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!
 A: 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.
A: Observe that, according to the property of the statement, we have that $d(x_3,x_2) \leq \alpha d(x_2,x_1)$, $d(x_4,x_3) \leq \alpha d(x_3,x_2) \leq \alpha^2 d(x_2,x_1)$, and so on. So, we conjeture that

$$d(x_{n+1},x_n) \leq \alpha^{n-1} d(x_2,x_1) \textrm{ for } n = 1,2,\dots \tag{$*$}$$

and in fact, this is easily proved by induction. Now, in order to use the completeness of the metric space, we will show that $(x_n)_{n=1}^\infty$ is a Cauchy sequence and we are done. So, let $m$ and $n$ two natural numbers, and suppose that $m<n$, that is, $n = m+k$ for some $k \in \mathbb Z^+$; next note that, using the triangle inequality,
\begin{align}
d(x_m,x_n) = d(x_m,x_{m+k}) &\leq d(x_m,x_{m+1}) + d(x_{m+1},x_{m+k}) \\
&\leq d(x_m,x_{m+1}) + d(x_{m+1},x_{m+2}) + d(x_{m+2},x_{m+k}) \\
& \; \; \vdots \\
&\leq \sum_{i=0}^{k-1} d(x_{m+i},x_{m+i+1}) = \sum_{i=0}^{k-1} d(x_{(m+i)+1},x_{m+i}) 
\end{align}
and, by $(*)$, this latter is less than or equal to $$\sum_{i=0}^{k-1} \alpha^{m+i-1}d(x_2,x_1) = \frac{\alpha^{m-1}(1-\alpha^k)}{1-\alpha} d(x_2,x_1).$$
Now, since $0<\alpha<1$, 
$$d(x_m,x_n) < \frac{\alpha^{m-1}}{1-\alpha} d(x_2,x_1)$$
and since $\alpha^{m-1}$ tends to zero as $m$ goes to infinity, for any $\varepsilon>0$ there must be exists some $N \in \mathbb Z^+$ such that the latter is less than $\varepsilon$ for every $m \geq N$.
