# radius of covergentia

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• July 9th 2011, 03:26 PM
capea
radius of covergentia
could you determinate for what values os q and x the serie is convergent.

$\sum _{z=1}^{\infty } \frac{(-1)^{-1+z} q^{-x (1+2 z)^2}}{(1+2 z)^2}\simeq \frac{5}{6}-\text{Catalan}-\frac{q^{-9 x}}{12}+\frac{q^{-x}}{4}+\frac{1}{4} \sqrt{\pi } \sqrt{x} \text{Erf}\left[\sqrt{x} \sqrt{\text{Log}[q]}\right] \sqrt{\text{Log}[q]}-\frac{1}{4} \sqrt{\pi } \sqrt{x} \text{Erf}\left[3 \sqrt{x} \sqrt{\text{Log}[q]}\right] \sqrt{\text{Log}[q]}-\frac{1}{3} x^2 \text{Log}[q]^2+\frac{34}{15} x^3 \text{Log}[q]^3-\frac{163}{84} x^4 \text{Log}[q]^4$
and check if the solution is good
thanks
• July 9th 2011, 04:12 PM
Tinyboss
Re: radius of covergentia
Nobody's going to solve that beast for you. Why don't you tell us what you tried and where you got stuck, and maybe we can give some hints.
• July 10th 2011, 02:32 AM
girdav
Re: radius of covergentia
Use Cauchy-Hadamard theorem and look at the cases $x<0$, $x=0$ and $x>0$.