Proof that a sequence converges. Let $(a_n)_{n=1}^{\infty}$ be a sequence, let $\; 0 < q < 1$ such that
$\;\forall n \in \mathbb{N}$ $\;n>2$ 
$ \left | a_{n+1} -a_n \right | < q\left | a_{n} -a_{n-1}\right |$
prove that $(a_n)_{n=1}^{\infty}$ converges.
Well, I thought about proving it this way:
$ \left | a_{n+1} -a_n \right | < q*\left | a_{n} -a_{n-1}\right |<\left | a_{n} -a_{n-1}\right | $ 
therefore $ \left | a_{n+1} -a_n \right | <\left | a_{n} -a_{n-1}\right | \;$  for each $n>2$ and that obviously shows us that the sequence converges because the gaps between each $2$ indexes reduces as $n$ grows.
I have a feeling however that this is not a formal way of proving this, it seems too easy.
 A: $|a_{n+1} - a_n| < |a_n - a_{n-1}|$ is not sufficient to prove convergence of $(a_n)$. A counterexample is $a_n := 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$.

For any $n$,
$$|a_{n+1} - a_n| < q |a_n - a_{n-1}| < q^2 |a_{n-1} - a_{n-2}| < \cdots < q^{n-1} |a_2 - a_1|.$$
Use this to show $(a_n)$ is Cauchy (and therefore convergent).

 For $m < n$ we have $$|a_n - a_m| \le \sum_{k=m}^{n-1} |a_{k+1} - a_k| \le |a_2 - a_1| \sum_{k=m}^{n-1} q^{k-1} \le |a_2 - a_1| \frac{q^{m-1}}{1-q}.$$

A: Choose an integer $m<n$. By the triangle inequality, 
$$|a_n-a_m|\leq |a_n-a_{n-1}|+|a_{n-1}-a_m|.$$
You can continue to apply the triangle inequality in this way until you obtain:
$$|a_n-a_m|\leq |a_n-a_{n-1}|+|a_{n-1}-a_{n-2}|+|a_{n-2}-a_{n-3}|+\cdots+|a_{m+1}-a_m|$$
And then from the problem statement, this is strictly less than:
$$q^{n}|a_2-a_1|+q^{n-1}|a_2-a_1|+\cdots+|q^{m-1}|a_2-a_1|=|a_2-a_1|(q^n+q^{n-1}+\cdots+q^m-1).$$
So, choosing a large enough $m$ (and hence $n$) will guarantee the distance between any terms beyond a certain point will be arbitrarily small, so the sequence is Cauchy. If you are in $\mathbb{R}$ or $\mathbb{C}$ with the usual metric, which you probably are, then your metric space is complete, and therefore convergence is implied.
EDIT: Your only mistake, and it is common, is that you must prove that after a certain integer, every pair $(a_n,a_m)$ has arbitrarily small distance, not just pairs of successive terms. I have made that mistake so many times. 
