I want to determine in what interval this series has uniform convergence
$$\sum_{n=0}^{\infty}(1-x)^n $$
Finding the radius of convergence is easy, since $|1-x|<1$
$$-1<1-x<1 $$ $$0<x<2 $$
The convergence radius is therefore $\rho=1$. To determine the range of uniform convergence, I use this proposition from my text book.
Proposition 5.30 Let $\sum_{n=0}^\infty c_nx^n$ be a power series with radius of convergence $\rho>0$. Then $\sum_{n=0}^\infty c_nx^n$ is uniformly convergent on any interval of the form $[-r,r]$, where $r\in]0,\rho[$.
It tells me to choose an $r$ from the interval, $]0,1[$. Then the series is convergent in $[-r,r]$, but I don't see how this can be true.
Say I choose $r=0.5$, then the series is unformly convergent in $[-0.5,0.5]$.
But now the series doesn't even converge, since $|1+0.5|>1$, and it becomes divergent.
Origin of this question
This question was originally a multiple choice question, asking in what interval $\sum_{n=0}^{\infty}(1-x)^n $ is uniformly convergent. There were, among others, these two choices
- $\frac{1}{4} \leq x \leq \frac{3}{4}$
- $0<x<1$
It turns out that the correct answer is $\frac{1}{4} \leq x \leq \frac{3}{4}$, but I don't see why the second choice isn't correct. We know that $x$ can come arbitrarily close to $0$, and still ensure uniform convergence. We have also seen (from comments on John's answer) that $0.1 \leq \leq 1.9 x$ ensures uniform convergence. So why can't we say that $0<x<1$ ensures uniform convergence?
Your series does not have the form indicated in the theorem -- the theorem involves powers of a single variable; your series involves powers of $(1-x)$.
If you let $y = 1-x$, then you can apply the theorem to the series $$ \sum y^n $$ to conclude an interval of uniform convergence in $y$; you can then use this to say for which $x$ the original series converges uniformly.