TITLE: Uniform convergence of geometric series
QUESTION [18 upvotes]: How do I show that the geometric series $\sum_{k=0}^\infty x^k$ converges uniformly on any interval $[a,b]$ for $-1 < a < b < 1$?
The Cauchy test says that $\sum_{k=0}^\infty x^k$ converges uniformly if, for every $\varepsilon>0$, there exists a natural number $N$ so that for any $m,n>N$ and all $x\in[a,b]$, $|\sum_{k=0}^m x^k - \sum_{k=0}^n x^k|<\varepsilon$.
The rightmost condition simplifies to $|\sum_{k=m}^n x^k|<\varepsilon$, but I don't see where to go from there. I realize this is probably a straightforward application of definitions, but I'm really lost here.

REPLY [2 votes]: For every $x$ such that $|x|\lt1$, $\sum\limits_{k=n}^{+\infty}x^k=\frac{x^n}{1-x}$ and $\left|\frac{x^n}{1-x}\right|\leqslant\frac{|x|^n}{1-|x|}$. Hence,
$$
\sup\limits_{x\in[a,b]}\,\left|\sum\limits_{k=n}^{+\infty}x^k\right|\leqslant\frac{r^n}{1-r}\underset{n\to\infty}{\longrightarrow}0,
$$
with $r=\max\{|a|,|b|\}\lt1$, which proves the uniform convergence on $[a,b]$.