Equivalent condition for sequence to be Cauchy in a metric space

A metric space $$(X,d)$$ is called ultrametric if strict triangle inequality holds: $$d(x,y)\le \max\{ d(x,z), d(z,y)\}.$$

In the book $$p$$-adic Lie groups by Schneider, the author states following theorem (p.6):

A sequence $$(x_n)$$ in $$X$$ is a Cauchy sequence if and only if $$\lim_{n\rightarrow \infty} d(x_n,x_{n+1})=0$$.

I confused here whether this is true only for ultrametric space $$(X,d)$$? With usual metric on $$\mathbb{R}$$, we can take $$x_n=1+\frac{1}{2}+\cdots + \frac{1}{n}$$, which is not a Cauchy. Am I missing something?

There are questions on mathstack where the condition in above statement is replaced by $$d(x_n,x_{n+1}) for some fixed $$C\in\mathbb{R}$$ and $$0 for all $$n\ge 1$$. Then $$(x_n)$$ is Cauchy.

It seems that this condition in terms of $$C$$ and $$r$$ is weaker than that in above quoted statement. Am I right? In other words, we have only one way implication below but not conversely: $$\Big{(}d(x_{n+1},x_n)

• In an ultrametric you will have $d(x_{n+p},x_n)\leq\max(d(x_{n+p},x_{n+p-1}),...,d(x_{n+1},x_n))$. If $d(x_{n+1},x_n)\to0$, then for $n>N$, $d(x_{n+1},x_n)<\epsilon$. Therefore, $d(x_{n+p},x_n)<\epsilon$. – logarithm May 6 '19 at 14:18
• It is true that the hypothesis of $d$ being an ultrametric is important because in general the condition needs not hold, as you've pointed out. The discussion in the last paragraph is not too easy to follow: in my opinion, the phrasing is a bit imprecise. You may want to share the the original. – Saucy O'Path May 6 '19 at 14:19
• Ah, I see. So it was $0<r<1$, not $0<r<n$. – Saucy O'Path May 6 '19 at 14:25
• Oh, sorry! It was my typing mistake. Thanks @Saucy – Beginner May 6 '19 at 14:26