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There is definition of Cauchy sequence in the book of Introduction to Calculus and classical analysis by Omar hijab and that is :

$\forall n,m\in \Bbb N \space e_n \ge 0 ,e_n\to 0 \space ,|a_{m+n}- a_n|\lt e_n $

$e_n$ is a error sequence for cauchy sequence .and there is a general definition which many books including Rudin use which is :

$\forall\epsilon\gt0 \space\space\exists N\in\Bbb N \space \space s.t \space \space\forall m,n\ge N \space\space |a_n-a_m|\lt\epsilon$

I want to prove that these definitions are equivalent.I don't have any idea how to prove it !

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  • $\begingroup$ Did you get what I wrote in my answer? It might seem a bit strange, but once you get used to manipulating $\epsilon$ when speaking about limit it's not very difficult $\endgroup$ – mvggz Nov 7 '14 at 14:19
  • $\begingroup$ Yeah,thanks so much ,I always have difficulties with proving clear and obvious facts that analysis wants to proves and how to write it,but you did it well and your answer is clear to me ! $\endgroup$ – haleh Nov 7 '14 at 14:45
  • $\begingroup$ excuse me !how could you conclude $|a_n-a_{n+p}|\lt e_n$? I think your last sentence is not enough to prove this? $\endgroup$ – haleh Nov 7 '14 at 15:18
  • $\begingroup$ I've edited my answer, and deleted my last comment. I think it's clearer this way, introducing the sup of $ |a_n-a_{n+p}| $ $\endgroup$ – mvggz Nov 10 '14 at 10:34
  • $\begingroup$ I don't know what Omar Hijab wrote, but your first formula line is inintelligible, resp., doesn't make sense with all interpretations I tried. $\endgroup$ – Christian Blatter Nov 10 '14 at 10:54
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1) => 2) is quite obvious: $(e_n) \rightarrow 0 $ => For any $ \epsilon >0 $, there is a N such as: $ n \geq N $ => $ e_n \leq \epsilon $

=> $\forall\epsilon\gt0 \space\space\exists N\in\Bbb N \space \space s.t \space \space \forall m\in\Bbb N ;\forall n\ge N \space\space |a_{n+m}-a_n|\lt\epsilon $

That is the same as: $\forall\epsilon\gt0 \space\space\exists N\in\Bbb N \space \space s.t \space \space\forall m,n\ge N \space\space |a_n-a_m|\lt\epsilon$

2) => 1) : Starting with : $\forall\epsilon\gt0 \space\space\exists N\in\Bbb N \space \space s.t \space \space\forall m,n\ge N \space\space |a_n-a_m|\lt\epsilon$

Since one can switch the role of n and m, we can assume that m> n . Let's write m=n+p

2) <=> $ \forall\epsilon\gt0 \space\space\exists N\in\Bbb N \space \space s.t \space \space\forall p\in\Bbb N,n\ge N \space\space |a_n-a_{n+p}|\lt\epsilon $

Here the last condition allows you to define : $\epsilon_n = sup(|a_n-a_{n+p}|;p\in\Bbb N)$ since this sequence of p is always bounded.

You then have by definition : $ n \leq N : |a_n-a_{n+p}| \leq \epsilon_n $ ; $ n > N : |a_n-a_{n+p}| \leq \epsilon_n \leq \epsilon $

Hence this sequence can be dominated by an other sequence $(\epsilon_n)$ that $\rightarrow 0$, which is your first definition.

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