# Math History: Cauchy Criterion for Sequence/Series

Shackleford
I know the Cauchy criterion for a convergent sequence. A Cauchy sequence is one in which the distance between successive terms becomes smaller and smaller. You can find a number N such that the terms after that, pairwise, have a a distance that is less than epsilon.

After looking at an example in the book, I was able to write this down. It would appear that the sequence converges to zero since the numerator is bounded by -1 and 1. It looks like the Cauchy criterion for convergent series is satisfied too since you can make m and n and a function of epsilon. However, I'm not too sure about my work.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111116_173313.jpg [Broken]

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Homework Helper
You could just explain what a Cauchy sequence is just by giving the definition, though your description is reasonable.

As for you working can you walk me through what you have done, its not clear in the photo. I find it easier if you type the work in herer, the nI can just cut paste s& edit as required

Shackleford
You could just explain what a Cauchy sequence is just by giving the definition, though your description is reasonable.

As for you working can you walk me through what you have done, its not clear in the photo. I find it easier if you type the work in herer, the nI can just cut paste s& edit as required

Yeah, when I write up the homework I'll simply put down the definition of a Cauchy sequence.

I wrote down the difference of the sm and sn terms with m > n. If I correctly setup the relation, it seems that the terms are less than (m-n)/2m since the numerator is bounded and the denominator becomes increasingly larger. If that's true, then I assert you can find an arbitrary epsilon by making m, n a function of epsilon.

kru_
If m > n, then the terms on the left would be strictly less than (m-n)/2^n, correct? Since making the denominator smaller makes the term bigger, you want the smaller of the two denominators to serve as your upper bound.

Shackleford
If m > n, then the terms on the left would be strictly less than (m-n)/2^n, correct? Since making the denominator smaller makes the term bigger, you want the smaller of the two denominators to serve as your upper bound.

Right. So I'm thinking this series is convergent.

Shackleford
Actually, I think I have it backwards. I'm changing n > m.

It looks like sn - sm < (n-m)/2m.

So, for an arbitrary epsilon I should be able to find an N such that m,n > N implies the difference is less than epsilon.

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Homework Helper
I know the Cauchy criterion for a convergent sequence. A Cauchy sequence is one in which the distance between successive terms becomes smaller and smaller.
This statement is untrue. For example, the series
$$\sum \frac{1}{n}$$
has "distance between successive termsj"
$$\frac{1}{n}- \frac{1}{n+1}= \frac{1}{n(n+1)}$$
which goes to 0 but is NOT a Cauchy sequence.

You can find a number N such that the terms after that, pairwise, have a a distance that is less than epsilon.
Okay, this is better. The distance between terms, pairwise, is $|a_n- a_m|$ and that must go to 0, for any m and n, not just $|a_{n+1}- a_n|$

After looking at an example in the book, I was able to write this down. It would appear that the sequence converges to zero since the numerator is bounded by -1 and 1. It looks like the Cauchy criterion for convergent series is satisfied too since you can make m and n and a function of epsilon. However, I'm not too sure about my work.

http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-1.png

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20111116_173313.jpg [Broken]

Last edited by a moderator:
Shackleford
This statement is untrue. For example, the series
$$\sum \frac{1}{n}$$
has "distance between successive termsj"
$$\frac{1}{n}- \frac{1}{n+1}= \frac{1}{n(n+1)}$$
which goes to 0 but is NOT a Cauchy sequence.

Okay, this is better. The distance between terms, pairwise, is $|a_n- a_m|$ and that must go to 0, for any m and n, not just $|a_{n+1}- a_n|$

The book works your example, too. Going by the definition for my problem

$|s_{n}- s_m|$ < (n-m)/2m

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Shackleford
The relation in my previous post looks wrong. Using some convergence tests, I'm fairly certainly this series does converge. I found my error. This should be correct.

$|s_{n}- s_m|$ < [1/2n] + [1/2m]

The sequence converges to zero which implies the series converges.

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