Given sequence $x_n, |x_{k+1}-x_k| < a^{-k}, \forall k \in \mathbb{N} $. show sequence is Cauchy,where $a > 1$.
I was doing rough work where i considered $N= 40$, and $m=52$ and $n= 41$. So i got $|x_{50}-x_{41}| < 11 a^{-k}$. It seems that $|x_m-x_n| < |m-n|a^{-k}$. Now as $k \to \infty, a^{-k} \to 0 < \epsilon $
I am not sure, something feels wrong. can anyone help ?
Write this clearly and you will find it easier.
$$|x_k-x_{k+1}|<a^{-k}$$ $$|x_{k+1}-x_{k+2}|<a^{-(k+1)}$$ $$\vdots$$ $$|x_{k+i}-x_{k+i+1}|<a^{-(k+i)}$$
Summing these up and using triangle inequality gives
$$|x_k-x_{k+i+1}|<\sum\limits_{n=k}^{k+i} a^{-n}<\sum_{n=k}^{\infty} a^{-n}.$$
Can you take it from here?