How to prove that $\left|\sum_{k=0}^{n-1}\sin(2k+1)x\right|$ is bounded

Question

I'm currently studying Fourier series using a textbook which unfortunately provides only partial solutions to its explanations. To show the uniform convergence of a Fourier series I need to prove the assumption from the title above. Any help is very much appreciated.

Answer 1

You can calculate this sum explicitly:

\begin{align} \sum_{k=0}^{n-1}\sin(2k+1)x &= \operatorname{Im}\left(\sum_{k=0}^{n-1}e^{(2k+1)ix}\right)\\ &= \operatorname{Im}\left(e^{ix}\sum_{k=0}^{n-1}\left(e^{2ix}\right)^{k}\right)\\ &= \operatorname{Im}\left(e^{ix}\frac{e^{2nix}-1}{e^{2ix}-1}\right)\\ &= \operatorname{Im}\left(\frac{e^{2nix}-1}{e^{ix}-e^{-ix}}\right)\\ &= \operatorname{Im}\left(\frac{\cos(2nix)-1+i\sin(2\pi ix)}{2i\sin x}\right)\\ &= \frac{\sin(2nx)}{2\sin x} \end{align}

Now it should be clear that it is bounded.