The series $\sum_{n=1}^\infty x(1-x)^n$ is not uniformly convergent on $[0,1]$ How can we show that $\sum_{n=1}^\infty x(1-x)^n$ is not uniformly convergent on $[0,1]$
I got the definition of uniform convergence  to $f$ for a set $E$ of values of $x$ if, for each $\epsilon>0$, there exists an integer $N$  such that
$$|f_n(x)-f(x)|<\epsilon$$ 
for every $x \in E$. 
***********Can someone prove by showing it's pointwise convergent?**********
 A: We have for the remainder with $x \in [0,1]$,
$$\sup_{x \in [0,1]}\sum_{k = n+1}^{\infty}x(1-x)^k \geqslant \sup_{x \in [0,1]}\sum_{k = n+1}^{2n}x(1-x)^k \geqslant \sup_{x \in [0,1]}nx(1 - x)^{n} \\ \geqslant \frac{n}{n}\left(1 - \frac{1}{n}\right)^{n} \to e^{-1} \neq 0$$
For the series to converge uniformly we must have
$$\lim_{n \to \infty} \sup_{x \in [0,1]} \sum_{k = n+1}^{\infty}x(1-x)^k = 0,$$
which is clearly not the case. 
A: 
The negation of uniform convergence of a sequence is given a sequence $f_n(x)$ that converges pointwise to $f(x)$, the convergence fails to be uniform for $x\in A$ if there exists an $\epsilon>0$ such that for all $N'$ there exists an $x\in A$ and a number $N>N'$ such that
$$|f_N(x)-f(x)|\ge \epsilon$$

For the given series, note that for $N\ge2$ and $x=1/N$
$$\left|\sum_{n=1}^N x(1-x)^n-(1-x)\right|=\left(1-\frac1N\right)^{N+1}\ge \frac18\tag 1$$
Therefore,  there exists an $\epsilon =\frac18 >0$ such that for all $N'\ge 1$, there exists a number $N>N'$ and there exists an $x=1/N \in[0,1]$ such that $(1)$ is true.
A: We will get the sum function,
$$S_n(x)=\sum_{i=1}^{n}[x(1-x)^i ]=x(\sum_{i=1}^{n}(1-x)^i ) $$
$$=x(\frac{1-x}{x})(1-(1-x)^{n+1})$$
$$=(1-x)(1-(1-x)^{n+1})$$
If $S_n(x)$ be uniformly convergent, then $\lim_{x\rightarrow x_0}(\lim_{n\rightarrow \infty}S_n(x))=\lim_{n\rightarrow \infty}(\lim_{x\rightarrow x_0}S_n(x))$, where $x_0$ is a limit point of $[0,1]$.
If $x_0=0$ then R.H.S=$1$ and L.H.S=$0$
Hence $S_n(x)$ is not uniformly convergent.
A: Write 
$$f_N (x)= \sum_{n=1}^N x (1-x)^n =x (1-x) \frac {1 - (1-x)^{N}}{x} = (1-x) - (1-x)^{N+1}.$$
(the last two equalities are derived assuming  $x\ne 0$, but the expression on the righthand side is correct also for $x=0$ because $f_N(0)=0$). 
Let's look at the pointwise limit: for $x\ne 0$, $f_N(x) \to 1-x$, and  $f_N(0)=0$. 
In other words,  the (point-wise) limit  $f$ is given by 
$$ f (x) = \begin{cases} 0 & x=0 \\ 1-x & x \in (0,1]\end{cases}$$ 
This function is not continuous, and this by itself implies that the convergence is not uniform (uniform limit of continuous functions on $[0,1]$ is continuous). 
However, we can also prove by definition. We use Bernoulli's inequality: 
$$(1+a)^n \ge 1+na,$$
for $n\in{\mathbb N}$ and $a>-1$. 
Take $x_N = \frac{1}{2(N+1)}$, we see that 
$$f_N (x_N) = (1-x_N) -(1 - \frac{1}{2(N+1)})^{N+1} \le 1-x_N - (1-\frac 12)=\frac 12 - X_N$$
(where the inequality was obtained by Bernoulli's inequality with $a=-\frac{1}{2(N+1)}$ and $n=N+1$). 
On the other hand, since $x_n\ne 0$, $f(x_N) = 1-x_N$. 
It follows then that for all $N$: 
$$ \sup_{x \in [0,1]} |  f(x)-f_N(x)| \ge |f(x_N) - f_N(x_N)| \ge 1-x_N - (1/2 - x_N)=1/2.$$ 
