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Let $\alpha$ be some positive noninteger real constant and $n$ be an arbitrary nonnegative integer. Consider a series $ S_{n}(x) = \sum\limits_{k=0}^{\infty} {\alpha \choose k} \frac{x^k}{\Gamma(1-\alpha+k)\Gamma(1+n\alpha - k)}\ $ Is it possible to find it's sum?

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    Well can you estimate the min value of (n^4 sec(n)^2 - 1) accurately ? Now add some variables and taylor series and binomium and ask for a closed form. You might want to wiki or google Flint Hill series if my questions seems weird or unfamiliar. Since gamma(-a*x) behaves equally 'difficult' as sec(a*n) and your taylor series is quite exotic looking + the fact that its derivative is probably not elementary either ( both with respect to alpha and/or x) , I doubt if there is a solution. And if there is I assume it to be very very general , like generalized hypergeo or such. If differentiable.2012-10-24

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Let's do the simple case $m=0$. The series is $ S_0(x)=\sum_{k=0}^\infty\frac{x^k}{\Gamma(1-\alpha+k)\Gamma(1-k)} $ Use the convention $1/\Gamma(z)=0$ when $z$ is a pole of $\Gamma$. Then all but the first term vanishes and we get $ S_0(x) = \frac{1}{\Gamma(1-\alpha)} $

Maple says $ S_1(x)=-\frac{\operatorname{sin} \bigl(\pi (\alpha + 1)\bigr)}{\pi \alpha} \; {}_2F_1\bigl([-\alpha,-\alpha],[1 - \alpha],x\bigr) $
I leave $m>1$ to the reader (ha ha).

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    Haha, I like your humor. Nevertheless $S_{n}(x)$ are coefficients of some series and they all are important.2012-10-24