Sharp upper bounds on hypergeometric function ${}_2F_1[a,b,c;z]$ when $|z|\geq1$

Question

Generally, hypergeometric function ${}_2F_1[a,b,c;z]$ is defined using Gauss series ${}_2F_1[a,b,c;z]=\sum_{n=0}^\infty\frac{(a)_n(b)_n}{(c)_nn!}z^n$ with $|z|<1$, and there seems to be a lot of literature relating ${}_2F_1[a,b,c;z]$ to power mean when $z\in(0,1)$ (see this for a survey).

I am wondering what is known about upper bounds on hypergeometric function ${}_2F_1[a,b,c;z]$ when $|z|\geq1$ (provided the series converges). Does anyone have any pointers to the literature?

Specifics of my problem

I am trying to upper-bound the following summation (which is known to be a probability, so trivial upper bound is one):

$$S(x,y)=\sum_{i=0}^y\sum_{j=0}^x\frac{i!j!}{x!y!}\gamma^{x-j}(1-\gamma)^{y+j}\left(\binom{x}{j}{}_2F_1\left[-y,-j,1+x-j,-\frac{\gamma}{1-\gamma}\right]\right)^2$$

where $x$ and $y$ are non-negative integers and $\gamma\in(0,1)$. The hypergeometric function inside the summation converges, and, I believe an upper bound on it would be helpful in upper-bounding the entire expression. Any help would be appreciated.

Answer 1

The Gauss hypergeometric function satisfies the following functional equation: $$ {}_2F_1(a,b;c;z) = (1-z)^{-a}{}_2F_1\left(a,c-b,c;\frac{z}{z-1}\right). $$ So, you can write your hypergeometric function as \begin{align} {}_2F_1\left(-y,-j;1+x-j;-\frac{\gamma}{1-\gamma}\right) &= \left(1+\frac{\gamma}{1-\gamma}\right)^{y}{}_2F_1\left(-y,1+x;1+x-j;\frac{-\frac{\gamma}{1-\gamma}}{-\frac{\gamma}{1-\gamma}-1}\right)\\ &= (1-\gamma)^{-y}{}_2F_1\left(-y,1+x;1+x-j;\gamma\right). \end{align} This should allow you to use the bounds you already have for $0<z<1$.

Answer 2

In this case series are not convergent. And the Gauss function is not defined by series outside the unit circle. But there are explicit analytical continuation formulas to return inside of the unit circle. And then you may use any inequalities you like for good values of argument. You reference is not the only and is not a complete survey.

Answer 3

For the asymptotic analysis of hypergeometric functions, perhaps the Barnes contour integral representation is a more powerful tool than the series expansion. The book The special functions and their approximations (part I) by Luke contains extensive discussion of the limiting behaviour of the (generalized) hypergeometric functions.