundedly convergent on $\mathbb{R}$?
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First, let's note that for $x\in(0,2\pi)$ $$\sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\int_{0}^{x}\sum\limits_{k=1}^{n}\cos kt\ dt= -\frac{x}{2}+\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{2\sin{\frac{t}{2}}}\ dt$$ $$=-\frac{x}{2}+\int_{0}^{x}\left(\frac{1}{2\sin{\frac{t}{2}}}-\frac{1}{t}\right)\sin \frac{(2n+1)t}{2}\ dt +\int_{0}^{x}\frac{\sin \frac{(2n+1)t}{2}}{t}dt.$$ Now, the first integral at the right-hand side tends to zero by the Riemann-Lebesgue lemma. The second one is equal to (via a substitution $s=(2n+1)t/2$) the integral $$\int_{0}^{\frac{(2n+1)x}{2}}\frac{\sin s}{s}\ ds\to\int_{0}^{\infty}\frac{\sin s}{s}\ ds=\frac{\pi}{2}.$$ Therefore $$\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=f(x),\qquad x\in(0,2\pi).$$ The series converges on $\mathbb R$ to the periodic extension of $f(x)$.
if you take $x=0$ then the left side is a summation of zeros and the right isn't zero. – Prometheus Dec 8 '10 at 6:51
The argument works for $x\neq 2\pi m$. – Andrey Rekalo Dec 8 '10 at 7:09
@Prometheus: At that point the series converges to $(\lim_{x\to0,x>0}f(x)+\lim_{x\to0,x<0}f(x))/2=$ $(\lim_{x\to0,x>0}f(x)+\lim_{x\to2\pi,x<2\pi}f(x))/2 =$ $(\frac{\pi}{2}+\frac{\pi-2\pi}{2})/2=0$. – AD. Dec 8 '10 at 11:41