Prove that a sequence satisfying:
$|x_k-x_{k+1}|< \frac{1}{a^k}$ for some $a>1$, for all $k$
is cauchy.
It follows from the criterion that:
$|a_n - a_m|<\sum_{i=n}^{m}{\frac{1}{a^i}}, m>n$
And it can be shown that
$\sum_{i=n}^{m}{\frac{1}{a^i}} < \frac{1}{a^{n-1}}$ if $a>2$
So now we know that a sequence satisfying $|x_k-x_{k+1}|< \frac{1}{a^k}$ for some $a>2$ is cauchy. However, I do not know how to proceed for cases where $1<a<2$.
You're close:
$$\begin{align*}\sum_{i=n}^m \frac{1}{a^i} &\le \frac{1}{a^{n}} \sum_{i=0}^{m-n+1} \frac{1}{a^i}\\ &\le \frac{1}{a^{n}} \sum_{i=0}^\infty \frac{1}{a^i}\\ &=\frac{1}{a^n} \frac{1}{1-(1/a)} \end{align*}$$
Note that your work for $a>2$ is a special case of this argument, which works for $a>1$.