# 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.

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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 $\left|\sum_{k=n}^{m}f_k(x)\right|$.

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$b$ may be negative, so $\sum_{k=n}^m b^k$ could be negative and thus less than $\left|\sum_{k=n}^m x^k\right|$ –  Orange Sep 26 '12 at 20:25
Did you mean $\max(|a|,|b|)$ in place of $b$? –  Orange Sep 26 '12 at 20:28
@Orange $\max(|a|, |b|)$ would do. Let me fix my answer. –  Ayman Hourieh Sep 26 '12 at 20:29
Does this all hold true for complex domains? –  ellya Apr 19 '14 at 16:56
@ellya Yes if you assume $|z| < |a| < 1$ for some $a$. –  Ayman Hourieh Apr 19 '14 at 18:13

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 $$\left|\sum_{k=0}^nx^k-\sum_{k=0}^mx^k\right|<\varepsilon.$$

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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 $[a,b]$.

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You seem to be doing fine, your way will work. The sum from $m$ to $n$ is a finite geometric series with first term $x^m$ and common ratio $x$. Its sum is equal to $$\frac{x^m(1-x^{n+-m+1})}{1-x}.\tag{1}$$ Now it is just a matter of making $|x^m|$ small. How small? Note that $1-x\gt 1-b$ and $0\lt 1-x^{n+1-m}\lt 2$.

Let $c=\max(|a|,|b|)$, and let $c=\frac{1}{1+d}$. If $m\ge 1$, then by the Binomial Theorem, or more simply by the Bernoulli Inequality, we have $(1+d)^m \ge 1+dm$.

So if $|x|\le c$, we get $|x^m|\lt \frac{1}{1+dm}$. It now should not be hard to find appropriate $m$.

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