• July 9th 2011, 02:26 PM
capea
$\sum _{z=1}^{\infty } \frac{(-1)^{-1+z} q^{-x (1+2 z)^2}}{(1+2 z)^2}\simeq \frac{5}{6}-\text{Catalan}-\frac{q^{-9 x}}{12}+\frac{q^{-x}}{4}+\frac{1}{4} \sqrt{\pi } \sqrt{x} \text{Erf}\left[\sqrt{x} \sqrt{\text{Log}[q]}\right] \sqrt{\text{Log}[q]}-\frac{1}{4} \sqrt{\pi } \sqrt{x} \text{Erf}\left[3 \sqrt{x} \sqrt{\text{Log}[q]}\right] \sqrt{\text{Log}[q]}-\frac{1}{3} x^2 \text{Log}[q]^2+\frac{34}{15} x^3 \text{Log}[q]^3-\frac{163}{84} x^4 \text{Log}[q]^4$
Use Cauchy-Hadamard theorem and look at the cases $x<0$, $x=0$ and $x>0$.