Show that $(x_n)^{\infty}_{n=1}$ converges. Let $(X, d)$ a complete metric space and $(x_n)^{\infty}_{n=1}$ a sequence such that $d(x_{n+1},x_n) \leq \alpha d(x_n, x_{n-1})$ for some $0<\alpha<1$  and for all $n\geq 2$. Show that $(x_n)^{\infty}_{n=1}$ converges.
Been dealing with this problem without success, any suggestions would be great!

Observe that 
$$d(x_{n+1},x_n) \leq \alpha d(x_n, x_{n-1})\le \alpha^2 d(x_{n-1}, x_{n-2})\le \cdots  \le \alpha^{n-1} d(x_2,x_1)$$
(Take $m>n$ and use triangle inequality)
$$\implies d(x_m,x_n)\le d(x_{m+1},x_m)+d(x_m,x_{m-1})+\cdots +d(x_{n+1},x_{n})\le\sum_{k=(n-1)}^{(m-1)}\alpha^k d(x_2,x_1) $$
Now you know that $\sum_{k\ge 1}\alpha^k$ converges so does $\sum_{k\ge 1}\alpha^k d(x_2,x_1)$.
$\therefore$ For every $\epsilon>0$ , $\exists N \in \mathbb N$ such that $\forall (n-1),(m-1)>N$ we have  $$\sum_{k=(n-1)}^{(m-1)}\alpha^k d(x_2,x_1)<\epsilon\implies d(x_m,x_n)<\epsilon , \forall (n-1),(m-1)>N.$$ 
Hence $(x_n)^{\infty}_{n=1}$ is a cauchy sequence.
So by completeness of $(X,d)$ we have $(x_n)^{\infty}_{n=1}$ converges.
