It is known that binomial coefficients can be generalized to the following: for $s\in\mathbb R$ and $k\in\mathbb N$, \begin{equation*} \binom{s}{k} := \prod_{i=0}^{k-1} \frac{s-i}{k-i} = \frac{s(s-1) \cdots(s-k+1)}{k !} = \frac{(s)_k}{k !} = \frac{\Gamma(s+1)}{\Gamma(s-k+1)k !}, \end{equation*} where $(s)_k := s(s-1) \cdots (s-k+1)$ denotes the falling factorial. For the generalized binomial coefficients, the following Newton's generalized binomial theorem holds: \begin{equation*} (1+x)^s = \sum_{k=0}^{\infty} \binom{s}{k} x^k. \end{equation*}
It is natural to further generalize to the case where two arguments are both non-integers in the following way: for $s,p\in\mathbb R$, \begin{equation*} \binom{s}{p} := \prod_{i=0}^{\lceil p \rceil-1} \frac{s-i}{p-i} = \frac{(s)_{\lceil p \rceil}}{(p)_{\lceil p \rceil}} = \frac{\Gamma(s+1) \Gamma(p-\lceil p \rceil+1)}{\Gamma(s-\lceil p \rceil+1) \Gamma(p+1)}. \end{equation*} But for such a generalization, do we have the counterpart of the binomial theorem? What is the following series: for $0<p<1$, \begin{equation*} \sum_{k=0}^{\infty} \binom{s}{k+p} x^k = ? \end{equation*} Or is there a good definition for $\binom{s}{k+p}$ so that the above series can have a meaning?
TIA...
Defined $\binom{\alpha}{\beta}:=\frac{\Gamma(\alpha+1)}{\Gamma(\beta+1)\Gamma(\alpha-\beta+1)}$, the series $\sum_{k=0}^\infty\binom{s}{k+p}z^k$ is a hypergeometric one, and has a closed form in terms of the incomplete beta function (elementary if $s$ and $p$ are half-integers).
Indeed, with $\mathrm{B}(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ the (complete) beta function, write $$ \binom{s}{k+p}=\frac{\mathrm{B}(p,k+1)}{\mathrm{B}(p,s-p+1)}\binom{s-p}{k}; $$ then for $\Re p>0$ and $0<|z|<1$ we get \begin{align*} \sum_{k=0}^\infty\binom{s}{k+p}z^k &=\frac1{\mathrm{B}(p,s-p+1)}\sum_{k=0}^\infty\binom{s-p}{k}z^k\int_0^1 t^{p-1}(1-t)^k\,dt\\ &=\frac1{\mathrm{B}(p,s-p+1)}\int_0^1 t^{p-1}\big(1+z(1-t)\big)^{s-p}\,dt\\ &=\boxed{z^{-p}(1+z)^s I_{z/(1+z)}(p,s-p+1)} \end{align*} (the principal branches taken). Note also that, after $t=\frac{(1+z)x}{1+zx}$, we get $$ \sum_{k=0}^\infty\binom{s}{k+p}z^k=\frac{(1+z)^s}{\mathrm{B}(p,s-p+1)}\int_0^1\frac{x^{p-1}\,dx}{(1+zx)^{s+1}}, $$ leading to $\mathrm{B}(-z;p,-s)$. All these things get analytically continued easily.