Series of Functions - Pointwise and Uniform Convergence. I was hoping for some help for the following questions.
Prove that the series $\sum_{n=1}^\infty x^n(1-x)$ converges pointwise but not uniformly on $[0,1]$.
Prove that the series $\sum_{n=1}^\infty (-1)^nx^n(1-x)$ converges uniformly on $[0,1]$.

$f(x)=\sum_{n=1}^\infty  x^n(1-x)=x-x^2+x^2-x^3+x^3-x^4+$..
Note that $f(x)=x $ except $f(1)=0$ and $|S_N-x|=x^{N+1}$ which has supremum $1$ on $[0,1)$ 
for all $N$.So we don't have uniform convergence .[Pointwise convergence follows from 
$x^{N+1} \to 0$ as $ N\to \infty$]
Or you may use the fact that uniform convergence of cont. functions is cont. 
$g(x)=\sum_{n=1}^\infty (-1)^nx^n(1-x)$
$=-x+x^2+x^2-x^3+x^4-x^3+x^4-x^5..$
$= -x+2x^2( 1-x+x^2-x^3+...) =-x+2x^2\frac1 {1+x}   = \frac {x^2-x}{1+x}               $
We again calculate $|S_{2N}-g(x)|=[-x+2x^2-2x^3+...+2x^{2N}-x^{2N+1} ]  -  \frac {x^2-x}{1+x}   =           $
$=\frac {x^{2N+1}-x^{2N+2}}{1+x} <  min (\frac {1-x}{1+x} , x^{2N+1})$ and  supremum of this function approaches $0$ as  $N \to \infty .  $    
[In view of $\lim_{x\to 1} \frac {1-x}{1+x}=0  $  and $x^k $ is uniformly convergent to $0$ on $[0,a]  $ for $a<1  $]   
Similarly , $|S_{2N+1}-g(x)|$  approaches $0$ as $N \to \infty$.
So we have uniform convergence to $g(x)$ on $[0,1]$.
