- Prove that if we take
\[ f (z) = \frac {1}{(1 - z) ^ {\alpha}}, \quad \mathrm{for} | z | < 1 \]
(defined in terms of the principal branch of the logarithm), where \(\alpha\) is a fixed complex number not in \(\{0,-1,-2,\ldots\}\), then
\[ f (z) = \sum_ {n = 0} ^ {\infty} a _ {n} (\alpha) z ^ {n} \]
with
\[ a _ {n} (\alpha) \sim \frac {1}{\Gamma (\alpha)} n ^ {\alpha - 1} \quad \text { as } n \to \infty . \]
If \(\alpha\) is a nonpositive integer, describe separately what happens to the coefficients.
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Solution
The generalized binomial theorem gives \[ (1-z)^{-\alpha}=\sum_{n=0}^\infty \frac{(\alpha)_n}{n!}z^n, \qquad |z|<1, \] where \[ (\alpha)_n=\alpha(\alpha+1)\cdots(\alpha+n-1) =\frac{\Gamma(n+\alpha)}{\Gamma(\alpha)}. \] Hence \[ a_n(\alpha)=\frac{\Gamma(n+\alpha)}{\Gamma(\alpha)\Gamma(n+1)}. \] For fixed \(\alpha\notin\{0,-1,-2,\ldots\}\), the gamma-ratio asymptotic \[ \frac{\Gamma(n+\alpha)}{\Gamma(n+1)}\sim n^{\alpha-1} \] gives \[ \boxed{a_n(\alpha)\sim \frac{n^{\alpha-1}}{\Gamma(\alpha)}}. \]
If \(\alpha=-m\) with \(m\in\mathbb Z_{\ge0}\), then \[ (1-z)^{-\alpha}=(1-z)^m \] is a polynomial of degree \(m\), so \(a_n(-m)=0\) for every \(n>m\). This is why the asymptotic statement requires excluding the nonpositive integers.