Cauchy sequence definition:
"$\forall \epsilon>0, \exists N$ such that $\forall n,m>N, |a_{n}-a_{m}|<\epsilon$".
I was told that it is not sufficient to consider $m=n+1$ but however, I thought, if we can show that consecutive difference is always less than $\epsilon$ for a sequence, then can't we use triangular inequality to show the sequence is Cauchy, for instance:
$$
|a_m-a_n|= |a_m-a_{m-1}+…-a_n|\leq|a_m-a_{m-1}|+…+|a_{n+1}-a_n|\leq (m-n+1) \epsilon,
$$
and since $\epsilon$ is arbitrary then the sequence is proved to be Cauchy.
Best Answer
While your equation is correct as stated, $m$ can be arbitrarily large, which can make the factor $m - n - 1$ preceding your $\epsilon$ arbitrarily large, in turn making $(m - n - 1) \epsilon$ arbitrarily large.
$a_n := \sum_{i = 1}^n \frac{1}{i}$ is a sequence obeying the alternative property you've given, but it is not convergent.