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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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Probably not. The reason is that the behaviour for gamma at negative numbers is not so simple. To many poles and some chaotic stuff related to good rational approximations I think (continued fractions for alpha and such , compare to n sec(n) ). This is not an answer of course. But Im not even sure S_n(x) is complex differentiable in all x and alpha. – mick Oct 24 '12 at 21:19
@mick thank you for comment, what did you mean by "compare to $n \sec(n)$"? Can you give me some reference please? – Nimza Oct 24 '12 at 21:26
What happens when $k > \alpha$? – Jacob Oct 24 '12 at 21:26
@Jacob do you mean ${ \alpha \choose k}$? We define it via Gamma functions, we are not afraid of negative arguments of Gamma. – Nimza Oct 24 '12 at 21:29
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(-ax) behaves equally 'difficult' as sec(an) 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. – mick Oct 24 '12 at 21:39

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. – Nimza Oct 24 '12 at 21:47

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