TITLE: How to show if $|a_{n+1} - a_{n}| \le \frac{1}{2^n}$ then the sequence is Cauchy.
QUESTION [6 upvotes]: Let $\{a_n\}$ be a sequence of real numbers such that $|a_{n+1} - a_n| \le \dfrac{1}{2^n}$.   I would like to show that this sequence is Cauchy.  
Letting $\epsilon > 0$, I said choose $N$ such that $1/2^N \le \epsilon$. However, I'm not sure if this is the right $N$, since $m, n > N$ doesn't seem to necessarily imply $|a_n - a_m| \le \epsilon$.
Can someone show me the light?

REPLY [11 votes]: Apply the triangle inequality repeatedly. If $n,m>N$ then we can assume WLOG that $m>n$ and we have
$$\begin{align}
|a_m-a_n|&\leq |a_{n+1}-a_n|+|a_{n+2}-a_{n+1}|+\cdots+|a_m-a_{m-1}|\\
&\leq \frac{1}{2^n}+\frac{1}{2^{n+1}}+\cdots+\frac{1}{2^{m-1}}\\
&< \sum\limits_{k=n}^\infty \frac{1}{2^k}=\frac{1}{2^n}\sum\limits_{k=0}^\infty\frac{1}{2^k}=\frac{1}{2^{n-1}}<\frac{1}{2^{N-1}}
\end{align}$$
and by making $N$ sufficiently large we have $\frac{1}{2^{N-1}}<\epsilon$.

REPLY [6 votes]: let $m>n$ then we have$$\begin{align}|a_m-a_n|&\le|a_m-a_{m-1}|+|a_{m-1}-a_{m-2}|+\dots+|a_{n+1}-a_n|\\&\le \frac{1}{2^{m-1}}+\frac{1}{2^{m-2}}+\dots\frac{1}{2^{n}}\end{align}$$ what can you say about this sum?