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.
 A: 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}
A: This is exactly that:
$$
x_{n+p} - x_n = \sum_{k=1}^p x_{n_k} - x_{n_{k-1}}
$$
$$
|x_{n+p} - x_n| \le  \sum_{k=1}^p |x_{n+k} - x_{{n+k-1}}| 
\le \sum_{k=1}^p r^{n+k-1} 
\\  \sum_{k=1}^\infty r^{n+k-1} 
= \frac{r^{n}}{1-r}
\to_{n\to\infty} 0
$$
So eventually, 
$$\sup_{p\in\mathbb N} |x_{n+p} - x_n| \to_{n\to\infty} 0
$$
A: You are definitely on the right track. Here's what you want to do: suppose without loss of generality that $m>n$ then
$$\begin{align}|x_m-x_n| &= |x_m-x_{m-1}+x_{m-1}-\cdots-x_n| \\ &\le |x_m-x_{m-1}|+\cdots+|x_{n+1}-x_n| \\ &= r^m+r^{m-1}+\cdots+r^n \\ &= r^n(1+r+\cdots+r^{m-n}) \\ &= r^n\frac{1-r^{m-n+1}}{1-r}.\end{align}$$
If you can make this less than $\varepsilon$, you'll be done.
