# 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 $$\sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n = xk ~_2F_1(1-k, 1-b;2;x).$$

• $b \in (0,1)$ ? Isn't $b$ a positive integer? Oct 25, 2016 at 5:55
• No. It lies between $0$ and $1$.
– user168556
Oct 25, 2016 at 5:58

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}

• Thanks for the detailed answer. I have a related question. How do we know that the infinite sum $\sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n$ converges?
– user389066
Jul 28, 2017 at 3:26
• @User31443: Since $$\sum_{n=1}^\infty\binom{k}{n}x^n=(1+x)^k-1$$ and $$\sum_{n=1}^\infty\binom{b-1}{n-1}x^n=x(1+x)^{b-1}$$ have radius of convergence of $1$, we know that $$\limsup_{n\to\infty}\binom{k}{n}^{1/n}=1$$ and $$\limsup_{n\to\infty}\binom{b-1}{n-1}^{1/n}=1$$ Therefore, $$\limsup_{n\to\infty}\left[\binom{k}{n}\binom{b-1}{n-1}\right]^{1/n}\le1$$ Thus, the series $$\sum_{n=1}^\infty\binom{k}{n}\binom{b-1}{n-1}x^n$$ has a radius of convergence $\ge1$.
– robjohn
Jul 28, 2017 at 7:45
• Thanks. How does $\sum_{n=1}^\infty\binom{k}{n}x^n$ have radius of convergence $1$? If $a_n = \binom{k}{n}x^n$, then we should have $\lim_{n \to \infty} |a_n|^{1/n} = \lim_{n \to \infty}\left|\binom{k}{n}x^n\right|^{1/n} < 1$. To have the radius of convergence of $1$, we should have $|x| < 1$. How do we reach from $\lim_{n \to \infty}\left|\binom{k}{n}x^n\right|^{1/n} < 1$ to $|x| < 1$? Second query: How $\limsup_{n\to\infty}\left[\binom{k}{n}\binom{b-1}{n-1}\right]^{1/n}\le1$ implies that $\sum_{n=1}^\infty\binom{k}{n}\binom{b-1}{n-1}x^n$ has a radius of convergence $\geq 1$?
– user389066
Jul 28, 2017 at 22:23
• Second comment due to limit on the characters: As the author of the question pointed out, we have $\sum_{n = 1}^{\infty} {k\choose n}{b-1 \choose n-1}x^n = xk ~_2F_1(1-k, 1-b;2;x).$ So can we directly say that the series always converges as it can be represented in the form of Guass hypergeometric function?
– user389066
Jul 28, 2017 at 22:40
• @User31443: The formula for the radius of convergence $\left(\limsup\limits_{n\to\infty}|a_n|^{1/n}\right)^{-1}$ is given in this article. Furthermore, we know that the power series of a function converges at points closer to the point of expansion than the closest singularity. If $k\not\in\mathbb{Z}^{\ge0}$, the function $(1+x)^k$ has a singularity at $x=-1$, so the radius of convergence is $1$.
– robjohn
Jul 29, 2017 at 5:11

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

\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}