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Math Help - Summation of 1/n

  1. #1
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    Summation of 1/n

    How do we prove that  \sum\limits_{n \leq x} \frac{1}{n} = \log{x} + C + O(\frac{1}{x})           .
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Are you familiar with the Euler-MacLaurin formula or partial summation?
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Let  S(x) = \sum_{n\leq x} 1 = \lfloor x \rfloor .

    By partial summation we get

     \sum_{n\leq x} \frac{1}{n} = \frac{S(x)}{x} + \int_{1}^{x} \frac{S(t)}{t^2} dt .

     = \frac{\lfloor x \rfloor}{x} + \int_{1}^{x} \frac{\lfloor t \rfloor}{t^2} dt .

     = 1-\frac{\{ x \} }{x} + \log(x) + \int_{1}^{x} \frac{\{ t \} }{t^2} dt .

     = 1-\frac{\{ x \} }{x} + \log(x) + \int_{1}^{\infty} \frac{\{ t \} }{t^2} dt - \int_{x}^{\infty} \frac{\{ t \} }{t^2} dt.

     = \log(x) + C + O\left(\frac{1}{x}\right) + O\left(\int_{x}^{\infty} \frac{1}{t^2} dt\right) .

     = \log(x) + C +O\left(\frac{1}{x} \right) .


    Just to clarify,  C = 1+\int_{1}^{\infty} \frac{\{ t \} }{t^2} dt

    and  O\left(\frac{1}{x} \right) = -\frac{\{ x \} }{x}- \int_{x}^{\infty} \frac{\{ t \} }{t^2} dt .
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    On a side note, this also proves  \int_{1}^{\infty} \frac{\{ t \} }{t^2} dt = 1-\gamma .
    Last edited by chiph588@; December 27th 2009 at 03:55 PM.
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