In my research I have run across the hypergeometric function $${}\_3F\_2(d,d,d;d+1,d+1;z)$$ where d is a positive integer and 0≤z≤1. When I plot this as a function of d on a semilog plot, it appears to be exponential (or close to it) for large d, but I haven't been able to prove it or find a formula for the exponential as a function of z. I've poked around on the web, but I'm a physicist rather than a mathematician and I'm a bit overwhelmed by the hypergeometric literature. Can anyone help? Thanks.



Exponential approximation for 3F2 hypergeometric function with repeated indices
Indeed the hypergeometric series is unimodal (the terms for fixed $z$ and $d$ first grow with $k$ then decrease again). The hypergeometric series is given by
$$
{}\_{3}F\_{2}(d,d,d;d+1,d+1;z)=\sum\_{k=0}^{\infty} \frac{(d)\_{k}(d)\_{k}(d)\_{k}}{(d+1)\_{k}(d+1)\_{k} k!} z^{k}=\sum\_{k=0}^{\infty} b(k,d) z^{k},
$$
with the Pochhammer symbol $(a)\_{k}=\frac{\Gamma(a+k)}{\Gamma(a)}$.


Therefore the ratio of two consecutive terms gives the nicely compact
$$
\frac{b(k+1,d) z}{b(k,d)}=\frac{(d+k)^{3}}{(d+k+1)^{2}(k+1)} z
$$
which should be equal to 1 for $k=k\_{\max}$. Expanding the ratio for large $d$ we find for the highest order in $d$ 
$$
k\_{\max}=\frac{z}{1-z} \ d + O(1).
$$
This expansion is valid for $d\ z$ and $d \ (1-z)$ well larger than 1, as can be seen by the following terms in the expansion.
Now we do as Noam Elkies suggested in his comment: make the index $k$ continuous and expand the terms of the hypergeometric sum to second order around $k\_{\max}$. Actually, it is the logarithm of the terms that gets expanded to afterwards have a nice Gaussian integral, when we switch from summation over $k$ to integral. We get
$$
\sum\_{k=0}^{\infty} b(k,d) z^{k}\approx \int\_{-\infty}^{\infty} \exp\{\ln(b(k\_{\max} + \tau,d) z^{k\_{\max}+\tau})\}\ d\tau.
$$
The exponent is expanded to second order in $\tau$ giving a Gaussian integral that can be calculated easily. (The factor in front of the $\tau^2$ term is indeed negative, the integral is converging.) The extension of the lower integration limit to $-\infty$ is okay, since at this stage the integrand is rapidly vanishing in that region as function of $\tau$. 


All this can be done nicely with your favourite computer algebra system. After tweaking the result a bit and expanding it for large $d$, the final result is
$$
{}\_{3}F\_{2}(d,d,d;d+1,d+1;z)\approx\exp\left\{\frac{(5 z +1)^2}{8 \ d \ z}\right\}\ (1-z)^{-d+2}.
$$
This approximation becomes bad if $d \ z \approx 1$ and smaller (i.e., for very small $z$). But in this regime the even simpler
$$
{}\_{3}F\_{2}(d,d,d;d+1,d+1;z)\approx (1-z)^{-d+2}
$$
can be used. For $d\ (z-1)$ smaller than one we get also invalid results.


As a byproduct I mention below the asymptotic expansion for hypergeometric functions with higher indices
$$
{}\_{p+1}F\_{p}(\overbrace{d,\dots,d}^{p+1};\underbrace{d+1,\dots,d+1}\_{p};z)\approx\exp\left\{\frac{(z (2 p +1) +1)^2}{8 \ d \ z}\right\}\ (1-z)^{-d+p}.
$$
Again for small $d\ z$ the exponential prefactor should be avoided for a good approximation. For $z$ approaching 1 the approximation breaks down.


All computations were conducted with Mathematica 10.


