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 $f(x)$. 