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Math Help - What does this sum converge to?

  1. #1
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    What does this sum converge to?

    Hi! I've stumbled upon a convergent sum of the form:

    S(k) = \sum_{n = 0}^{\infty}{k^{n^2}}

    This sum converges for every k with |k| < 1. I've checked properties of the function numerically (for real values of k) and found that for |k| close to zero the function goes approximately as S(k) = k + 1. I need to find an expression for the function for k close to 1. I've already found that it goes as S(k) = C(k)(1 - k)^{\beta}, where \beta \approx -\frac{1}{2} and C(k) is another function that goes to some finite constant of approximately 0.89 as k goes to 1. I'd be happy if anyone knows an analytic expression of the function (exact or approximate for k close to 1).

    Thanks in advance.

    /Simon
    Last edited by sitho; February 23rd 2009 at 04:49 PM.
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  2. #2
    Junior Member
    Joined
    Feb 2009
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    Ok, so I've found the approximate behaviour of the function to be S(k) \approx \frac{1}{2}\sqrt{-\frac{\pi}{\ln(k)}} in the limit k \rightarrow 1^{-} by approximating the sum with an integral. I'm not sure however how valid this approximation is in this limit. It fits with numerical estimations though. The constant \approx 0,89 is hence \frac{\sqrt{\pi}}{2} \approx 0,8862... I think though that such a fundamental sum should have been studied in detail. Are there any more exact solutions to the problem?
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