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\sum 1/log(n) goes like ?

  1. Aug 14, 2011 #1

    In my probability course it is used that [itex]\frac{1}{n} \sum_{k=3}^n \frac{1}{\log k} \propto \frac{1}{\log n}[/itex].

    I hope I'm not being daft: can somebody enlighten me on why this is so?
  2. jcsd
  3. Aug 15, 2011 #2

    a method to find a equivalent is to border the sum by integrals (in attachment)

    Attached Files:

  4. Aug 15, 2011 #3


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    Are you sure jjacquelin that this series 1+1/x+2/x^2+6/x^3 converges? I take it to be sum_k k!/x^k.

    I took a similar approach by integrating by parts, but sadly also ended up with a divergent series. I think this is a non-trivial question to prove from scratch.
  5. Aug 15, 2011 #4
    We don't consider the infinite series which doesn't converge. We consider a limited number of the first terms in order to derive an equivalent. This is the usal way for asymptotic series.
    In fact, in my previous attachment, where a typographic character which is not exactly the good one. In the asymptotic formula the "equal" should be the "tilde". (see attachment)

    Attached Files:

  6. Aug 15, 2011 #5
    Thank you
  7. Aug 15, 2011 #6


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    I have not seen that before, thank you for the explanation.
  8. Sep 12, 2011 #7

    what is the summation of 1/log(n) + 1/log(n-2) + 1/log(n-4) ..

  9. Sep 13, 2011 #8


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    Please do not "hijack" someone elses thread to ask a different questions- start your own thread.

    Also, for fixed n, n- i will eventually be 0 so if you mean an infinite sum, that will not exist. I presume you mean
    [tex]\sum_{i= 0}^{n-1} \frac{1}{ln(n- i)}[/tex]
  10. Sep 13, 2011 #9
    sorry abt that ..
    actually i wanted to know the summation
    ∑ 1/lg(n−2i)
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