Power series of a function related to Gamma function

Question

I'm having trouble solving the following exercise:

Let $\alpha \in \mathbb C$ fixed and $f: D \to \mathbb C$ defined by $$f(z) := \frac{1}{(1-z)^\alpha}.$$ Let $$f(z) = \sum_{n=0}^\infty a_n(\alpha) z^n $$ be the power series representation of $f$. Show that for the coefficients $a_n(\alpha)$ the following holds: $$\lim_{n\to \infty}\frac{a_n(\alpha)}{n^{\alpha - 1}} = \frac{1}{\Gamma(\alpha)}$$

Where $D$ denotes the unit ball and $\Gamma$ the Gamma function.
Attempt:
I assume I need to find out explicit representation of the coefficients $a_n(\alpha)$ and then show the identity using the Gauß representation of the Gamma function: $$\Gamma(s) = \lim_{n\to \infty} \frac{n! n^s}{s(s+1) \cdot \cdot \cdot (s+n)}.$$ However, I was unable to find a useful one. Expanding the function $f$ into a Taylor series didn't work out (if I did not do it wrong). Is there an easier way to find out the coefficients?

Answer 1

Using the binomial series expansion and applying the binomial identity $$\binom{-\alpha}{n}=\binom{\alpha+n-1}{n}(-1)^n$$

we derive for $|z|<1$ and $\alpha\in\mathbb{C}$ \begin{align*} f(z)=\frac{1}{(1-z)^\alpha}&=\sum_{n=0}^\infty\binom{-\alpha}{n}(-z)^n\\ &=\sum_{n=0}^\infty\binom{\alpha+n-1}{n}z^n\\ \end{align*}

It follows \begin{align*} a_n(\alpha)&=\binom{\alpha+n-1}{n}=\frac{1}{n!}(\alpha+n-1)(\alpha+n-2)\cdots\alpha\\ &=\frac{1}{n!}(\alpha+n-1)^{\underline{n}} \end{align*} with $z^{\underline{n}}=z(z-1)\cdots(z-n+1)$ the falling factorial.

The following is valid \begin{align*} \lim_{n\to\infty}\frac{a_n(\alpha)}{n^{\alpha-1}}=\lim_{n\to\infty}\frac{(\alpha+n-1)^{\underline{n}}}{n!n^{\alpha-1}}=\frac{1}{\Gamma(\alpha)}\tag{1} \end{align*}

In order to show (1) we use a representation of the Gamma function $\Gamma(\alpha)$ according to C.F. Gauss. We obtain \begin{align*} \color{blue}{\Gamma(\alpha)}&=\lim_{n\to\infty}\frac{n^\alpha n!}{\alpha(\alpha+1)\cdots(\alpha+n)}\\ &=\lim_{n\to\infty}\frac{n^\alpha n!}{(\alpha+n)^{\underline{n+1}}}\\ &=\lim_{n\to\infty}\left(\frac{n}{\alpha+n}\cdot\frac{n^{\alpha-1} n!}{(\alpha+n-1)^{\underline{n}}}\right)\\ &=\lim_{n\to\infty}\frac{1}{\frac{\alpha}{n}+1}\cdot\lim_{n\to\infty}\frac{n^{\alpha-1} n!}{(\alpha+n-1)^{\underline{n}}}\\ &\color{blue}{=\lim_{n\to\infty}\frac{n^{\alpha-1} n!}{(\alpha+n-1)^{\underline{n}}}}\\ \end{align*} and the claim follows.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}{\pars{-1}^{n}{-\alpha \choose n} \over n^{\alpha - 1}} & = \lim_{n \to \infty}\bracks{{\pars{-1}^{n} \over n^{\alpha - 1}}\, {\pars{-\alpha}! \over n!\pars{-\alpha - n}!}} = \lim_{n \to \infty}\bracks{{\pars{-1}^{n} \over n^{\alpha - 1}} \,{\Gamma\pars{1 - \alpha} \over \Gamma\pars{n + 1}\,\Gamma\pars{1 -\alpha - n}}} \\[5mm] & = \lim_{n \to \infty}\braces{{\pars{-1}^{n} \over n^{\alpha - 1}\,\Gamma\pars{n + 1}}\, {\pi \over \Gamma\pars{\alpha}\sin\pars{\pi\alpha}}\,{\Gamma\pars{n + \alpha}\sin\pars{\pi\bracks{n + \alpha}} \over \pi}} \\[5mm] & = {1 \over \Gamma\pars{\alpha}}\lim_{n \to \infty}\bracks{{1 \over n^{\alpha - 1}} {\Gamma\pars{n + \alpha} \over \Gamma\pars{n + 1}}} \\[5mm] & = {1 \over \Gamma\pars{\alpha}}\lim_{n \to \infty}\bracks{% {1 \over n^{\alpha -1}}\, {\root{2\pi}\pars{n + \alpha}^{n + \alpha + 1/2}\expo{-n - \alpha} \over \root{2\pi}\pars{n + 1}^{n + 1 + 1/2}\expo{-n - 1}}} \\[5mm] & = {1 \over \Gamma\pars{\alpha}}\lim_{n \to \infty}\bracks{% {1 \over n^{\alpha -1}}\,{n^{n + \alpha + 1/2}\,\pars{1 + \alpha/n}^{n + \alpha + 1/2} \over n^{n + 3/2}\,\pars{1 + 1/n}^{n + 3/2}}\,\expo{1 - \alpha}} \\[5mm] & = {1 \over \Gamma\pars{\alpha}}\lim_{n \to \infty}\bracks{% {\pars{1 + \alpha/n}^{n} \over \pars{1 + 1/n}^{n}}\,\expo{1 - \alpha}} = \bbx{1 \over \Gamma\pars{\alpha}} \end{align}