$d(x_{n+2},x_{n+1})≤rd(x_{n+1},x_n)$ gives convergent sequence

Question

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?

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 $m>n\geq N$,

$d(x_m,x_n)$

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

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

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

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

$\leq\epsilon$.

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