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Math History: Cauchy Criterion for Sequence/Series

  • #1
1,654
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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]
 
Last edited by a moderator:

Answers and Replies

  • #2
lanedance
Homework Helper
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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
 
  • #3
1,654
2
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.
 
  • #4
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0
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.
 
  • #5
1,654
2
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.
 
  • #6
1,654
2
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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  • #7
HallsofIvy
Science Advisor
Homework Helper
41,808
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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
[tex]\sum \frac{1}{n}[/tex]
has "distance between successive termsj"
[tex]\frac{1}{n}- \frac{1}{n+1}= \frac{1}{n(n+1)}[/tex]
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 [itex]|a_n- a_m|[/itex] and that must go to 0, for any m and n, not just [itex]|a_{n+1}- a_n|[/itex]

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:
  • #8
1,654
2
This statement is untrue. For example, the series
[tex]\sum \frac{1}{n}[/tex]
has "distance between successive termsj"
[tex]\frac{1}{n}- \frac{1}{n+1}= \frac{1}{n(n+1)}[/tex]
which goes to 0 but is NOT a Cauchy sequence.


Okay, this is better. The distance between terms, pairwise, is [itex]|a_n- a_m|[/itex] and that must go to 0, for any m and n, not just [itex]|a_{n+1}- a_n|[/itex]
The book works your example, too. Going by the definition for my problem

[itex]|s_{n}- s_m|[/itex] < (n-m)/2m
 
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  • #9
1,654
2
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.

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

The sequence converges to zero which implies the series converges.
 
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