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Math Help - Prove that the sequence is not bounded above

  1. #1
    MHF Contributor alexmahone's Avatar
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    Prove that the sequence is not bounded above

    Prove that a_n=\frac{1\cdot 3\cdots(2n+1)}{2\cdot 4\cdots(2n)} is not bounded above.
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    Super Member girdav's Avatar
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    Re: Prove that the sequence is not bounded above

    We use \ln(1+x)\geq x-\frac{x^2}2 for x\geq 0 to get \ln a_n=\sum_{j=1}^n\ln\left(1+\frac 1{2j}\right)\geq \sum_{j=1}^n\frac 1{2j}-\frac 12\sum_{j=1}^n\frac 1{(2j)^2}=\frac 12\sum_{j=1}^n\frac 1j-\frac 18\sum_{j=1}^n\frac 1{j^2}.
    Since \lim_{n\to \infty}\sum_{j=1}^n\frac 1j=+\infty and \lim_{n\to \infty}\sum_{j=1}^n\frac 1{j^2}<+\infty, we get the result.
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