Assume $n$ is a positive odd integer, I need to find the asymptotic as $n$ goes to infinity of the sum $$s(n,x)=\frac1x\sum_{k=0}^n (-1)^k\binom{-x-\frac12}{k}\binom{x-\frac12}{n-k},$$ where the binomial coefficients is used as polynomials, which is defined by $$ \binom{x}{n}=\frac{x(x-1)(x-2) \cdots(x-n+1)}{n(n-1)(n-2) \cdots 2 \cdot 1}. $$
Some special value for $x=\pm\frac12$ and $x=0$ is solved by analytic approach, but I can't evaluate the general $x$.
For special values:
- $s(n,\pm\frac12)$ is constant.
- $s(n,0)\sim \ln(n)n^{-\frac12}$.
Note that, if $x\ne 0$, $$ (1 - t)^{ - x - 1/2} (1 + t)^{x - 1/2} = \sum\limits_{n = 0}^\infty x\,s(n,x)\,t^n . $$ We can employ singularity analysis on the generating function. If $ \pm 2x \notin \{ 0,1,3,5, \ldots\}$, then $$ x\,s(n,x) \sim \frac{{(2n)^{x - 1/2} }}{{\Gamma (x + 1/2)}} + ( - 1)^n \frac{{(2n)^{ - x - 1/2} }}{{\Gamma (1/2 - x)}} $$ as $n\to +\infty$. If $2x \in \{ -1,-3,-5, \ldots\}$, then $$ x\,s(n,x) \sim ( - 1)^n \frac{{(2n)^{ - x - 1/2} }}{{\Gamma (1/2 - x)}} $$ as $n\to +\infty$. If $2x \in \{ 1,3,5, \ldots\}$, then $$ x\,s(n,x) \sim \frac{{(2n)^{x - 1/2} }}{{\Gamma (x + 1/2)}} $$ as $n\to +\infty$.
The limiting case $x\to 0$ of the first asymptotics gives $$ s(2n + 1,0) \sim \frac{{2\log (4n + 2)}}{{\sqrt {\pi (4n + 2)} }} \sim \frac{{\log n}}{{\sqrt {\pi n} }} $$ as $n\to +\infty$. Note that $s(2n,x)\to \pm \infty$ as $x\to 0$.