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Sum of a finite series

  1. Jun 20, 2012 #1
    Just wondering, is there a way to sort of "collapse" a finite series (to get the sum) that isn't classified as arithmetic, geometric or a p-series.
     
  2. jcsd
  3. Jun 20, 2012 #2

    Mark44

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    Not that I'm aware of.
     
  4. Jun 21, 2012 #3

    HallsofIvy

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    Since you use the word "collapse", there is always the "collapsing" or "telescoping series". For example, it is easy to show that the series [tex]\sum_{i=1}^n \frac{1}{n^2+ n}[/tex] sums to [itex]1- 1/(n+1)[/itex]. That is because, using "partial fractions", we can rewrite [tex]\frac{1}{n^2+ n}= \frac{1}{n}- \frac{1}{n+1}[/tex] so that each "1/k" term reappears as "-1/(k+1)" and cancels. The only terms that survive are the first, 1, and the last, -1/(n+1).
     
  5. Jun 21, 2012 #4
    What kind of a series are you solving???
     
  6. Jun 21, 2012 #5
    I'm solving a series with just one coefficient, n, and an exponent. I'm aware that I can bring the coefficient "through" the integral and solve from there but I don't know what sort of formula to use.
    I also know I can just write it out.
     
  7. Jun 21, 2012 #6
    I was hoping you would bring that up. For some reason, I really like telescoping series, but my series doesn't fall under it's description.
     
  8. Jun 21, 2012 #7

    Ray Vickson

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    What, exactly, is the series? Is it [itex]S_p = \sum_{n=1}^N n^p,[/itex], or is it something else? It the series I wrote is the one you want, you are out of luck: for general p there is NO formula (although, of course, for integers p = 0,1,2,3, and a few more, you can derive results). However, the results get increasingly complicated as p increases. For example, when p = 3 we get
    [tex] S_3 = \frac{1}{4} N^2 (n+1)^2,[/tex]
    but for p = 10 we get
    [tex] S_{10} = \frac{1}{66} N(N+1)(2N+1)(N^2-N-1)(3N^6+9N^5+2N^4 -11N^3 +3N^2 +10N-5),[/tex]
    and it gets worse for larger p.

    RGV
     
  9. Jun 21, 2012 #8

    Ray Vickson

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    What, exactly, is the series? Is it [itex]S_p = \sum_{n=1}^N n^p,[/itex], or is it something else? It the series I wrote is the one you want, you are out of luck: for general p there is NO formula (although, of course, for integers p = 0,1,2,3---that are not really large---you can derive results). However, the results get increasingly complicated as p increases. For example, when p = 3 we get
    [tex] S_3 = \frac{1}{4} N^2 (N+1)^2,[/tex]
    but for p = 10 we get
    [tex] S_{10} = \frac{1}{66} N(N+1)(2N+1)(N^2-N-1)(3N^6+9N^5+2N^4 -11N^3 +3N^2 +10N-5),[/tex]
    and it gets worse for larger p. For fractional values of p, such as p = 1/2, etc., there are no known formulas. Basically, when we have such a series with known numerical values of N and p we just evaluate the sum numerically.

    RGV
     
  10. Jun 21, 2012 #9
    Yes, it is the series you wrote: p stays the same, n is the term.
    Thanks everyone.
     
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