When does $d(x_n,x_{n+1})\rightarrow 0$ imply $(x_n)$ is a Cauchy sequence? http://math.stackexchange.com/q/107336/21436 answers the question as to wether or not this "weaker" Cauchy criteria is equivalent to the actual cauchy criteria. I've seen the examples of why that's not, in general, the case. My question is under what conditions do they become equivalent. (Or what sort of metric would a space have to have, for the equivalence to occur).
I figure with the trivial metric it must be, because it means after a certain order every consecutive term will be equal. Also, one comment mentioned it was equivalent in $\mathbb{Q}_{p}$. Can you give me other examples of when and how this two definitions are equivalent? 
Update (16/8): I think they are equivalent given an ultrametric space.
Proof: Assume $d(x_n,x_{n+1})\rightarrow 0$. Then,
$\forall \epsilon >0 \exists N\in\mathbb N_+$such that
$\forall n>N d(x_{n+1},x_n)<\epsilon$
Let m>n > N. 
$$d(x_m,x_n) <= max(d(x_{n+1},x_n),d(x_{n+1},x_m))$$
(Ultrametric)
Applying this m-n-1 times we get $d(x_m,x_n) <\epsilon$ as required by the criteria. Is this proof correct? 
Thanks 
 A: The Cauchy sequence definition is $\forall \varepsilon \in \mathbb{R} \ \exists N \in \mathbb{N}\  \forall n,m>N \ d(x_n,x_m) < \varepsilon$, while the "weak" one is $\forall \varepsilon \in \mathbb{R} \ \exists N \in \mathbb{N}\  \forall n>N \ d(x_n,x_{n+1}) < \varepsilon$. Obviously a Cauchy sequence is a "weak" Cauchy sequence.
As hinted in a comment in the linked question, the two definitions are equivalent if $d$ is an ultrametric: you can then find some other examples in the Wikipedia page.
To prove this equivalence, recall the definition of ultrametric space: is a metric space where $d(x,y)<\max(d(x,z),d(y,z))$ holds. Now consider a weak Cauchy sequence $(x_n)_{n \in \mathbb{N}}$ and fix $\varepsilon$. Then there exists $N$ such that $\forall n>N \ d(x_n,x_{n+1}) < \varepsilon$. Then if $n,m>N$ (wlog $N<n<m$) $d(x_n,x_m)<\max(d(x_n,x_{n+1}),d(x_{n+1},x_{m}))<\max(\varepsilon,d(x_{n+1},x_{m}))$ and doing the same trick* for $d(x_{n+1},x_{m})$ you can get $d(x_n,x_m)<\varepsilon$.
*the trick, while obvious, is actually an (easy) induction.
