Is this enough for a epsilon N proof? Cauchy sequence Assume that $ C \in \mathbb{R}, q < 1$ and $|x_{n+1} - x_n | < Cq^n$ for all n. Prove that this is a Cauchy-sequence...
What I do is that I assume without loss of generality that $m < n$ so $n = m + k$. 
Then by the triangle inequality, $$|x_{m+k} - x_m| \le |x_{m+k} - x_{m+k-1}|  + |x_{m+k-1} - x_{m+k-2}| + \ldots + |x_{m-1} - x_m|$$
By using some formulas that we prove in a previous exercise, I get that above is less than $$\le C \sum_{i=m}^{m+k-1} q^i = C\frac{q^m - q^n}{1-q}$$
Is above (apart from this final step that you might not know of?) the correct method? And what more can I now do? 
It's my first attempt at a epsilon proof for sequences, and I'm not sure if I can just say "pick N large so this stuff is less than epsilon", or if more is needed. 
EDIT: q $\in (0,1)$

You have done well and are almost there. Note that
$$\frac{q^m - q^n}{1-q} = q^m \frac{1-q^k}{1-q} \leq q^m\frac{1}{1-q} $$
for all $k \geq 1$. Define a new constant 
$$ C' := \frac{C}{1-q}. $$
You have shown that for all $m \in \mathbb{N}$ and all $k \geq 1$ that 
$$|x_{m+k} - x_{m}| \leq C' q^m. $$
Given $\epsilon > 0$ pick $N \in \mathbb{N}$ such that $q^N < \frac{\epsilon}{2C'}$. Then for all $m,n > N$ we have 
$$|x_m-x_n| \leq |x_m - x_N| + |x_n - x_N| \leq  2C'q^N < \epsilon. $$ 
