If $d(x_n, x_{n+k}) < \epsilon$ for all $n \ge N\in \mathbb{N}$, and $k \in \mathbb{N^*}$, then $(x_n)$ is a Cauchy sequence?

Question

I've tried to prove this is as: If the condition holds for any k, then

$\sum_{i=1}^{\infty} d(x_i, x_{i+1}) < \infty$, by the comparison test ($\epsilon < 1$).

Then the sequence of the partial sums of this series has to be Cauchy so that for all $n > m \ge N$ :

$|S_n - S_m| < \epsilon \implies \sum_{i=m}^{n-1} d(x_i, x_{i+1}) < \epsilon$ (are the limits of summation right?). By the triangle inequality, then, $d(x_m, x_n) < \epsilon.$

I want to know if this is correct?

Answer 1

Consider the sequence $x_n:={1\over n}$ ($n$ odd), and $:=0$ ($n$ even). This sequence fulfills the intended condition in the title of the question, but the sum $\sum_{k=1}^\infty d(x_{k-1},x_k)$ diverges. This shows that such sums are of no help in solving this problem.

As announced in Simione's first comment the claim is utterly obvious, and has nothing to do with completeness: Assume that the sequence $(x_n)_{n\geq0}$ satisfies the intended condition in the title of your question, and let an $\epsilon>0$ be given. Then there is an $N$ such that $d(x_n,x_{n+k})<\epsilon$ for all $n\geq N$ and all $k\geq1$. Assume that both $n$ and $m$ are $\geq N$. We may assume $m\geq n$. If $m=n$ then $d(x_n,x_m)=0<\epsilon$. Otherwise $m=n+k$ with $k\geq1$, and we have $d(x_n,x_m)=d(x_n,x_{n+k})<\epsilon$. It follows that the sequence $(x_n)_{n\geq0}$ is Cauchy according to the definition.