Proof of $\sum_\limits{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n \sim \frac{x^b k^b}{\Gamma(1+b)} \quad \text{as}~k \to \infty$ I want to prove the following aymptotic result:
$$\sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n \sim \frac{x^b k^b}{\Gamma(1+b)} \quad \text{as}~k \to \infty,$$
where $ k \in \mathbb{C}$, $b \in [0,1]$, and $x \in [0,1].$
I tried using some results of infinite series for combinatorics, but could not prove it. One of the results I tried is
$$\sum_{n = 0}^{\infty} {k \choose n}x^n = (1+x)^k.$$
Another way I tried is to use
$$ \sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n = xk ~_2F_1(1-k, 1-b;2;x).$$
 A: Define
$$
f_k(x)=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}x^{n-1}\tag{1}
$$
then we are looking for the asymptotic expansion of $x\,f_k(x)$.

Vandermonde's Identity and Gautschi's Inequality say
$$
\begin{align}
f_k(1)
&=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}\\
&=\binom{k+b-1}{k-1}\\
&=\frac{\Gamma(k+b)}{\Gamma(k)\Gamma(b+1)}\\[4pt]
&=\frac{k^b}{\Gamma(b+1)}\left(1+O\!\left(\frac1k\right)\right)\tag{2}
\end{align}
$$
Let $(n)_m$ be the Falling Factorial, then the $m^{\text{th}}$ derivative of $f_k$ at $1$ is
$$
\begin{align}
f_k^{(m)}(1)
&=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-1}{n-1}(n-1)_m\\
&=\sum_{n=1}^\infty\binom{k}{k-n}\binom{b-m-1}{n-m-1}(b-1)_m\\
&=\binom{k+b-m-1}{k-m-1}(b-1)_m\\
&=\frac{k^b(b-1)_m}{\Gamma(1+b)}\left(1+O\!\left(\frac1k\right)\right)\tag{3}
\end{align}
$$
For $|x-1|\lt1$, Taylor's Theorem and the Generalized Binomial Theorem say
$$
\begin{align}
x\,f_k(x)
&=x\sum_{m=0}^\infty\frac{f_k^{(m)}(1)}{m!}(x-1)^m\\
&=\frac{k^bx}{\Gamma(1+b)}\left(1+O\!\left(\frac1k\right)\right)\sum_{m=0}^\infty\frac{(b-1)_m}{m!}(x-1)^m\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{k^bx^b}{\Gamma(1+b)}\left(1+O\!\left(\frac1k\right)\right)}\tag{4}
\end{align}
$$

Asymptotic Expansion of Fractional Binomial Coefficients
Using the Euler-Maclaurin Sum Formula, we get
$$
\begin{align}
\log\binom{n+\alpha}{n}
&=\sum_{k=1}^n\log\left(1+\frac\alpha{k}\right)\\
&=\alpha\sum_{k=1}^n\frac1k-\frac{\alpha^2}2\sum_{k=1}^n\frac1{k^2}+\frac{\alpha^3}3\sum_{k=1}^n\frac1{k^3}-\frac{\alpha^4}4\sum_{k=1}^n\frac1{k^4}+\cdots\\
&=\alpha\left(\log(n)+\gamma+\frac1{2n}-\frac1{12n^2}+O\!\left(\frac1{n^4}\right)\right)\\
&-\frac{\alpha^2}2\left(\zeta(2)-\frac1n+\frac1{2n^2}+O\!\left(\frac1{n^3}\right)\right)\\
&+\frac{\alpha^3}3\left(\zeta(3)-\frac1{2n^2}+O\!\left(\frac1{n^3}\right)\right)\\
&-\frac{\alpha^4}4\left(\zeta(4)+O\!\left(\frac1{n^3}\right)\right)\\
&=\alpha\log(n)-\log(\Gamma(1+\alpha))+\frac{\alpha+\alpha^2}{2n}-\frac{\alpha+3\alpha^2+2\alpha^3}{12n^2}+O\!\left(\frac1{n^3}\right)\tag{5}
\end{align}
$$
Therefore, as $n\to\infty$,
$$
\binom{n+\alpha}{n}=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{\alpha+\alpha^2}{2n}-\frac{2\alpha+3\alpha^2-2\alpha^3-3\alpha^4}{24n^2}+O\!\left(\frac1{n^3}\right)\right)\tag{6}
$$
and so, for fixed $m$,
$$
\begin{align}
\binom{n-m+\alpha}{n-m}
&=\frac{(n-m)^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{\alpha+\alpha^2}{2n}+O\!\left(\frac1{n^2}\right)\right)\\
&=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1-\frac mn\right)^\alpha\left(1+\frac{\alpha+\alpha^2}{2n}+O\!\left(\frac1{n^2}\right)\right)\\
&=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{(1-2m)\alpha+\alpha^2}{2n}+O\!\left(\frac1{n^2}\right)\right)\tag{7}
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{n = 1}^{\infty}{k\choose n}{b - 1 \choose n - 1}x^{n}\ \sim\
{x^{b}k^{b} \over \Gamma\pars{1 + b}} \quad \mbox{as}\ k \to \infty:\ ?}$.

