# 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 $\mathbb{R}$?

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. – Ofir 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
For any $$n\geq 1$$, we know where the stationary points of $$F_n(x)$$ occur since $$F_n'(x)$$ has a simple closed form.
It follows that $$\sup_{x\in\mathbb{R}} |F_n(x)| = \sum_{k=1}^{n}\frac{1}{k}\sin\left(\frac{2\pi k}{2n+1}\right)$$ and over $$[0,\pi]$$ we have $$\sin(x)\leq \frac{4}{\pi^2}x(\pi-x)$$ by concavity, therefore $$\sup_{x\in\mathbb{R}} |F_n(x)| \leq \frac{8n^2}{(2n+1)^2} <2.$$ We may also prove that the sequence $$A_n = \sum_{k=1}^{n}\frac{1}{k}\sin\left(\frac{2\pi k}{2n+1}\right)$$ is increasing and convergent to $$\int_{0}^{\pi}\frac{\sin x}{x}\,dx = \text{Si}(\pi) \approx 1.85194.$$ Ultimately we may check that $$\sum_{k\geq 1}\frac{\sin(kx)}{k}$$ is the Fourier series of the sawtooth wave, i.e. the $$2\pi$$-periodic extension of $$\frac{\pi-x}{2}$$ defined over $$(0,2\pi)$$. This is enough to ensure convergence in $$L^2$$, plus we have a uniform bound for $$|F_n(x)|$$.