Proving that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence. Let $(x_n)_{n=1}^{\infty}$ be a sequence such that $|x_{n+1} - x_n| < r^n$ for all $n \geq 1$, for some $0 < r < 1$. Prove that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence.
I understand that a Cauchy sequence means that for all $\varepsilon > 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. But this one is really giving me a headache.
I tried doing something like: let $ m > n$. Therefore $x_n - x_m$ =  $(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m) $ and then somehow using the triangle inequality to compute some sum such that $x_n - x_m$  < sum which would be epsilon?
any help is appreciated, thank you.

For every $\epsilon>0$, take a natural number $N$ such that $r^N <(1-r)\epsilon$, for example by taking $N=\lfloor\frac{\ln (1-r)\epsilon}{\ln r}\rfloor+1$. Then, for all $m,n\geq N$, assume $m<n$, we have
\begin{align}
|x_n - x_m|&=|(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m)|\\
&\leq |(x_n - x_{n-1})| + |(x_{n-1} - x_{n-2})| + ... + |(x_{m+1} - x_m)|\\
&< r^{n-1}+\dots+r^m\\
&=r^m(1+r+r^2+\dots+r^{n-m-1})\\
&<\frac{r^m}{1-r}\\
&\leq\frac{r^N}{1-r}\\
&<\epsilon
\end{align}
