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Math Help - MAX prove question..

  1. #1
    MHF Contributor
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    MAX prove question..

    An is a bounded sequence which converges to 0.
    An>0 for every n.
    prove that in this sequence we have a maximal member
    ??

    again its obvious because if a sequence is descending and it converges from a positive number
    to 0.
    so 0 is the larger lower bound
    and our first member must be the larger

    how to transform it to math?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by transgalactic View Post
    and our first member must be the larger
    A sequence which is convergent and bounded can be non decreasing. For example the maximal member of (A_n)=\left(\mathrm{e}^{-(n-10)^2}\right)_{n\in\mathbb{N}} is A_{10}=\mathrm{e}^{-(10-10)^2}=1>A_0...
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  3. #3
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    If A_n has no maximal term then \left( {\exists N_1 } \right)\left[ 0<{A_1 < A_{N_1 } } \right].
    Moreover, \left( {K > 1} \right)\left( {\exists N_k } \right)\left[ {A_{N_{K - 1} } < A_{N_k } } \right].
    That simply means that the sub sequence \left(A_{N_n}\right) cannot converge to 0.
    That is a contradiction.
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