Please help prove that $(x_n)$ is a Cauchy sequence if $|x_{n+1} - x_n| \leq Cr^n$ Past final exam question for an intro to Real Analysis course:
Let $C > 0$, $0<r<1$ and suppose that $\forall n\in \mathbb N, |x_{n+1} - x_n| \leq Cr^n$. Please help me prove that $(x_n)$ is a Cauchy sequence. (We can assume the $\lim\limits_{n\to\infty} r^n=0,$ for $0<r<1$)
So, I know that a Cauchy series must satisfy $|x_{n}-x_m| < \epsilon$ for any $\epsilon>0, \in \mathbb R$ and for all $n,m \gt H(\epsilon) \in \mathbb N$ Note that there can't be any conditions on n and m (I saw somewhere else someone required $m>n$ which you can only do if you're proving its not at Cauchy sequence, right?)
Another way of doing this is showing that it is contractive (and thus a Cauchy series) if there is a constant $a$ such that $|x_{n+1}-x_n| \leq a|x_n-x_{n-1}|$ 
Clearly I'm supposed to make use of $\lim\limits_{n\to\infty} r^n=0,$ for $0<r<1$... But I don't even know how where to start with this. As I am bumbling through this problem, a more thorough answer would be much appreciated. Thanks! 
 A: Let $\epsilon>0$ be given. Then we can find $N$ such that $r^N\frac C{\color{red}{1-r}}<\frac\epsilon2$ because $r^n\color{red}\to0$ as $n\to \infty$.
Then for any $n>N$ we have
$$\begin{align} |x_n-x_N|&\le|x_{N+1}-x_N|+\ldots+ |x_n-x_{n-1}|\\&\le Cr^N+Cr^{N+1}+\ldots +Cr^{n-1}
\\&=Cr^N\cdot(1+r+\ldots + r^{n-N-1})
\\&\color{red}<Cr^N\sum_{k=0}^\infty r^k \color{red}= Cr^N\cdot\frac1{1-r}<\frac\epsilon2,\end{align}$$
hence for $n,m>N$
$$ |x_n-x_m|\le|x_n-x_N|+|x_m-x_N|<\frac\epsilon2+\frac\epsilon2=\epsilon.$$
Remark: I've marked all places in red where we made use of $0<r<1$.
A: Hint: $|x_n-x_m|\le |x_n- x_{n-1}|+|x_{n-1}-x_{n-2}|+\dots+|x_{m+1}-x_{m}|$.
A: We don't need to assume $r^n\xrightarrow{n\to\infty}0$. Since $|r|<1$, $$\sum_{n=1}^\infty r^n=\frac r{1-r}$$ and hence $r^n\xrightarrow{n\to\infty}0$.

The desired claim can easily be generalized: Let $(E,d)$ be a metric space, $(x_n)_{n\in\mathbb N}\subseteq E$ and $(\varepsilon_n)_{n\in\mathbb N}\subseteq\mathbb R$ be summable with $$d(x_n,x_{n+1})<\varepsilon_n\;\;\;\text{for all }m,n\ge N\tag1$$ for some $N\in\mathbb N$.
We show that $(x_n)_{n\in\mathbb N}$ is Cauchy: Let $\varepsilon>0$. Since $(\varepsilon_n)_{n\in\mathbb N}$ is summable, $\left(\sum_{i=1}^n\varepsilon_i\right)_{n\in\mathbb N}$ is Cauchy and hence $$\sum_{i=m+1}^{n}\varepsilon_i=\left|\sum_{i=1}^m\varepsilon_i-\sum_{i=1}^n\varepsilon_i\right|<\varepsilon\;\;\;\text{for all }n\ge m\ge N_1\tag2$$ and hence $$d(x_m,x_n)=\sum_{i=m}^{n-1}d(x_i,x_{i+1})<\sum_{i=m}^{n-1}\varepsilon_i<\varepsilon\;\;\;\text{for all }n\ge m>\max(N,N_1).\tag3$$
