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Math Help - radius of covergentia

  1. #1
    Junior Member
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    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
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  2. #2
    Senior Member Tinyboss's Avatar
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    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.
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  3. #3
    Super Member girdav's Avatar
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    Re: radius of covergentia

    Use Cauchy-Hadamard theorem and look at the cases x<0, x=0 and x>0.
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