Proof that the sum $\sum _{n=1} ^{\infty} (-1)^n \sin (nx)$ is bounded

Question

How can I prove that there is some constant $M>0$, such that for all $N\in\mathbb{N}$ and $x\in [0,\pi]$, $$\left|\sum _{n=1} ^{N} (-1)^n \sin (nx)\right| < M\text{?}$$

Answer 1

Clearly $$ (-1)^n \sin nx = \mathrm{Im}\, \mathrm{e}^{nxi+n\pi i}, $$ and thus, for $k\ne m\pi$, $$ \sum_{k=1}^n(-1)^k \sin kx = \mathrm{Im}\, \sum_{k=1}^n\mathrm{e}^{k(x+\pi) i}= \mathrm{Im}\,\frac{\mathrm{e}^{(n+1)(x+\pi) i}-1}{\mathrm{e}^{(x+\pi) i}-1}, $$ and hence $$ \left|\sum_{k=1}^n(-1)^k \sin kx\,\right| \le \left|\frac{\mathrm{e}^{(n+1)(x+\pi) i}-1}{\mathrm{e}^{(x+\pi) i}-1}\,\right|\le \frac{2}{2|\cos (x/2)|}, $$ since $$ \mathrm{e}^{(x+\pi) i}-1=\mathrm{e}^{(x+\pi) i/2}\left(\mathrm{e}^{(x+\pi) i/2}-\mathrm{e}^{-(x+\pi) i/2}\right)=2i\sin((x+\pi)/2)\mathrm{e}^{(x+\pi) i/2}-2i\cos(x/2)\mathrm{e}^{(x+\pi)}. $$ Also, if $x=m\pi$, then this sum vanishes.

Note. The sum is bounded for every $x$, but it is not uniformly bounded.