Limit involving a hypergeometric function

OK, let's start with an integral representation of that hypergeometric:

$$_2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) \\= \frac{1}{B(-n,-m)} \int_0^1 dx \: x^{-(n+1)} (1-x)^{-(m+1)} \left[1-\left(1-\frac{r}{z}\right)x\right]^m$$

where $B$ is the beta function. Please do not concern yourself with poles involved in gamma functions of negative numbers for now: I will address this below.

As $z \rightarrow 0^+$, we find that

$$\begin{align}\lim_{z \rightarrow 0^+} z^m\ _2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) &= \frac{r^m}{B(-n,-m)} \int_0^1 dx \: x^{-(n-m+1)} (1-x)^{-(m+1)}\\ &= r^m \frac{B(-(n-m),-m)}{B(-n,-m)}\\ &= r^m \lim_{\epsilon \rightarrow 0^+} \frac{\Gamma(-n+m+\epsilon) \Gamma(-n-m+\epsilon)}{\Gamma(-n+\epsilon)^2} \end{align} $$

Note that, in that last line, I used the definition of the Beta function, along with a cautionary treatment of the Gamma function near negative integers, which are poles. (I am assuming that $n>m$.) The nice thing is that we have ratios of these Gamma function values, so the singularities will cancel and leave us with something useful.

I use the following property of the Gamma function (see Equation (41) of this reference):

$$\Gamma(x) \Gamma(-x) = -\frac{\pi}{x \sin{\pi x}}$$

Also note that, for small $\epsilon$

$$\sin{\pi (n-\epsilon)} \approx (-1)^{n+1} \pi \epsilon$$

Putting this all together (I leave the algebra as an exercise for the reader), I get that

$$\lim_{z \rightarrow 0^+} z^m \ _2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) = r^m \frac{n!^2}{(n+m)! (n-m)!}$$

which I believe is equivalent to the stated result.

That last statement is readily seen from writing out the product above:

$$\prod_{i=1}^m \frac{n-(i-1)}{n+1} = \frac{n}{n+1} \frac{n-1}{n+2}\frac{n-2}{n+3}\ldots\frac{n-(m-1)}{n+m}$$

The numerator of the above product is

$$\frac{n!}{(n-m)!}$$

and the denominator is

$$\frac{(n+m)!}{n!}$$

The result follows.

Author by

M.B.M.

Updated on August 01, 2022