Prove that a sequence is a Cauchy sequence. The question is this.

Let $(s_n)$ be a sequence such that
  $$\left|s_{n+1}-s_n\right| < 2^{-n}, \forall n \in\mathbb N$$
  Prove that $(s_n)$ is a Cauchy sequence and hence convergent.

My proof is below.
Proof.
Let $\epsilon > 0 $ and $N = \dfrac{\ln\left(\frac{1}{\epsilon}\right)}{\ln(2)}$. Then, $\forall n > N = \dfrac{\ln(\frac{1}{\epsilon})}{\ln(2)}$ implies 
$$n\ln(2) > \ln\left(\frac{1}{\epsilon}\right)$$
$$2^n > \frac{1}{\epsilon}$$
$$\epsilon > \frac{1}{2^n} > \left|s_{n+1}-s_n\right|$$Q.E.D.
Is this ok??
 A: No. It does not work. You have to show that given $\epsilon>0$ then you can find $N$ big enough so that if $n,m\geq N$ then $|s_n-s_m|<\epsilon$. You showed that $s_n$ and $s_{n+1}$ have a difference of less than $\epsilon$ but you want to show that $s_n$ and $s_m$ have a difference of less than $\epsilon$, see the difference?
Try this. Fix $\epsilon>0$. Choose $N$ big enough an integer so that $1/2^N<\epsilon$ (this is possible, why?). Then if $n,m> N$ (say wlog that $m\geq n$, then $$|s_n-s_m|=|s_n-s_{n+1}+s_{n+1}-...+s_{m+1}-s_{m}|$$Apply the triangle inequality, and try to go from here. Let me know if you get stuck. 
Added:
So you can finish it like this: 
$$
|s_n-s_m|\leq |s_n-s_{n+1}|+...+|s_{m-1}-s_m|\leq 1/2^n+...+1/2^{m-1}$$$$=(1/2^n)(1+1/2+...+1/2^{m-n-1})<(1/2^n)(2)=1/2^{n-1}
$$
Since $n>N$ then $n-1\geq N$ so $2^{n-1}\geq 2^N$ which implies that $1/2^{n-1}\leq 1/2^N<\epsilon$ just as we wanted. 
A: The following result is useful

if $\sum_{n=1}^{\infty}|a_{n+1}-a_n|< \infty$, then ${a _n} $ is a Cauchy sequence. 

For a proof of this result, see here.
A: \begin{align}
&\mbox{Let's}\quad m >n
\\
\left\vert s_{m} - s_{n}\right\vert
&=
\left\vert
\left(s_{m} - s_{m - 1}\right) + \left(s_{m - 1} - s_{m - 2}\right)
+
\cdots
+
\left(s_{n + 1} - s_{n}\right)
\right\vert
\leq
\sum_{i = n}^{m - 1}\left\vert s_{i + 1} - s_{i}\right\vert
<
\sum_{i = n}^{m - 1}2^{-i}
\\[3mm]&=
{2^{-n}\left\lbrack 2^{-\left(m - n\right)} - 1\right\rbrack
 \over
1/2 - 1}
=
2^{1 -n}\left\lbrack 1 - 2^{-\left(m - n\right)}\right\rbrack
<
2^{1 - n}
\end{align}
Given $\displaystyle{\epsilon > 0}$, just take any
$\displaystyle{N > 1 - {\ln\left(\epsilon\right) \over \ln\left(2\right)}}$. Then,
$\displaystyle{\quad m, n > N\quad\Longrightarrow\quad
\left\vert s_{m} - s_{n}\right\vert < \epsilon}$.
