1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook