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Math Help - Another summation (or geometric?) Series

  1. #1
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    Another summation (or geometric?) Series

    Hello everyone,

    I need the formula of the following series:

    i = 0k - 1 n/[(log n) - i] = ? for all n != 0 and n != 2i

    Thanks!
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  2. #2
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    Re: Another summation (or geometric?) Series

    Quote Originally Posted by mohamedennahdi View Post
    Hello everyone,

    I need the formula of the following series:

    i = 0k - 1 n/[(log n) - i] = ? for all n != 0 and n != 2i

    Thanks!
    Just so we're clear, is this the sum you're trying to evaluate?

    \displaystyle \begin{align*} \sum_{i = 0}^{k - 1}{\frac{n}{\log{(n)} - i}} \end{align*} for \displaystyle \begin{align*} n \neq \left\{ 0, 2^i \right\} \end{align*}
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  3. #3
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    Re: Another summation (or geometric?) Series

    Yes.
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  4. #4
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    Re: Another summation (or geometric?) Series

    Wolfram gets this.
    Thanks from mohamedennahdi
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  5. #5
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    Re: Another summation (or geometric?) Series

    This one is tough!
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  6. #6
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    Re: Another summation (or geometric?) Series

    Is there any one to agree that:

    i=0(log n - 1) n / ((log2 n) - i) = n (j=1log n 1 / j) ?
    Last edited by mohamedennahdi; July 26th 2012 at 07:33 PM.
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  7. #7
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    Re: Another summation (or geometric?) Series

    Quote Originally Posted by mohamedennahdi View Post
    Is there any one to agree that:

    i=0(log n - 1) n / ((log2 n) - i) = n (j=1log n 1 / j) ?
    No I completely disagree. The counter in the sum can only take integer values.
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  8. #8
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    Re: Another summation (or geometric?) Series

    You are right, the last suggestion is correct only when N is a power of the base (base 2 in my case).
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