Is $\sum_{k=1}^n \sin((2k+1)x)$ uniformly bounded on $\mathbb{R}$?

Question

I want to show that $$\sum_{k=1}^n \sin((2k+1)x)=\frac{1-\cos(2nx)}{2\sin x}$$ is uniformly bounded on $\mathbb{R}$.

Since $$\left\vert \frac{1-\cos(2nx)}{2\sin x} \right\vert \leq \left\vert \frac{1}{\sin x}\right\vert$$ it is uniformly bounded on any interval $[a,b]$ containing no $m\pi$ $(m \in \mathbb{Z})$.
But I have no idea whether the series is uniformly bounded on $\mathbb{R}$.
Intuitively, I think the answer is NO.
Is there any nice proof to show this?
Or my guess is wrong, so that it is uniformly bounded on $\mathbb{R}$?
Then how should I treat the points of $x=m\pi$?
Since I am just a beginner with this chapter, there are so many questions......
Would you help me, please?

Answer 1

Suppose $|\frac {1-\cos (2nx)} {2\sin x}| \leq M$ for all $n$ and $x$. Then $|\frac {2\sin^{2} {nx}} {2\sin x}| \leq M$ and letting $x \to \frac {\pi} {2n}$ leads to a contradiction for $n> \frac {M\pi} 2$ (since $\sin x \leq x$ ).