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.

1$\begingroup$ Looks like this is a polynomial with unimodal terms, so you can probably do this by bare hands by locating the maximal term(s) and approximating the sum by a Gaussian integral. $\endgroup$ – Noam D. Elkies Jul 22 '16 at 16:29
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 $d$ $$ k_{\max}=\frac{z}{1z} \ d + O(1). $$ This expansion is valid for $d\ z$ and $d \ (1z)$ 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 $$ {}_{3}F_{2}(d,d,d;d+1,d+1;z)\approx\exp\left\{\frac{(5 z +1)^2}{8 \ d \ z}\right\}\ (1z)^{d+2}. $$ This approximation becomes bad if $d \ z \approx 1$ and smaller (i.e., for very small $z$). But in this regime the even simpler $$ {}_{3}F_{2}(d,d,d;d+1,d+1;z)\approx (1z)^{d+2} $$ can be used. For $d\ (z1)$ smaller than one we get also invalid results.
As a byproduct I mention below the asymptotic expansion for hypergeometric functions with higher indices $$ {}_{p+1}F_{p}(\overbrace{d,\dots,d}^{p+1};\underbrace{d+1,\dots,d+1}_{p};z)\approx\exp\left\{\frac{(z (2 p +1) +1)^2}{8 \ d \ z}\right\}\ (1z)^{d+p}. $$ 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.

$\begingroup$ @Scott Hill  There's a very detailed example of this type of computation in Bender and Orszag, "Advanced Mathematical Methods for Scientists and Engineers," pages 304305. (Nice answer, Johannes.) $\endgroup$ – Tom Dickens Jul 28 '16 at 22:28

$\begingroup$ @Tom Dickens Thanks, Tom. For the computation in Bender&Orszag, see my answer here: mathoverflow.net/questions/185822/…. $\endgroup$ – Johannes Trost Jul 29 '16 at 18:26