Thread: What does this sum converge to?

1. 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).

/Simon

2. 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?