I am stuck and posting this out of desperation. Is there an approximation to the following product of Gamma functions: $$\sum_{h=0}^{M}\frac{1}{\Gamma(h+1)\Gamma(M-h+1)}\frac{\Gamma(N-hd)}{\Gamma(-hd)}$$
where $d$ is a real value number, $N$ and $M$ are integers both growing to a large number (no where near $\infty$ though), and $N>M$.
Note: I have already tried replacing the gamma ratio using the Stirling approximation, but it requires adding two more $\Gamma$ terms: $\frac{\Gamma(N-hd)}{\Gamma(-hd)}\Rightarrow \frac{\Gamma(N-hd)}{\Gamma(N)}\frac{\Gamma(N)}{\Gamma(-hd)}\approx\frac{1}{N^{hd}}\frac{\Gamma(N)}{\Gamma(-hd)}$ while replacing gamma terms ratio with this approximation.
Disregarding convergence issues at the moment, the sum can be written as $$ \frac{1}{\Gamma(-N)}\sum_{h=0}^{M}\frac{B(-N,N-hd)}{h!(M-h)!}=\frac{1}{M!\Gamma(-N)}\int_{0}^{1}\sum_{h=0}^{M}\binom{M}{h}x^{N-hd-1}(1-x)^{-N-1}\,dx $$ that by the binomial theorem equals $$ \frac{1}{\Gamma(M+1)\Gamma(-N)}\int_{0}^{1}\frac{x^{N-1}}{(1-x)^{N+1}}(1+x^{-d})^M\,dx $$ or $$\boxed{\frac{1}{\Gamma(M+1)\Gamma(-N)}\int_{1}^{+\infty}\frac{(1+x^d)^M}{(x-1)^{N+1}}\,dx} $$ If $d=1$, that boils down to $-\frac{2^{M-N}\Gamma(N-M)\sin(M\pi)}{\pi}.$
Otherwise, hypergeometric functions are involved.