# Cauchy's convergence criterion

Prove Cauchy's Convergence Criterion.

So I need to show that the partial sum $s_{n}=\sum^{n}_{i=0}a_{i}$ is Cauchy. $\rightarrow\forall\epsilon>0,\exists N$ s.t. $\forall n,m>N |s_m-s_n|<\epsilon$.

My guess is that I look at the fact that $s_n=\sum^{n} |s_n|=|a_{1}+...+a_{n}|$ and find a way to get $|s_m-s_n|<\epsilon$.

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I honstly do not understand what you are asking. You start by ordering us to prove Chauchy's convergence criterion, but then your second paragraph does not seem related to that at all. – Mariano Suárez-Alvarez Jan 25 '12 at 5:15
In any case, Cauchy's convergence criterion is proved in pretty much every textbook that deals with series and/or sequences. Have you tried looking at one and/or asking your instructor? While this site can be an immensely useful resource, you have to do a great deal of work yourself for it to be actually useful and —as far as I can tell, at least—that does not seem to be the case. – Mariano Suárez-Alvarez Jan 25 '12 at 5:16
What do you mean by $s_n=\sum^{n} |s_n|=|a_{1}+...+a_{n}|$? – leo Jan 25 '12 at 5:25

Consider the partial sum in $\mathbb{R}$

$$s_n = \sum_{i=1}^n a_i.$$

Since $(\mathbb{R},|\cdot|)$ is complete, the partial sums converge iff they are Cauchy sequences. So for a convergent series this means that for any given $\epsilon > 0$

$$\exists N_0 \in \mathbb{N} : \forall n,m > N_0 \implies \left|\sum_{i=0}^n a_i -\sum_{i=0}^m a_i\right| < \epsilon.$$

Then all you have to do is observe that the difference for $n>m$ say is

$$\sum_{i=0}^n a_i -\sum_{i=0}^m a_i = \sum_{i=m+1}^n a_i$$

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