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Let $(X, d)$ be a complete metric space, $r∈ (0,1)$ 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 $n∈ℕ$. Show that $\{x_n\}$ is a convergent sequence.

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closed as off-topic by Did, wythagoras, avid19, anomaly, Mike Pierce Jun 12 at 18:13

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2 Answers 2

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.

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HINT: Show that the sequence is Cauchy. What can you say about $\sum_{k\ge 0}d(x_{n+k},x_{n+k+1})$? You’ll be using the triangle inequality and looking at the sum of a geometric series.

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