Cauchy sequnce and convergence of a non-monotonic sequence.

[kapitan90]

Homework Statement

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$$N\ni$$

Homework 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?

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?

 

[clamtrox]

So let's take the example of a_n=√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.