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

Question

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 $\mathbb{R}$?

Answer 1

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 $f(x)$.

Answer 2

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 $|F_n(x)|$.

Answer 3

Here is an intuitive and easy-to-think method: $$ \begin{aligned} f(x):=\sum_{n=1}^{+\infty} \frac{\sin n x}{n} & =\sum_{n=1}^{+\infty} \frac{\exp (\mathrm{i} n x)-\exp (-\mathrm{i} n x)}{2 \mathrm{i} n} \\ & =\frac{1}{2 \mathrm{i}} \sum_{n=1}^{+\infty} \frac{\left(\mathrm{e}^{\mathrm{i} x}\right)^n-\left(\mathrm{e}^{-\mathrm{i} x}\right)^n}{n} \\ & =\frac{1}{2 \mathrm{i}}\left(-\ln \left(1-\mathrm{e}^{\mathrm{i} x}\right)+\ln \left(1-\mathrm{e}^{-\mathrm{i} x}\right)\right) \\ & =\frac{1}{2 \mathrm{i}} \ln \frac{1-\mathrm{e}^{-\mathrm{i} x}}{1-\mathrm{e}^{\mathrm{i} x}} \\ & =\frac{1}{2 \mathrm{i}} \ln \left(-\mathrm{e}^{-\mathrm{i} x}\right)=\frac{1}{2 \mathrm{i}} \ln \left(\mathrm{e}^{\mathrm{i}(\pi-x)}\right) \\ & =\frac{\pi-x}{2} \end{aligned} $$ The last step here should involve the knowledge of the multi-valued property of complex functions. I may not be strict in writing it this way, but I can get the correct result. One of my imagined solutions is to determine its single value based on $f(\pi/2)=1-1/3+1/5+\cdots=\pi/4$.