$\alpha \cdot (\alpha+1) \cdots (\alpha + n -1)\frac{1}{n!} \sim \frac{1}{\Gamma(\alpha)}n^{\alpha-1} \; \text{as} \; n \to \infty$

Question

This is Exercise 4 from Chapter 6 of Stein and Shakarchi's Complex Analysis.

Prove that if we take $$f(z) = \frac{1}{(1-z)^\alpha}, \; \text{for } |z|<1$$ (defined in terms of the principal branch of the logarithm), where $\alpha$ is a fixed complex number, then $$f(z)= \sum_{n=0}^\infty a_n(\alpha)z^n$$ with $$a_n(\alpha) \sim \frac{1}{\Gamma(\alpha)}n^{\alpha-1} \; \text{as} \; n \to \infty.$$

It is easy to see that $a_n(\alpha) =\alpha \cdot (\alpha+1) \cdots (\alpha + n -1)\frac{1}{n!}$.

But I can't figure out how to show that grows as $\frac{1}{\Gamma(\alpha)}n^{\alpha - 1}$ as $n\to \infty$. I would greatly appreciate any help.

Answer 1

I assume that you know the solution to Exercise 1 from the same chapter. The proof below uses its result that $\Gamma(s) = \lim_{n\rightarrow\infty}\frac{n^s s!}{s(s+1)\cdots(s+n)}$.

By taking the n-th derivative of $\frac{1}{(1-z)^\alpha} = \sum_{n=0}^\infty a_n(\alpha) z^n$ on both sides and setting $z=0$, we have

$a_n(\alpha) = \alpha \cdot (\alpha+1) \cdots (\alpha+n-1) \frac{1}{n!}.$

Noticing how similar this looks to the formula for $\Gamma(s)$ given above, we have

$a_n(\alpha) = \frac{\alpha \cdot (\alpha+1) \cdots (\alpha+n)n^\alpha}{(\alpha+n) n! n^\alpha} \sim \frac{n^\alpha}{\Gamma(\alpha)(\alpha+n)} \sim \frac{1}{\Gamma(\alpha)} n^{\alpha-1}$ as $n\rightarrow\infty$.