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Homework Help: Cauchy sequnce and convergence of a non-monotonic sequence.

  1. May 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello,
    I have a question concerning convergence of the non-monotonic sequences which takes place when the Cauchy criterion is satisfied.
    I understand that [tex]|a_n - a_m| <ε[/tex] for all n,m[tex]N\ni[/tex]

    2. Relevant equations
    What I don't see is how [tex](a_{n+1} - a_n) →0[/tex]is not equivalent/enough. Doesn't the fact that the difference tends to zero mean that it can eventually be smaller than any ε?

    3. The attempt at a solution
    I know the examples when this doesn't work, like [tex](\sqrt{n})[/tex] but I don't understand the difference between these two criteria.

    Could anyone explain why they aren't equivalent?
     
  2. jcsd
  3. May 10, 2012 #2
    So let's take the example of an=√n. For large n, you have then [itex]a_{n+1}-a_n \sim n^{-1/2} \rightarrow 0[/itex] when n→∞. We can write the Cauchy criterion as (let p = n-m)
    [tex] \epsilon > a_{m+p}-a_m = \sum_{n=m}^{m+p} a_{n+1}-a_n [/tex]
    So now when n is very large, we have
    [tex] \epsilon > \sum_{n=m}^{m+p} a_{n+1}-a_n \sim \sum_{n=m}^p n^{-1/2} \sim \int_m^{m+p} dn n^{-1/2} = (\sqrt{m+p}-\sqrt{m})/2 [/tex]
    and this grows without limit if you let p to grow, so it can't be smaller than ε for all p.
     
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