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!  
 A: Hint: Let   $ 1> \lambda > L$ then you have  (for large enough $n $)
$$|X_{n+1} - x| < \lambda |X_n -x|  < \lambda^2 |X_{n-1} - x| < .... < \lambda^{n} |X_1 -x|$$
A: Since
$$\lim_{x\to\infty} \frac{|X_{n+1} - x|}{|X_n - x|}=L<1,$$
for $\varepsilon\in(0,1-L)$, $\exists N\in\mathbb{N}$ such that
$$ \frac{|X_{n+1} - x|}{|X_n - x|}\le L+\varepsilon, \forall n\ge N$$
So for $n\ge N$, one has
$$ |X_{n}-x|<(L+\varepsilon)|X_{n-1}-x|\le(L+\varepsilon)^2|X_{n-2}-x|\le \cdots\le(L+\varepsilon)^{n-N}|X_N-x| $$
Since $L+\varepsilon<1$, letting $n\to\infty$, one has
$$ \lim_{n\to\infty}|X_N-x|=0$$
or
$$ \lim_{n\to\infty}X_N=x.$$
A: For $0<\ell< 1-L<1$ There exists N such that if $n>N$
then
$$\left|\frac{|X_{n+1} - x|}{|X_n - x|} -L\right|\le \ell \implies |X_{n+1} - x| <(\ell +L)|X_{n} - x|~~\forall n>N $$
But $$0<\ell< 1-L<1 \implies 0<(\ell +L)<1$$
and by induction we have for $n>N$
$$  |X_{n+1} - x| <(\ell +L)^{n+1-N}|X_{N} - x|\to 0 $$
Since 
$$|X_{n+1} - x| <(\ell +L)|X_{n} - x|< (\ell +L)|X_{n} - x|\\<(\ell +L)^2|X_{n-1} - x|\\<\cdots<(\ell +L)^{n+1-N}|X_{N} - x|$$
A: 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$,,
$\rightarrow:$
$\lim_{n \rightarrow \infty} X_n =x.$
