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Math Help - Limits regarding sequences

  1. #1
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    Limits regarding sequences

    I have proven that if {an} and {bn} are 2 sequences such that an >= bn for each n>n0 , for some n0 belongs to positive integers with limit of an when n goes to infinity is L and limit of bn when n goes to infinity is M , then L >= M by using a contradiction proof.

    My problem is I think that we can say, if an > bn for each n > n0 ,for some n0 belongs to positive integers, then L > M. But I'm having a difficulty in proving that claim. I have tried to prove it in the same way by using the contradiction proof but I get stuck because then I assume L <= M and as in earlier proof when I get (M-L)/2 as epsilon I get an epsilon which is >= 0 but epsilon should be strictly greater than 0.

    So, is there another way of proving this? Please can somebody help?
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  2. #2
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    Re: Limits regarding sequences

    Quote Originally Posted by Kristen111111111111111111 View Post
    I have proven that if {an} and {bn} are 2 sequences such that an >= bn for each n>n0 , for some n0 belongs to positive integers with limit of an when n goes to infinity is L and limit of bn when n goes to infinity is M , then L >= M by using a contradiction proof.
    To prove this by contradiction assume that M>L. Then you do know that \epsilon= \frac{M-L}{2}>0.

    So almost all b_n>a_n .
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  3. #3
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    Re: Limits regarding sequences

    Quote Originally Posted by Kristen111111111111111111 View Post
    My problem is I think that we can say, if an > bn for each n > n0 ,for some n0 belongs to positive integers, then L > M.
    This is false. Consider a_n = 1/n and b_n = 1/n^2 (ignoring n = 1).
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  4. #4
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    Re: Limits regarding sequences

    Well I guess what I thought is wrong when considering your example.. Thank you so much for helping
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  5. #5
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    Re: Limits regarding sequences

    In general, the best you can say for limits is \le or \ge. That is, if a_n< b_n (strictly less than) for all n> n_0, then, using the indirect proof that Plato suggested, we can show that \lim a_n\le \lim b_n (less than or equal to).
    Last edited by HallsofIvy; August 28th 2013 at 07:13 AM.
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