Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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

share|cite|improve this answer
$b$ 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 $$ \left|\sum_{k=0}^nx^k-\sum_{k=0}^mx^k\right|<\varepsilon. $$

share|cite|improve this answer

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 $[a,b]$.

share|cite|improve this answer

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 $m$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.