# Cauchy sequence confusion

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.

• No, you cannot find a factor to replace $\varepsilon$ by so that it will be $<\varepsilon$ for all $m,n$. – user10354138 Jun 11 at 14:42

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.