Prove that $\sum x^n (1-{2x \over \pi})^n \tan^{n} x$ converges uniformly. $\forall \alpha:0<\alpha<{\pi \over 2}$.
Prove:$$\sum_{n=1}^{\infty} x^n \left(1-{2x \over \pi}\right)^n \tan^{n} x$$ converges uniformly in $[0,\alpha]$.
It's easy to get $$S_n(x)=\sum_{k=1}^n x^k \left(1-{2x \over \pi}\right)^k \tan^{k} x={x\left(1-{2x \over \pi} \right)\tan x\left[1- x^n \left(1-{2x \over \pi}\right)^n \tan^{n} x\right] \over 1-x\left(1-{2x \over \pi}\right)\tan x}$$.
But I can't figure out $S(x)$.

Hint
Use that $$f(x)=x \left(1-{2x \over \pi}\right) \tan x$$
is nonnegative and increasing in $[0,\alpha]$ and that $f(\alpha)<1$ (where $\alpha<\pi/2).$
Thus
$$\left|x^n \left(1-{2x \over \pi}\right)^n \tan^{n} x\right|=\left|x \left(1-{2x \over \pi}\right) \tan x\right|^n\le (f(\alpha))^n.$$
