Evaluate series sum

Question

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?

Answer 1

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).