Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,d)$ be a complete metric space, $r\in (0,1)$ and $(x_n)$ be a sequence in $X$ such that $d(x_{n+2},x_{n+1})≤rd(x_{n+1},x_n)$ for every $n\in N$. How can we show that $(x_n)$ is a convergent sequence?

share|improve this question

1 Answer 1

If $x_1=x_2$ we are done. Otherwise, for every $\epsilon>0$, there is a positive integer $N$ such that $r^{N-1}\leq\epsilon(1-r)\frac{1}{d(x_2,x_1)}$. Then for any $m>n\geq N$,


$\leq d(x_m,x_{m-1})+...+d(x_{n+1},x_n)$

$\leq d(x_2,x_1)(r^{m-2}+...+r^{n-1})$

$\leq d(x_2,x_1)\frac{r^{n-1}}{1-r}$

$\leq d(x_2,x_1)\frac{r^{N-1}}{1-r}$


This shows that the sequence is Cauchy so that it converges since $X$ is complete.

share|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.