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Cauchy Sequence

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    3. 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: Oct 6, 2008
  2. jcsd
  3. Oct 6, 2008 #2


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    Homework Helper

    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?
  4. Oct 6, 2008 #3
    Thanks, now I understand that definition/concept. Incidentally, I also understand the contractive sequence concept now.
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