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 $[0,1]$.

• Both series of functions converge uniformly on $[0,1]$ to $0$. – Pedro Tamaroff Apr 19 '13 at 18:10
• I think he means the series whose terms are given. @Dominic, these are geometric series whose sums you can determine explicitly. – Umberto P. Apr 19 '13 at 18:14
• – Martin Sleziak Apr 25 '17 at 8:17

$f(x)=\sum_{n=1}^\infty x^n(1-x)=x-x^2+x^2-x^3+x^3-x^4+$..

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 $[0,1]$.

• Thank you so very much guys. I wish I had 15+ points so I could upvote your answer. – Dominic Jake King Lin Apr 20 '13 at 1:16