Hi, I think I am close to the answer for this but I am missing something:

Suppose $\displaystyle x_{n}$ is a sequence satisfying $\displaystyle |x_{n+1}-x_{n}|<\frac{1}{2^n}$ for all n. Show that $\displaystyle x_{n}$ is a convergent sequence.

So to prove it is convergent, I need to prove it is a Cauchy sequence, ie for all $\displaystyle \epsilon$ there is an N such that $\displaystyle |x_{m}-x_{n}|<\epsilon$ for all $\displaystyle n,m>N$.

Now $\displaystyle |x_{m}-x_{n}|= |x_{m}-x_{m-1}+x_{m-1}-...+x_{n+1}-x_{n}|$ and the triangle inequality gives $\displaystyle |x_{m}-x_{n}|\leq|x_{m}-x_{m-1}|+|x_{m-1}-x_{m-2}|+...+|x_{n+1}-x_{n}|$.

Then using the inequality given in the question I get $\displaystyle |x_{m}-x_{n}|<\frac{1}{2^{m-1}}+\frac{1}{2^{m-2}}+...+\frac{1}{2^n}<\frac{1}{2^n}+\frac{1}{2^n}+ ...+\frac{1}{2^n}=\frac{m-n}{2^n}$.

This is where I get stuck, I think I need to get rid of that m-n somehow.