# Homework Help: Cauchy sequnce and convergence of a non-monotonic sequence.

1. May 9, 2012

### kapitan90

1. The problem statement, all variables and given/known data
Hello,
I have a question concerning convergence of the non-monotonic sequences which takes place when the Cauchy criterion is satisfied.
I understand that $$|a_n - a_m| <ε$$ for all n,m$$N\ni$$

2. Relevant equations
What I don't see is how $$(a_{n+1} - a_n) →0$$is not equivalent/enough. Doesn't the fact that the difference tends to zero mean that it can eventually be smaller than any ε?

3. The attempt at a solution
I know the examples when this doesn't work, like $$(\sqrt{n})$$ but I don't understand the difference between these two criteria.

Could anyone explain why they aren't equivalent?

2. May 10, 2012

### clamtrox

So let's take the example of an=√n. For large n, you have then $a_{n+1}-a_n \sim n^{-1/2} \rightarrow 0$ when n→∞. We can write the Cauchy criterion as (let p = n-m)
$$\epsilon > a_{m+p}-a_m = \sum_{n=m}^{m+p} a_{n+1}-a_n$$
So now when n is very large, we have
$$\epsilon > \sum_{n=m}^{m+p} a_{n+1}-a_n \sim \sum_{n=m}^p n^{-1/2} \sim \int_m^{m+p} dn n^{-1/2} = (\sqrt{m+p}-\sqrt{m})/2$$
and this grows without limit if you let p to grow, so it can't be smaller than ε for all p.