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

Let $(x_n)_{n=1}^{\infty}$ be a sequence such that $|x_{n+1} - x_n| < r^n$ for all $n \geq 1$, for some $0 < r < 1$. Prove that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence.

I understand that a Cauchy sequence means that for all $\varepsilon > 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. But this one is really giving me a headache.

I tried doing something like: let $ m > n$. Therefore $x_n - x_m$ = $(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m) $ and then somehow using the triangle inequality to compute some sum such that $x_n - x_m$ < sum which would be epsilon?

any help is appreciated, thank you.

share|cite|improve this question
If you can show that the sequence is convergent, then that is synonymous with a Cauchy sequence. – George1811 Mar 4 '14 at 16:35
You can bound $|x_m-x_n|$ by a geometric sum. – Martín-Blas Pérez Pinilla Mar 4 '14 at 16:38
up vote 5 down vote accepted

For every $\epsilon>0$, take a natural number $N$ such that $r^N <(1-r)\epsilon$, for example by taking $N=\lfloor\frac{\ln (1-r)\epsilon}{\ln r}\rfloor+1$. Then, for all $m,n\geq N$, assume $m<n$, we have \begin{align} |x_n - x_m|&=|(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m)|\\ &\leq |(x_n - x_{n-1})| + |(x_{n-1} - x_{n-2})| + ... + |(x_{m+1} - x_m)|\\ &< r^{n-1}+\dots+r^m\\ &=r^m(1+r+r^2+\dots+r^{n-m-1})\\ &<\frac{r^m}{1-r}\\ &\leq\frac{r^N}{1-r}\\ &<\epsilon \end{align}

share|cite|improve this answer

You are definitely on the right track. Here's what you want to do: suppose without loss of generality that $m>n$ then

$$\begin{align}|x_m-x_n| &= |x_m-x_{m-1}+x_{m-1}-\cdots-x_n| \\ &\le |x_m-x_{m-1}|+\cdots+|x_{n+1}-x_n| \\ &= r^m+r^{m-1}+\cdots+r^n \\ &= r^n(1+r+\cdots+r^{m-n}) \\ &= r^n\frac{1-r^{m-n+1}}{1-r}.\end{align}$$

If you can make this less than $\varepsilon$, you'll be done.

share|cite|improve this answer

This is exactly that: $$ x_{n+p} - x_n = \sum_{k=1}^p x_{n_k} - x_{n_{k-1}} $$ $$ |x_{n+p} - x_n| \le \sum_{k=1}^p |x_{n+k} - x_{{n+k-1}}| \le \sum_{k=1}^p r^{n+k-1} \\ \sum_{k=1}^\infty r^{n+k-1} = \frac{r^{n}}{1-r} \to_{n\to\infty} 0 $$ So eventually, $$\sup_{p\in\mathbb N} |x_{n+p} - x_n| \to_{n\to\infty} 0 $$

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.