Asymptotic behaviour of $\binom{n+\alpha-1}{n}$

Question

The numbers $\binom{n+\alpha-1}{n}$, $n\geq0$; $\alpha\in\mathbb{R}$, arise as the coefficients of the Taylor series of the holomorphic function $(1-z)^{-\alpha}$ around $z=0$. When $\alpha\not\in\{0,-1,-2,\ldots\}$ we can appeal to the Gamma function $\Gamma$ to show that $$ \binom{n+\alpha-1}{n}\sim \frac{1}{\Gamma(\alpha)}\cdot\frac{1}{n^{1-\alpha}}\,,\ \text{ when }n\to\infty, $$ where, for $x\in\mathbb{R}$ and $k\geq0$ integer,$$\binom{x}{k}:=\frac{x(x-1)(x-2)\cdots(x-k+1)}{k!}.$$

Using the Gamma function seems kind of an overkill to me. Are there a more elementary way to obtain this, maybe just up to the constant $1/\Gamma(\alpha)$?

Any contribution is appreciated, thank you!!

Answer 1

Avoiding the $\Gamma$ function: $$\binom{n+\alpha-1}{n} = \frac{1}{n!}\prod_{k=0}^{n-1}(n+\alpha-1-k) = \prod_{k=1}^{n}\left(1+\frac{\alpha-1}{k}\right) $$ and if we approximate $\left(1+\frac{\alpha-1}{k}\right)$ with $\exp\left(\frac{\alpha-1}{k}\right)$ the last product turns into $$\exp\left((\alpha-1) H_n\right)\approx e^{\gamma}\cdot n^{\alpha-1}. $$

Answer 2

Binomial coefficients:

$$\binom{n+\alpha-1}n=\frac{(n+\alpha-1)!}{n!(\alpha-1)!}=\frac1{(\alpha-1)!}(n+1)(n+2)\dots(n+\alpha-2)(n+\alpha-1)\\\sim\frac1{(\alpha-1)!}n^{\alpha-1}$$

if $\alpha\in\mathbb R$,

$$\binom{n+\alpha-1}n=\frac{(\alpha)(\alpha+1)\dots(\alpha+n-1)}{n!}\sim\frac1{n!}\alpha^{n-1}$$