Showing $|x_n -x|$ converges to x I'm given a generic sequence $x_n$, and I know that $\lim_{n ->\infty} \frac{|x_{n+1}-x|}{|x_n-x|} < 1$ and exists. I'm trying to show that $x_n$ converges to $x$. 
I tried a bunch of approaches, but the farthest I got was a proof by contradiction where I try show that $\lim_{n->\infty} |x_n-x|$ cannot be greater than 0, but instead must equal 0, resulting in $x_n$ converging to $x$. 
Can I get some assistance?
 A: Hint:
It is equivalent to show tha $|x_n-x|\to 0$. Now, since $\lim_{n\to\infty} \frac{|x_{n+1}-x|}{|x_n-x|} $ exists and is $<1$, there exists $k<1$ and $N_0$ such that
$$\frac{|x_{n+1}-x|}{|x_n-x|}\le k\quad\forall n \ge N_0.$$
Deduce that, for all $n\ge N_0$, one has $\;|x_n-x|\le k^{n-N_0}\,|x_{N_0}-x|$ (use induction on $n$) and conclude.
A: If we define $a_n=|x_n-x|$, then proving that $x_n$ converges to $x$ is equivalent to proving that $a_n$ converges to $0$.
This is clear since 


*

*$x_n$ converges to $x$ if and only if


For each $\epsilon>0$ there exists some $N>0$ such that for all $n>N$, $$|x_n-x|<\epsilon$$


*$a_n$ converges to $0$ if and only if


For each $\epsilon > 0$ there exists some $N>0$ such that for $n>N$, $$|a_n-0|<\epsilon$$

The two conditions are clearly identical, since $|a_n-0|=|a_n|=||x_n-x||=|x_n-x|$.

To prove that $a_n$ converges to $0$, let's first see what we know about $a_n$. We know that $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} < 1$$
Here's what I would advise you to do next:


*

*Prove that there exists some $c<1$ such that for a large enough $N$, it is true that if $n>N$, then $\frac{a_{n+1}}{a_n} < c$

*Then, use the fact that that same inequality translates to $a_{n+1}<c\cdot a_n$ (again, for $n>N$).

*From that, you can see that $a_{n+1}<c\cdot a_n<c\cdot(c\cdot a_{n-1})<c\cdot(c\cdot(c\cdot a_{n-2}))<\cdots$ if $n$ is large enough.

*Can you continue from here?

