Can anybody see how to deduce an asymptotic formula for the hypergeometric function
$$ \_3F\_2\left(\frac{1}{2},x,x;x+\frac{1}{2},x+\frac{1}{2} \hskip2pt\bigg|\hskip2pt 1\right), \quad\mbox{ as } x\to\infty?$$


For the standard definition of the hypergeometric series, see [here](https://dlmf.nist.gov/16.2).


**Remark**: I've tried to combine the numerous transformations and integral representations which $\_3F\_2$-functions fulfill. I've also tried to get inspired by existing works on asymptotic expansions of similar (but simpler) $\_2F\_1$-functions but never ended up with the desired result.



Asymptotic behavior of a hypergeometric function
$\newcommand\Ga\Gamma$Let $f(x)$ denote your hypergeometric expression. Then
$$f(x)\sim\sqrt{\pi x}$$
(as $x\to\infty$).


Indeed,
$$f(x)=\sum\_{k\ge0}\frac{(1/2)\_k}{k!}r\_k(x)^2,$$
where $(a)\_k:=a(a+1)\cdots(a+k-1)=\Ga(a+k)/\Ga(a)$ and
$$r\_k(x):=\frac{(x)\_k}{(x+1/2)\_k}=\frac{\Ga(x+k)}{\Ga(x)}\Big/\frac{\Ga(x+1/2+k)}{\Ga(x+1/2)}
\sim\sqrt{\frac x{x+k}}$$
uniformly in $k\ge0$ (as $x\to\infty$).
So,
\begin{equation\*}
 f(x)\sim x\,\sum\_{k\ge0}\frac{(1/2)\_k}{k!} \frac1{x+k}
=\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)}
\sim\sqrt{\pi x} \tag{1}\label{1}
\end{equation\*}
(as $x\to\infty$).




---


**Details on the equality in \eqref{1}:**
\begin{equation\*}
\begin{aligned}
 \sum\_{k\ge0}\frac{(1/2)\_k}{k!} \frac1{x+k}
& =\sum\_{k\ge0}\frac{(1/2)\_k}{k!} \int\_0^1 dt\,t^{x+k-1} \\ 
& =\int\_0^1 dt\,t^{x-1}\sum\_{k\ge0}\frac{(1/2)\_k}{k!} t^k \\ 
& =\int\_0^1 dt\,t^{x-1}(1-t)^{-1/2}, 
\end{aligned}
\end{equation\*}
in view of the Maclaurin series for $(1-t)^{-1/2}$.
[Therefore](https://en.wikipedia.org/wiki/Beta_function#Properties),
\begin{equation\*}
\begin{aligned}
 x\,\sum\_{k\ge0}\frac{(1/2)\_k}{k!} \frac1{x+k}
=x\,B(x,1/2)= x\,\frac{\Ga(1/2)\, \Ga(x)}{\Ga(x+1/2)}
=\frac{\sqrt{\pi }\, \Ga(x+1)}{\Ga(x+1/2)}, 
\end{aligned}
\end{equation\*}
as claimed.


