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Homework Help: 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 http://plus.maths.org/issue12/features/harmonic/index.html" [Broken], "there is no simple formula, akin to the formulae for the sums of arithmetic and geometric series, for the sum."
     
    Last edited by a moderator: May 3, 2017
  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]

    [tex]S_2=1+\frac{1}{2}=\frac{2+1}{2}[/tex]

    [tex]S_3=1+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}[/tex]

    [tex]S_4=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{12+6+4+3}{12}[/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

    Dick

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    Homework Helper

    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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    Staff Emeritus
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    Gold Member

    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

    Mark44

    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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