Uniform convergence and boundedness of $\sum_{k=1}^n \frac{\sin kx}{k}$ Why is $\sum_{k=1}^n \frac{\sin kx}{k}$ uniformly convergent on $[\delta,\pi]$ for all $0<\delta<\pi$ from this nice property of Fourier series? 
Furthermore, why is the partial sum $\sum_{k=1}^n \frac{\sin kx}{k}$ uniformly bounded for all $n$ and $x \in (0,\pi)$ as discussed in this question.
 A: Note that for $x > 0$ we have $|\sin (nx)| \leqslant |nx| = nx$ and 
$$\left|\sum_{k=1}^M\frac{\sin (nx)}{n}\right| \leqslant \sum_{k=1}^M\frac{|\sin (nx)|}{n} \leqslant  \sum_{k=1}^M \frac{nx}{n} = Mx,$$
and since $M = \min(N, \lfloor \frac{\pi}{x}\rfloor)\leqslant \lfloor \frac{\pi}{x}\rfloor \leqslant \frac{\pi}{x}$ we get
$$\left|\sum_{k=1}^M\frac{\sin (nx)}{n}\right| \leqslant \frac{\pi}{x} x = \pi$$
For the second sum in the linked question it was shown that
$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| \leqslant \frac{1}{|\sin(x/2)|}\left(\frac{1}{N} + \frac{1}{M+1} + \sum_{n=M+1}^{N-1}\left(\frac{1}{n} - \frac{1}{n+1} \right) \right) = \frac{2}{(M+1)|\sin(x/2)|}$$
For $z \in (0,\pi/2)$ we have the inequality $\frac{2z}{\pi} \leqslant \sin z < z$ and, thus, for $x \in (0,\pi)$ we have $\sin(x/2) \geqslant x /\pi$. Hence, 
$$\frac{2}{(M+1)|\sin(x/2)|} < \frac{2\pi}{(M+1)x}$$
In bounding this sum we can assume that $M < N$ (otherwise there are no terms) and, hence, $M = \lfloor \frac{\pi}{x} \rfloor$ implying that $M \leqslant \frac{\pi}{x} < M+1$ and $\frac{1}{(M+1)x} < \frac{1}{\pi}$.
Therefore,
$$\left|\sum_{n=M+1}^N \frac{\sin nx}{n} \right| \leqslant\frac{2}{(M+1)|\sin(x/2)|} < \frac{2\pi}{(M+1)x}< \frac{2\pi}{\pi} = 2$$
