# A sequence is Cauchy?

1. Mar 2, 2009

### soopo

1. The problem statement, all variables and given/known data
What does it mean when a sequence is Cauchy?

2. Mar 2, 2009

### HallsofIvy

A sequence of real numbers is a "Cauchy sequence" if and only if |an- am| goes to 0 as m and n go to infinity independently: given $\epsilon> 0$ there exist N such that if m and n are both > N, then $|a_n- a_m|< \epsilon$.

The advantage of working with Cauchy sequences is that (1) even if our sequence is of points in some abstract space, the "distance between points", here |p- q|, is a real number so we are now working with sequences of real numbers and (2) we know what we want the sequence to converge to.

Of course, for that to be useful, we have to know that the "Cauchy Criterion", that every Cauchy sequence converges, holds in our space- that our space is complete. That has to be proven separately. For exampe the set of real numbers is complete but the set of rational numbers is not. The sequence 3, 3.14, 3.141, 3.1415, 3.14159, ..., where each number contains one more digit in the decimal expansion of $\pi$ is a sequence of rational numbers (each number is a terminating decimal) and a Cauchy sequence (if m,n> N, am and an are identical for at least the first N decimal places so |am- an|< 10-N which goes to 0 as N goes to infinity) but does not converge to any rational number.

Last edited by a moderator: Mar 3, 2009
3. Mar 2, 2009

### soopo

So Cauchy sequence occurs when
If $$\forall \epsilon > 0 \exists N, m > N$$ and $$n > N$$, then
$$|a_{n} - a_{m}| < \epsilon.$$

Last edited: Mar 3, 2009
4. Mar 2, 2009

### fluidistic

I think you meant as m and n go to infinity.

5. Mar 2, 2009

### JG89

Halls, I think you mean |a_n - a_m| < epsilon

6. Mar 3, 2009

### HallsofIvy

Yes, of course. I'll go back and edit so I can pretend I never made those mistakes!