Om which interval $I\in \mathbb{R}$ does $\sum_{k\geq 0} \sin^{k}(x)$ uniformly converge?

Question

I know that $$\sum_{k\geq 0} \sin^{k}(x)=\sum_{k= 0}^{\infty} \sin^{k}(x)=\lim_{n\rightarrow\infty}\sum_{k=0}^{n} \sin^{k}(x)=\lim_{n\rightarrow\infty}\frac{1-(\sin(x))^{n+1}}{1-\sin(x)}=\frac{1}{1-\sin(x)}$$ when $|\sin(x)|<1$, thus when $|x|\neq \cfrac{\pi}{2}$.
I now want to show that $\sum_{k\geq 0} \sin^{k}(x)$ is uniformly convergent thus that:
$$\lim_{n\rightarrow \infty}\|\sum_{k=0}^{n} \sin^{k}(x)-\frac{1}{1-\sin(x)}\|=0\Leftrightarrow \lim_{n\rightarrow \infty}\|\frac{1-(\sin(x))^{n+1}}{1-\sin(x)}-\frac{1}{1-\sin(x)}\|=0\Leftrightarrow $$$$ \lim_{n\rightarrow \infty}\|\frac{-(\sin(x))^{n+1}}{1-\sin(x)}\|=0\Leftrightarrow\lim_{n\rightarrow \infty}\sup\{|\frac{(\sin(x))^{n+1}}{1-\sin(x)}|:x\in[0,2\pi]\}=0$$
I don't really know where to go from here I know that $x$ can't be $\pi/2$ or $-\pi/2$. I don't really know what the supremum of this function is. Can anyone help me with this?

Answer 1

Weirstrass M-test is useful for series, particularly geometric ones.

In your case, you can take $$ x\in (-\pi/2+\epsilon,\pi/2-\epsilon) $$ which then yields the inequality $$ |\sin^k(x)|\leq \sin^k(\pi/2-\epsilon) $$ with $$ \sum_{k\in \mathbb{N}}(\sin(\pi/2-\epsilon)^k=\frac{1}{1-\sin(\pi/2-\epsilon)} $$ as a geometric series with common ratio smaller than 1.