Uniform convergence of $\sum_{n \ge 0} (-1)^n(1-x)x^n$ Let $f_n: [0,1] \to \mathbb{R}$, $f_n(x)=(-1)^n (1-x)x^n, \forall x\in [0,1], n \ge 0$. Prove that $\sum\limits_{n \ge 0} f_n(x)$ convergenes uniformly.
Let $S_n(x) = \sum\limits_{k = 0}^n f_k(x), \forall x \in [0,1], n \ge 0$.
Then, 
$S_n(x)=1-x-x+x^2+x^2-x^3-x^3+x^4+...+(-1)^{n-1}x^{n-1}+(-1)^nx^n+(-1)^nx^n+(-1)^{n+1}x^{n+1}$
so
$S_n(x)=1-2x+2x^2-...+(-1)^n\cdot 2x^n+(-1)^{n+1}x^{n+1}, \forall x\in [0,1], n \ge 0$.
We know that $1+x^{2p+1}=(1+x)(x^{2p}-x^{2p-1}+x^{2p-2}-....+x^2-x+1)$, so, for $n=2p$, $p \ge 0$, we may write:
$$ S_{2p}= 2(1-x+x^2-...+x^{2p})-1-x^{2p+1}=2 \cdot \frac{1+x^{2p+1}}{1+x}-1-x^{2p+1}$$
Now, it's easy to see that $\lim\limits_{p \to \infty}S_{2p}(x)=\frac{1-x}{1+x}, \forall x\in [0,1]$.
For $n=2p+1$, $p \ge 0$, we have:
$$ S_{2p+1}(x)=2(1-x+x^2-...+x^{2p+2})-1-x^{2p+2}=2\cdot \frac{1+x^{2p+3}}{1+x} -1 -x^{2p+2} $$
In this case we have: $\lim\limits_{p \to \infty} S_{2p+1}(x) = \frac{1-x}{1+x}, \forall x\in [0,1]$.
So, $\lim\limits_{n \to \infty} S_n(x) =\frac{1-x}{1+x}, \forall x\in [0,1]$.
Now, it remains to prove that $\lim\limits_{n\to \infty} \sup\limits_{x \in [0,1]} |S_n(x)-\frac{1-x}{1+x}|=0$.
It's easy to see that $g_n(x)=|S_n(x)-\frac{1-x}{1+x}|=x^{n+1} \cdot \frac{1-x}{1+x}$.
How can I continue?

Hints : with your notation, $g_n'(x) = \frac{ x^{n-1}(-nx^2+n-2x)}{(x+1)^2}$. The sign of $g_n'(x)$ on $[0,1]$ is the same as the sign of $-nx^2+n-2x$ and consequently is non-negative on $\left[0, \frac{\sqrt{n^2+1}-1}{n}\right]$ and non-positive on $\left[\frac{\sqrt{n^2+1}-1}{n},1\right]$. Consequently, the maximum of $g_n$ on $[0,1]$ is reached at $x= \frac{\sqrt{n^2+1}-1}{n}$. A study of the convergence of $g_n\left( \frac{\sqrt{n^2+1}-1}{n} \right)$ should lead to a solution.
