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Cauchy Sequence

  • Thread starter Kudaros
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  • #1
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Homework Statement



Define the Sequence an= n1/2 where n is a natural number.

Show that |an+1-an| -> 0 but an is not a cauchy sequence

Homework Equations





The Attempt at a Solution



(Ignore this paragraph)Well, unfortunately I am stuck on the very first part. How exactly do I evaluate the limit as n -> infinity of |(n+1)^(1/2) - n^(1/2)| ? any hint at a trick would be most welcome (unless of course I am not seeing something that is obvious).

As for the rest, I need to show an is not a Cauchy sequence. The definition of a Cauchy sequence uses two sequences with different subscripts, m and n. In this case, can I take n+1 to be m and keep n as itself?

I think I need to show that the distance between the two sequences, an+1 and an is not decreasing as n becomes large.

edit: found limit.

Also, re-reading the question I see a flaw in my above statement. I just need to show that for any given m and n, as I vary them independently, they do not meet the cauchy criterion. The problem itself states that looking only at the n+1 term appears to meet the criterion, but in fact does not.

Confirm or deny?
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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To evaluate the limit multiply by (sqrt(n+1)+sqrt(n))/(sqrt(n+1)+sqrt(n)) and simplify. And to show it's not Cauchy, no, don't take m=n+1. You're going to show that goes to zero. Take m MUCH bigger than n. Say 4 times n?
 
  • #3
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Thanks, now I understand that definition/concept. Incidentally, I also understand the contractive sequence concept now.
 

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