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Is there a closed form for the harmonic series?

  1. Feb 28, 2009 #1
    Taking from Euler's offering that:

    [tex]\Sigma[/tex]1/n= ln(n) + [tex]\gamma[/tex]

    could you say that there is a closed form of the harmonic series?

    Does Euler's offering qualify?
     
  2. jcsd
  3. Feb 28, 2009 #2

    lurflurf

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    Your = is not the usual =, but the asymptomic =. As such it is not a closed form in the usual sense. To talk of a closed form one needs to define which functions and operations are to be acceptable and as in this case in what sense two forms are to be equal. The harmonic series in usually not considered to have a closes form as it can not be written in terms of the usual function with the usual =.
     
  4. Feb 28, 2009 #3
    Could you please explain a little more about this asymptotic equivalence?

    Would it be more accurate to include the summation of a "Kth" Bernoulli number, divided by the product of "K" and (n raised to the power of "K"), as K goes from one to infinity?

    But, including this summation would not constitute closed form, correct?
     
  5. Mar 1, 2009 #4

    HallsofIvy

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    No, Euler never said any such thing. The equation you write is obviously false. On the left, you have a sum with n varying over some unstated domain so the sum itself does NOT depend on n, but the right hand side does.

     
  6. Mar 1, 2009 #5

    lurflurf

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    H(n)= ln(n) + Euler–Mascheroni
    where Hn is the nth harmonic number
    would better be written
    H(n)~ln(n) + Euler–Mascheroni
    to avoid confusion
    interpeted as
    limit [-H(n)+ ln(n) + Euler–Mascheroni]=0

    That is to say that while the two sides are never equal the get closer the large n gets

    and no a integral or a sum (finite or infinite) would not be considered a closed form for harmonic numbers
     
  7. Mar 1, 2009 #6
    Thank you! That clarifies.
     
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