Let $(X, d)$ be a metric space.
I'm trying to show fast Cauchy sequences are Cauchy. My textbook defines $(x_k)_k$ is a fast Cauchy sequence if $\sum_{k=1}^{\infty}d(x_k,x_{k+1}) < \infty$, but I'm not really sure how to work with this definition.
Is this also equivalent to $d(x_k,x_{k+1}) \leq (\frac{1}{2})^k$ for all $k \in \mathbb{N}$? If so, why is that the case?
Suppose that $\{x_k\}$ is a fast Cauchy sequence and take $\epsilon >0$. As $\sum d(x_k,x_{k+1})$ is a converging series with positive terms, it exists $N \in \mathbb N$ such that $\sum_{k=n+1}^\infty d(x_k,x_{k+1}) < \epsilon$ for all $n >N$. But then for $p >n>N$, you have
$$d(x_n,x_p) \le d(x_n,x_{n+1}) +\dots +d(x_{p-1},x_p)<\epsilon $$ using the triangular inequality.
That proves that the series $\{x_n\}$ is Cauchy.