Let $s_n$ be a sequence such that
$|s_{n+1} - s_n| < 2^{-n}$
for all $n \in N$. Prove $s_n$ is a Cauchy sequence and hence a convergent sequence.
Here's what I've started with:
Proof:
Take $\epsilon > 0$. Let $N = -\log_2{\epsilon/2}$. Thus,
$ |s_{n+1} - s_n| < 2^{\log_2{\epsilon/2}} = \epsilon/2$
I can't seem to show from this that:
$ |s_m - s_n| <\epsilon$
Some ideas relating to geometric series of $\epsilon/2 + \epsilon/4 ... \epsilon/{2^n} < \epsilon$ are springing up in my mind, but I can't really pin it to the ground (and, I'm not sure if its fair play to use the infinite series formula in this case).
Here's a sketch; check the details carefully. You need to show that for all $\epsilon > 0$ that there is an $N$ such that for $n,m > N$, $|s_n - s_m| < \epsilon$. Let $\epsilon > 0$ be given. We have to find $N$.
Assume that $m > n$ without loss of generality. Then by the triangle inequality and the given property of the sequence, $$|s_m - s_n| < \frac{1}{2^m} + \cdots + \frac{1}{2^n} = \frac{1}{2^m}\left(1+\frac{1}{2} + \cdots + \frac{1}{2^{n-m}}\right)$$ and so by the formula for a geometric series, $$|s_m - s_n| < \frac{1}{2^m} \frac{1-\left(\frac{1}{2}\right)^{n-m}}{1 - \frac{1}{2}} < \frac{1}{2^{m-1}}$$
Now just take $N = \log_2{\epsilon}$. Then for $m > n > N$, we have that $$|s_m - s_n| < \frac{1}{2^{m-1}} \leq \frac{1}{2^N} = \epsilon$$