show that the sequence ${a_n}$ is a cauchy sequence Show that any sequence $\{a_n\}$ that has the property $|a_{n+1} -a_n|< b^n$ for $b<1$ is a Cauchy sequence. I'm having problems giving a formal proof of why this holds. 
 A: $$
u_{n+p}-u_n = \sum_{k=1}^p u_{n+k}-u_{n+k-1} \\
|u_{n+p}-u_n| \le \sum_{k=0}^{p-1} |u_{n+k+1}-u_{n+k}|
\le  \sum_{k=0}^{p-1} b^{n+k} \le \frac {b^n}{1-b}
$$
so $$
\sup_{p\ge 0} |u_{n+p}-u_n|\to 0
$$when $n\to \infty$
, that is $u$ is a Cauchy sequence.
A: $|a_m-a_n|=\left|\sum\limits_{k=n}^{m-1}{\left(a_{k+1}-a_k\right)}\right|<\sum\limits_{k=n}^{m-1}{\left|a_{k+1}-a_k\right|}<\frac{b^n}{1-b}$
Now $a_n$ is a cauchy sequence follows from the convergence of $b^n$ to zero.
A: It follows immediately from the following proposition
Proposition: Let $\langle a_n: n\in \mathbb{N} \rangle$ be a sequence such that $|a_{n+1}-a_n|< Mr^n$ for all $n\in \mathbb{N}$, where $M\in \mathbb{R}^{>0}$ and $r\in (0,1)$. Then  is a Cauchy sequence.
Proof: Given $\varepsilon>0$, choose $N$ such that $r^N<\frac{1-r}{M}\varepsilon$ note that this is possible since $r^N \to 0$. Without loss of generality suppose that $m>n\ge N$. Then 
\begin{align}|a_{m}-a_{n}|= \bigg|\sum_{i=n}^{m-1}a_{i+1}-a_i\bigg|\le\sum_{i=n}^{m-1}|a_{i+1}-a_i|<\sum_{i=n}^{m-1}Mr^i\\
\le \sum_{i=n}^{\infty}Mr^i =\frac{Mr^n}{1-r}\\
\le  \frac{Mr^N}{1-r}< \varepsilon  \end{align} 