\begin{align}
\mbox{As}\ k \to \infty\,,\quad{k! \over \pars{k - n}!} & \sim
{\root{2\pi}k^{k + 1/2}\expo{-k} \over
\root{2\pi}\pars{k - n}^{k - n + 1/2}\expo{-k + n}} =
{k^{n}\expo{-n} \over \pars{1 - n/k}^{k}\pars{1 - n/k}^{-n + 1/2}} \sim k^{n}
\end{align}

\begin{align}
\sum_{n = 1}^{\infty}{k\choose n}{b - 1 \choose n - 1}x^{n} & \sim
\sum_{n = 0}^{\infty}{k^{n} \over n!}{b - 1 \choose b - n}x^{n} =
\sum_{n = 0}^{\infty}{k^{n} \over n!}\,x^{n}
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{b - 1} \over z^{b - n + 1}}
\,{\dd z \over 2\pi\ic}
\\[5mm] & =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{b - 1} \over z^{b + 1}}
\sum_{n = 0}^{\infty}{\pars{kxz}^{n} \over n!}
\,{\dd z \over 2\pi\ic} =
\oint_{\verts{z}\ =\ 1^{-}}{\pars{1 + z}^{b - 1}\expo{kxz} \over z^{b + 1}}
\,{\dd z \over 2\pi\ic}
\end{align}

In order to evaluate the last integral, we choose the $\ds{z^{-b - 1}}$ branch cut along $\ds{\left(\,-\infty,0\,\right]}$ with
$\ds{-\pi < \mrm{arg}\pars{z} < \pi}$. The contour is indented 'around'
$\ds{ z = 0}$ with a semi-circumference of radius
$\ds{\epsilon\ \mid\ 0 < \epsilon < 1}$. As usual, the limit
$\ds{\epsilon \to 0^{+}}$ is taken at the very end.
\begin{align}
\left.\sum_{n = 1}^{\infty}{k\choose n}{b - 1 \choose n - 1}x^{n}\,
\right\vert_{\ 0\ <\ \epsilon\ <\ 1} & \sim
-\int_{-1}^{-\epsilon}{\pars{1 + \xi}^{b - 1}\expo{kx\xi} \over
\pars{-\xi}^{b + 1}\expo{\ic\pars{b + 1}\pi}}\,{\dd\xi \over 2\pi\ic} -
\int_{\pi}^{-\pi}{1 \over \epsilon^{b + 1}\expo{\ic\pars{b + 1}\theta}}
\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 2\pi\ic}
\\[5mm] & -
\int_{-\epsilon}^{-1}{\pars{1 + \xi}^{b - 1}\expo{kx\xi} \over
\pars{-\xi}^{b + 1}\expo{-\ic\pars{b + 1}\pi}}\,{\dd\xi \over 2\pi\ic}
\\[1cm] & =
\expo{-\ic b\pi}\int_{\epsilon}^{1}{\pars{1 - \xi}^{b - 1}\expo{-kx\xi} \over
\xi^{b + 1}}\,{\dd\xi \over 2\pi\ic} +
{\sin\pars{\pi b} \over \pi b}\,\epsilon^{-b}
\\[5mm] & -
\expo{\ic b\pi}\int_{\epsilon}^{1}{\pars{1 - \xi}^{b - 1}\expo{-kx\xi} \over
\xi^{b + 1}}\,{\dd\xi \over 2\pi\ic}
\\[1cm] & =
-\,{\sin\pars{b \pi} \over \pi}
\int_{\epsilon}^{1}\pars{1 - \xi}^{b - 1}\expo{-kx\xi}\xi^{-b - 1}\,\dd\xi +
{\sin\pars{\pi b} \over \pi b}\,\epsilon^{-b}
\end{align}

As $\ds{k \to \infty}$, the main contribution to the integral comes from
values of $\ds{\xi\ \mid \xi \gtrsim 0}$. Then,
\begin{align}
\left.\sum_{n = 1}^{\infty}{k\choose n}{b - 1 \choose n - 1}x^{n}\,
\right\vert_{\ 0\ <\ \epsilon\ <\ 1} & \sim
{\sin\pars{b \pi} \over \pi b}
\int_{\xi\ =\ \epsilon}^{\xi\ \to\ \infty}
\expo{-kx\xi}\,\dd\pars{\xi^{-b}} +
{\sin\pars{\pi b} \over \pi b}\,\epsilon^{-b}
\end{align}
Integrating by parts and taking the $\ds{\pars{~\epsilon \to 0^{+}~}}$-limit
\begin{align}
\sum_{n = 1}^{\infty}{k\choose n}{b - 1 \choose n - 1}x^{n}\,
 & \sim
{\sin\pars{\pi b} \over \pi b}\,kx\int_{0}^{\infty}\xi^{-b}\expo{-kx\xi}\,\dd\xi
=
{\sin\pars{\pi b} \over \pi b}\,\pars{kx}^{b}\,\Gamma\pars{1 - b}
\\[5mm] & =
{\sin\pars{\pi b} \over \pi b}\,\pars{kx}^{b}\,
{\pi \over \Gamma\pars{b}\sin\pars{\pi b}} =
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
k^{b}x^{b} \over \Gamma\pars{1 + b}}}
\end{align}
