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Is it possible to sumate this series?

  1. Sep 3, 2009 #1
    [tex]\sum^{\infty}_{n=0}(2n+1)e^{-Cn(n+1)}[/tex]

    [tex]C[/tex] - constant


    Is it possible to sumate this series?
     
  2. jcsd
  3. Sep 3, 2009 #2

    quasar987

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    If C is negative or 0, then no, obviously. If C>0, then yes...
     
  4. Sep 4, 2009 #3
    Yes [tex]C>0[/tex]. But I don't ask you if it convergate but can you summate analyticaly? Or you must use some numerical methods?
     
  5. Sep 4, 2009 #4

    quasar987

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    This is called finding a "closed form" for the series. I don't think one exists. So numerical approximation is the best one can do.
     
  6. Sep 4, 2009 #5
    A theta function or relative ... because of the quadratic in the exponent.
     
  7. Sep 4, 2009 #6

    Mute

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    That looks like the sum you need to do if you were to calculate the Partition function for a quantum mechanical dumbell. That sum cannot be computed in closed form. The typical approximations are to assume either low temperature ( [itex]C \rightarrow \infty[/itex]), in which case the sum is approximately

    [tex]1 + 3e^{-2C} + \ldots .[/tex]

    The other limit is high temperature, [itex]C \rightarrow 0[/itex], in which case the sum is approximately an integral,

    [tex]\int_0^\infty dn~(2n+1)e^{-Cn(n+1)},[/itex]

    which is easily solved by substitution.

    Systematic corrections to the integral form can be computed using the Euler-Maclaurin formula:

    http://en.wikipedia.org/wiki/Euler-Maclaurin_formula
     
  8. Sep 5, 2009 #7
    Yes that is my problem. Sum is

    [tex]Q_1=\sum^{\infty}_{l=0}(2l+1)e^{-\frac{\theta_r}{\theta}l(l+1)[/tex]

    where

    [tex]\theta_r=\frac{\hbar^2}{2I}[/tex]

    Why we can say that if [tex]\theta>>\theta_r[/tex]

    [tex]Q_1=\int^{\infty}_0dl(2l+1)e^{-\frac{\theta_r}{\theta}l(l+1)[/tex]?
     
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