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Cauchy sequence proof

  1. Jun 14, 2010 #1
    1. The problem statement, all variables and given/known data

    For each n \in N, let s_n = 1 + 1/2 + ... + 1/n. By considering s_2n - s_n, prove that {s_n} is not Cauchy.

    2. Relevant equations

    3. The attempt at a solution

    I know that s_2n - s_n = (1 + 1/2 + ... + 1/n + 1/(n+1) + ... + 1/2n) - (1 + 1/2 + ... + 1/n)
    = (1/(n+1) + 1/(n+2) + ... + 1/2n
    > 1/2n + 1/2n + ... + 1/2n
    = n * 1/2n = 1/2

    Let \epsilon > 0. Then I need to find N \in N such that m, n >= N. So if I let N = 1/2...

    And that's where I lose it. These don't work like regular epsilon proofs!
  2. jcsd
  3. Jun 14, 2010 #2
    Wait, are you sure of what you need to find? For any epsilon you need to find an N such that IF m,n > N, THEN | s_m - s_n | < epsilon.

    You just proved that for any m, s_2m is more than 1/2 away from s_m. So you win. Just choose epsilon < 1/2.
  4. Jun 14, 2010 #3
    Thank you so much; I think that was definitely part of my problem. I am also a bit unsure how to write it up in a formal way. Given what I already had, would the following finish it off?

    Let 0 < epsilon < 1/2. Then, for any m, n >=N, |s_2n - s_n| >= 1/2. Therefore {s_n} is not Cauchy.

    And I don't really need to say anything about the value of N, right?
  5. Jun 14, 2010 #4
    Well you should say something like "for any N" since you haven't identified N in your proof. Basically you are saying: no matter what N I choose, this relationship always exists for m,n greater than N. Thus NO N satisfies the condition for the given epsilon, hence the sequence isn't Cauchy.

    In doing these proofs I often like to think of it in an adversarial way. I'm trying to prove that a sequence isn't Cauchy. So now I get to pick an epsilon. Then my opponent, a very mean man who hates undergraduate math students, gets to pick a number N. Then I get to pick m and n > N. If | s_m - s_n | > epsilon, I win. If not, I lose.

    Now if I can play this game such that my opponent can never win, then the sequence isn't Cauchy. On the contrary, if he can always respond to my pick of some epsilon with an N that beats me, then the sequence is Cauchy.

    I'm a game theory kind of guy though, so maybe others prefer to just "for any" and "there exists".
  6. Jun 15, 2010 #5
    You should show:

    [tex]\exists\ \epsilon = (put\ here\ the\ right\ value)\ so\ \forall\ N\ there\ are\ m>n>N\ so\ that:\ |\sum_{k=n+1}^{m}a_k|\geq\epsilon [/tex]

    The most important thing is to understand and "feel" definitions and their meaning. You can't progress further before fully grasping them.

    This one is good :rofl:
    Last edited: Jun 15, 2010
  7. Jun 15, 2010 #6
    Thanks so much for your analogy! That really helps a lot.
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