Suppose that there is an $a\in(0,1)$ such that $|x_{n+1}-x_n| \leq a^n$ for all $n\in\mathbb N$. Prove that $x_n \to x$ for som $x\in \mathbb R$.

Question

Let $\{x_n\}$ be a sequence. Suppose that there is an $a\in(0,1)$ such that $$|x_{n+1}-x_n| \leq a^n$$ for all $n\in\mathbb N$. Prove that $x_n \to x$ for som $x\in \mathbb R$.

I need some help on this one. Or a hint in the right direction. I have I need to use the fact that it is cauchy. Because $a^n, \forall n\in \mathbb N$, seems very to saying $\forall \epsilon$.

Any hint is appreciated!

NOTE: This is very similar to this post but not is not.