Uniform convergence of geometric series

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.

Let $c = \max(|a|, |b|)$. We have:

$$ \left|\sum_{k=n}^{m}x^k\right| < \sum_{k=n}^{m}c^k $$

Since $\sum_{k=0}^{\infty}c^k$ converges ($|c| < 1$), we can make $\sum_{k=n}^{m}c^k$ as small as we want:

$$ \left|\sum_{k=n}^{m}x^k\right| < \sum_{k=n}^{m}c^k < \epsilon $$

Now per the Cauchy criterion, we have uniform convergence.

The Cauchy criterion says that a sequence of functions converges uniformly if and only if:

$$ \forall \epsilon > 0, \exists N\in\mathbb{N}:\forall n, m > N, \, |f_m(x) - f_n(x)| < \epsilon $$

For series, $|f_m(x) - f_n(x)|$ becomes $\left|\sum_{k=n}^{m}f_k(x)\right|$.

Given $m, n \in \mathbb{N}$, with $m>n$ we have for every $x \in \mathbb{R}\setminus\{-1,1\}$: $$ \left|\sum_{k=0}^nx^k-\sum_{k=0}^mx^k\right|\le \sum_{k=n+1}^m|x|^k=|x|\frac{|x|^n-|x|^m}{1-|x|}. $$ Let $a,b \in \mathbb{R}$ such that $[a,b] \subset (-1,1)$. Then for every $x \in [a,b]$ we have $$ \left|\sum_{k=0}^nx^k-\sum_{k=0}^mx^k\right|\le q\frac{q^n-q^m}{1-q} \le \frac{q^n}{1-q}, $$ where $q=\max\{|a|,|b|\}$. Let $\varepsilon>0$, and let $N$ be the smallest $n \in \mathbb{N}$ such that $$ \frac{q^n}{1-q}<\varepsilon, $$ i.e. $$ n>\frac{\ln((1-q)\varepsilon)}{\ln q}. $$ One may choose, e.g., $$ N=\lfloor\frac{\ln((1-q)\varepsilon)}{\ln q}\rfloor+1. $$ Then for every $m>n>N$ we have $$ \left|\sum_{k=0}^nx^k-\sum_{k=0}^mx^k\right|<\varepsilon. $$

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 $[a,b]$.