Showing a sequence of real numbers converges I need to show that for $0 < q < 1$ that the sequence ${x_n} = q^n$ converges, and to find its limit.
So I need to show that 
$\forall \epsilon > 0$   there is $N \in \mathbb{N}$ such that $\forall n>N, |x_n - x_m | < \epsilon$ , or $|q^n - q^m| < \epsilon \\$
Once I show convergence it should be pretty easy to find its limit. Don't know where to start with this though.

Hint: You are trying to prove that the sequence is a Cauchy sequence. It is easier to prove directly from the definition of limit that $\lim_{n\to\infty} q^n=0$. 
We have $q=\frac{1}{1+p}$ for some positive $p$. Now by the Binomial Theorem, we have $(1+p)^n \gt np$ if $n\ge 1$.  This inequality is enough for a proof based on the $\epsilon$-$N$ definition of limit.
The inequality is also enough to show that the sequence $(q^n)$ is Cauchy, if you want to take that path.
Added: For the inequality $(1+p)^n \ge np$, we can use the Binomial Theorem
$$(1+p)^n =1+\binom{n}{1}p+\binom{n}{2}p^2+\cdots +p^n.$$
In particular, $(1+p)^n\ge np$, since $np$ is one of the terms in the above expansion.
Now suppose we are given $\epsilon\gt 0$. We have 
$$|q^n-0|=q^n =\frac{1}{(1+p)^n}\le\frac{1}{pn}.$$
To make $q^n$ less than $\epsilon$, it is enough to make $\frac{1}{np}\lt \epsilon$, that is, make $n\gt \frac{1}{p\epsilon}$. Thus if $N$ is say the smallest integer greater than $1/(p\epsilon)$, and $n\ge N$, then we will have $q^n\lt \epsilon$.
