Proving that $f_n(x)=\sin^{n}(x)$ is uniformly convergent.

Question

Consider $$f_n\left(x\right)\:=\:\sin ^n\left(x\right)$$ How to prove that this sequence is uniformly convergent in $\left[0,b\right]\: \text{for}\:\frac{\pi }{2}>b>0$?

Answer 1

The function $x\mapsto \sin x$ is increasing on the interval $\left[0,\frac\pi2\right]$ so $$\left|\sin^n(x)\right|\le \sin ^nb\xrightarrow{n\to\infty}0,\quad\forall x\in[0,b]$$ so the sequence is uniformly convergent on $[0,b]$.

Answer 2

For all $x\in[0,b]$, $$ |f_n(x)|\le\sin^n(b) $$ and since $0\le\sin(b)\lt1$, we have that $f_n(x)\to0$ uniformly; that is, given an $\epsilon\gt0$, we can find an $N$ so that for $n\ge N$, $|f_n(x)|\le\epsilon$ independent of $x\in[0,b]$. The following will do

$$N=\left\lceil\frac{\log(\epsilon)}{\log(\sin(b))}\right\rceil$$