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Math Help - Prove that xn<= L for all n in N

  1. #1
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    Prove that xn<= L for all n in N

    Assume that (x_n) is a sequence satisfying: For every n∈N there is an integer m>n such that xm >xn. Also assume that (x_n) converges to L.

    Prove that xn≤L for all n∈N.

    I started out stating what I had:

    Have: ( \foralln \inN) \existsk such that x_m>x_n and (xn)->L which means: \forall \epsilon>0 \existsk such that for all n \geqk we have |x_n-L|< \epsilon

    and we need: x_n≤L


    How would I put that info into a proof and would it be by contradiction?
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  2. #2
    Senior Member Tinyboss's Avatar
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    Assume there is an n such that xn>L. Then show there are infinitely many m>n such that xm>xn. This contradicts convergence of xn to L.
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  3. #3
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    Quote Originally Posted by Tinyboss View Post
    Assume there is an n such that xn>L. Then show there are infinitely many m>n such that xm>xn. This contradicts convergence of xn to L.
    How would this be shown if we wanted strictly less than?
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  4. #4
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    If we know that (x_n)\to L then for any \delta>0 almost all the terms are in (L-\delta,L+\delta).
    Suppose that x_N>L for some N.
    Then let \delta=\frac{x_N-L}{2}.
    Now apply the almost all property and get a contradiction.
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