Given the functional series $$\sum_{k\geq 0} f_k,\qquad\left(f_k(x) :=\sin^k(x)\right),$$ on what interval(s) $I\subset\mathbb{R}$ does this converge uniformly to a limit function $g$?
Working this out, we start with some notation: $$f_n = \sin^n(x),\qquad s_n(x) = \sum_{i\geq 0}^{n}f_i(x).$$ Note how this is a geometric series when $x$ is constant, and the infinite geometric series converges if and only if $|\sin(x)|<1$. This means that the series converges pointwise to $$\frac1{1-\sin(x)}$$ for all $x\neq k\pi +\frac12 \pi$, $\;\;k = 1, 2, 3, ...$, let's call this set $E:=\{x\in\mathbb{R}|x\neq k\pi +\frac12 \pi\}$.
But we also know that if $|f_n(x)| \leq M_n, \quad (x\in E,\; n=1, 2, 3,...)$, then $\sum f_n$ converges uniformly on $E$ when $\sum M_n$ converges.
Now this is where I run in to trouble. Because I don't know how I would define the intervals of convergence. Of course I can get arbitrarily close to the excluded $x\in \mathbb{R}$, say $x=\frac12\pi+\epsilon$ and this would converge uniformly for all $x\in[\frac12\pi+\epsilon, \frac32\pi-\epsilon]$ since$f_n(x)$ is bounded by $M_n=\sin^n(\frac12\pi+\epsilon)<1$. But by the denseness of $\mathbb{Q}$ there is some other $x$ for which $f_n$ is not bounded by $M_n$. This means that if $E$ is defined as above, $$M_n = \sup_{x\in E}\left|f_n(x)\right|$$ does not converge to $0$ as $n\rightarrow\infty$, for $M_n = 1$.
Basically, I'm confused. Can anyone help me out?