Show that $X_n\to x$ if $\lim\limits_{n\to\infty} \frac{|X_{n+1} - x|}{|X_n - x|}<1.$ Suppose $\{X_n\}$ is a sequence and suppose for some $x \in \mathbb{R}$, the limit
$$L:=\lim_{n\to\infty} \frac{|X_{n+1} - x|}{|X_n - x|}$$
exists and $L <1$.  Show that $\{X_n\}$ converges to $x$.  
So far I have noted that since each the denominator and numerator must be positive then $L \geq 0$ and so $0 \leq L < 1$.  Also the following must hold,
$$|X_{n+1} - x| < |X_n -x|$$ since the quotient is between $0$ and $1$.  
Now i'm unsure how to proceed.  Any help is appreciated, thanks!  

Correct me if wrong:
$a_n:= X_n-x.$
Show:
$ \sum a_n$ is absolutely convergent using the ratio test.
Given : 
$\lim_{n \rightarrow \infty }|\dfrac{a_{n+1}}{a_n}| = L \lt 1$.
$\rightarrow:$
There is a $n_0$ such that $a_n \ne 0$ for $n \ge n_0.$
There is a $n_1 (\ge n_0) $ such that for $n \ge n_1$
$|\dfrac{a_{n+1}}{a_n}| \lt \Theta \lt 1,$
where $L \lt \Theta \lt 1$.
$\rightarrow:$ $\sum a_n $ absolutely convergent.
$\rightarrow:$ $\lim_{n \rightarrow \infty } |a_n| = 0$, and
$ \lim_{n \rightarrow \infty} (a_n) = 0$,,
$\rightarrow:$
$\lim_{n \rightarrow \infty} X_n =x.$
