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Math Help - Inequalities

  1. #1
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    Inequalities

    QUESTION

     \forall x > 2

    Use the fact that  \frac{1}{x} < \log x - \log (x-1) < \frac{1}{x-1}

    to show that  s_n - 1 < \log x < s_{n-1} < s_n

    where s_n = \sum^n_{n=1} \frac{1}{n}

    i.e. s_n = 1 + .... + \frac{1}{n}




    __________________________________________________ ________


    MY THOUGHTS:


    Clearly  s_n - s_{n-1} = \frac{1}{n} > 0 \Rightarrow s_n > s_{n-1}

    Also,  s_n - 1 = \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} < \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n-1} + 1 = s_{n-1}  \Rightarrow s_n - 1 < s_{n-1}

    But I don't know how to show that log x fits in the middle of that inequality. Please help.
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  2. #2
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    Have you noticed that \sum\limits_{k = 2}^n {\left[ {\log (k) - \log (k - 1)} \right]}  = \log (n)?
    Think collapsing sums.
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  3. #3
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    Not until now.

    Excellent, thank you very much. Solved it.
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