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.

  • $\begingroup$ No, you cannot find a factor to replace $\varepsilon$ by so that it will be $<\varepsilon$ for all $m,n$. $\endgroup$ – 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.


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