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Series convergence problem

  1. Dec 27, 2008 #1
    what are the conditions on [tex]\Gamma_{2n+1} [/tex] so as for the series to converge ?

    [tex]m\pi=\sum^{\infty}_{n=0} \frac{(-1^{n}) \Gamma_{2n+1}}{(2n+1) r^{2n+1}}[/tex]

    m = 0,1,2,....

    r is real number

    is there an explicit expression for [tex]\Gamma_{2n+1} [/tex] in terms of n ?
  2. jcsd
  3. Dec 27, 2008 #2


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    The gamma function for positive integer arguments is essentially the same as factoral, i.e. gamma(n+1)=n!.
  4. Dec 27, 2008 #3


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    [itex]\Gamma_{2n+1}[/itex] need not have anything to do with the gamma function! It's just a coefficient labelled with a capital gamma.

    As for the form of gamma... it will obviously depend on m, and it will have to depend on r, too. I'm guessing that you'll need to add more conditions on the form you want, otherwise you may not get a unique answer. (And I don't know if you can get an answer independent of r).

    (Following formulae lifted from wikipedia. check external reference to ensure correctness).

    For instance,

    [tex]\frac{\pi}{4} = \sum_{n=0}^\infty = \frac{(-1)^n}{2n+1}[/tex],

    so we could to choose [itex]\Gamma_{2n+1} = 4 m r^{2n+1}[/itex].

    Another series is

    [tex]\frac{\pi}{2} = \sum_{n=0}^\infty \frac{n!}{(2n+1)!!}[/tex]

    (where the (2n+1)!! = 1*3*5*...(2n+1) ), so we could choose [itex]\Gamma_{2n+1} = 2(-1)^n m r^{2n+1} (2n+1)n!/(2n+1)!![/itex]. (Admittedly, though, the factor of (-1)^n in there sort of breaks with the subscript notation of the coefficient).

    As for general conditions on Gamma, I guess I would run through the convergence tests for series, like the ratio test. http://en.wikipedia.org/wiki/Ratio_test
    Last edited: Dec 27, 2008
  5. Dec 27, 2008 #4
  6. Dec 31, 2008 #5
    i think i sorted it out .

    [tex]tan^{-1}(\frac{s}{r})=\sum^{\infty}_{n=0} \frac{(-1^{n}) s^{2n+1} }{(2n+1) r^{2n+1}}[/tex]



    => [tex]\Gamma_{2n+1} = s^{2n+1}[/tex]
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