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Bounded Sequence, Cauchy

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove that if {a[tex]_{n}[/tex]} is a sequence of rational numbers such that {a[tex]_{n+1}[/tex]} > {a[tex]_{n}[/tex]} for all n [tex]\in[/tex] [tex]\textbf{N}[/tex] and there exists an M[tex]\in[/tex] [tex]\textbf{Q}[/tex] such that {a[tex]_{n}[/tex]} [tex]\leq[/tex] M for all n [tex]\in[/tex] [tex]\textbf{N}[/tex], then {a[tex]_{n}[/tex]} is a Cauchy sequence of rational numbers.

    2. Relevant equations
    Do not use the least upper bound property.

    A sequence is Cauchy in the rational numbers if [tex]\exists[/tex] an N [tex]\in[/tex][tex]\textbf{N}[/tex], such that |{a[tex]_{n}[/tex]} - {a[tex]_{m}[/tex]} | < [tex]\epsilon[/tex] for all n, m [tex]\geq[/tex] N.

    If a sequence converges, it is Cauchy.

    3. The attempt at a solution

    I understand why this is true, but I am having trouble formulating the math to do a proof behind it. I can see that if the sequence never gets bigger than M, and that it is strictly increasing, the sequence must start converging and be Cauchy, but I'm kind of confused at how to start doing the epsilon stuff.


    Edit: I'm not sure why those are showing up as superscripts. They are supposed to be subscripts.
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2
    someone please heeeeeeelp
  4. Sep 18, 2007 #3
    try a proof by contradiction.
    suppose [tex] a_n [/tex] is no cauchy sequence.For a fixed [tex] \epsilon > 0 [/tex]show there is a subsequence [tex] b_n [/tex] of [tex] a_n[/tex] such that
    [tex] b_{n+1} > b_n + \epsilon [/tex]
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