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Explicit Formula for Sum of Series

  1. Apr 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine an explicit formula for the sum of n terms for the given series:

    1, [tex]\frac{1}{2}[/tex], [tex]\frac{1}{3}[/tex], [tex]\frac{1}{4}[/tex], [tex]\frac{1}{5}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I calculated the first 5 terms for the sum sequence and got:

    1, [tex]\frac{3}{2}[/tex], [tex]\frac{11}{6}[/tex], [tex]\frac{25}{12}[/tex], [tex]\frac{137}{60}[/tex]

    but I can't find a pattern to determine an explicit formula.
  2. jcsd
  3. Apr 20, 2008 #2
    Maybe I'm misunderstanding your question, but wouldn't it just be [tex]\sum_{n=1}^\infty 1/n[/tex] ?
  4. Apr 20, 2008 #3
    The question asks for n terms, not an infinite number of terms
  5. Apr 20, 2008 #4
    Oh, sorry. I read the original post too quickly.
  6. Apr 20, 2008 #5
    That's alright, nothing to bash heads in about! As for the problem, it's a doozy...
  7. Apr 20, 2008 #6
    You're right, it is a doozy! This series is known as a harmonic series, and according to this site, "there is no simple formula, akin to the formulae for the sums of arithmetic and geometric series, for the sum."
  8. Apr 20, 2008 #7
    Well then wouldn't it just be:

    [tex]\sum_{k=1}^n \frac{1}{k}[/tex]

    [tex] S_1=1[/tex]




    blah, that doesn't actually seem to work!!
  9. Apr 20, 2008 #8
    I know! It's whack!

    I started out trying to find formulas for numerator and denominator separately. In cases where it doesn't simplify, the denominator is n! . As for the numerator, aside from the obvious (n! + n!/2 + n!/3+...+n!/n) I do not see how to arrive at a formula.
  10. Apr 20, 2008 #9


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    According to http://mathworld.wolfram.com/HarmonicSeries.html you can write it as a sum of the Euler-Mascheroni constant and a digamma function. I'm guessing if there were an easier expression they would have mentioned it. I think steelphantom is right, there is no elementary formula.
  11. Mar 2, 2009 #10
    I think all it is is sigma from k=2 to n of 1/n. very simple that would start off with one half then one third then one quarter and so on.
  12. Mar 2, 2009 #11

    Tom Mattson

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    That much is obvious. The trouble is finding a formula for the nth partial sum.
  13. Feb 20, 2010 #12
    I think it's impossible, infinity would be the common denominator.
  14. Feb 20, 2010 #13


    Staff: Mentor

    Not for the sum of n (a finite number) terms.
  15. Feb 21, 2010 #14
    The denominator ought to be the lcm(1,2,...,n) whereas the numerator is the puzzle.
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