Boundedness of $\sum_{k=1}^n (-1)^{k-1}\sin(kx)$ How do I show that the series $\sum\limits_{k=1}^n (-1)^{k-1}\sin(kx)$ is bounded?
In other words:

Show that there exists a number $B$ such that $\left|\sum\limits_{k=1}^n (-1)^{k-1}\sin(kx)\right|\leq B$ for all $n\in\mathbb{N}$ and all $x\in I$, where $I$ is a compact interval of $\mathbb{R}$.


In the case of $\sum\limits_{k=1}^n\sin(kx)$ we can find a telescope sum which delivers a closed form but the $(-1)^{k-1}$ term prevents us from applying this telescope sum technique to this particular series. Maybe there is another approach?
 A: $(-1)^k\sin(kx)$ is the imaginary part of $(-1)^k e^{kxi} = (-e^{xi})^k$, thus it's enough to show $|\sum_{k=1}^n (-e^{xi})^k| = |-e^{xi}\frac{1-(-e^{xi})^n}{1+e^{xi}}|= |\frac{1-(-e^{xi})^n}{1+e^{xi}}|$ is bounded.
Update: We can bound the above uniformly when $x$ is bounded away from $\pi$ (or rather $(2k+1)\pi$), so that $1+e^{xi}$ is bounded away from $0$.
I thought it sould be easy to bound the function around $\pi$, as the function is $0$ around $\pi$. It just occurred to me that this is actually false!
First we can establish the Fourier series expansion $\frac{x}{2} = \sum_{k=1}^\infty \frac{(-1)^{k-1}\sin(kx)}{k}$ over the interval $[-\pi, \pi]$. By Dirichlet's theorem for the convergence of Fourier series, the equality holds for any $x\in (-\pi, \pi)$, while $x=\pi$, the series is obviously $0$. That is, we know the function defined by the series is not continuous over $I = [-\pi, \pi]$. If $\sum_{k=1}^\infty (-1)^{k-1}\sin(kx)$ is uniformly bounded over $I$, then by Dirichlet's test, $\sum_{k=1}^\infty \frac{(-1)^{k-1}\sin(kx)}{k}$ converges uniformly on $I$, thus must be continuous.
So the correct statement should be

For any compact subset $I$ of $\mathbb{R}$ that doesn't include any odd multiple of $\pi$, there exists a number $B_I$ such that $\left|\sum\limits_{k=1}^n (-1)^{k-1}\sin(kx)\right|\leq B_I$ for all $n\in\mathbb{N}$ and all $x\in I$.


