Prove $\sum_{n = 1}^{\infty} 2^{-n} x^n$ does not converge uniformly on $(-2, 2)$

Question

How can one go about proving this? (I understand that the said series does converge uniformly on all $[-a, a]$ where $0 \leq a < 2$.)

I am especially interested in knowing if there is a way to prove this using the definition of uniformly Cauchy series.

Answer 1

If the given series converges uniformly on $(-2,2)$ then it converges uniformly on $[-2,2]$ since the functions $f_n\colon x\mapsto 2^{-n}x^n$ are continuous on $[-2,2]$ and $$\sup_{x \in (-2,2)} \lvert S_k(x) - S_m(x)\rvert = \sup_{x\in [-2,2]} \lvert S_k(x) - S_m(x)\rvert$$ where $$S_k(x)=\sum_{n=1}^kf_n(x)$$ and it's clear that the series doesn't converge for $x=2$ or $x=-2$.

Answer 2

Another idea, if it converges at some point $x$ the limit is

$$ f(x)=\frac{1}{1-x/2} $$

Consider $x \in (1,2)$. You need to find a sequence $x_n$ converging to $2$ such that $f_n(x_n)$ is bounded away from zero for large enough n. An example is

$$ x_n=2\left(1-\frac1n\right) $$

Then

$$ f_n(x_n)=\left(1-\frac1n\right)^n \to \frac1e $$

disproving uniform convergence in $(1,2]$. Similarly for $x \to -2$.