I am looking for an analytic continuation of the function $f(z)$ when $1/2<z<1$. Here is the function in a few different forms. $$ \begin{aligned} f(z)% &=(1-z)^2F_2(\alpha;1,1;\alpha,\alpha;z,z)\\ &=(1-z)F_1(1;\alpha-1,1;\alpha;z,z(1-z)^{-1})\\ &={_2F_1}(1,1;\alpha,z^2(1-z)^{-2}), \end{aligned} $$ where $\alpha>2$ is an arbitrary positive real parameter, $F_1$ and $F_2$ are the first and second Appell hypergeometric functions and $_2F_1$ is the Gauss hypergeometric function.
All of these forms are defined by absolutely convergent series when $z\in (0,1/2)$. I know $f(z)$ is real and strictly increasing for $z\in(0,1)$.
Resources detailing the analytic continuation for the Appell function in this special case would also be acceptable.
Additional Information:
The function $f(z)$ is defined by the limit $f(z):=\lim_{n\to\infty}f_n(z)$ where $$ f_n(z)=\left(\frac{1-z}{1-z^{n+1}}\right)^2\sum_{k,\ell=0}^n\frac{(\alpha)_{k+\ell}(1)_k(1)_\ell}{(\alpha)_k(\alpha)_\ell k! \ell !}z^{k+\ell}. $$ Here is a plot for various $n$ and $\alpha=10$ along with the limiting solution using $_2F_1$ form (black).
Another definition for $f_n(z)$ is $$ \tag{1} f_n(z):=\frac{\frac{1}{\Gamma(\alpha)}\int_0^\infty \left(\sum_{k=0}^n\frac{(zx)^k}{(\alpha)_k}\right)^2 x^{\alpha-1} e^{-x}\,\mathrm dx}{\left(\frac{1}{\Gamma(\alpha)}\int_0^\infty \sum_{k=0}^n\frac{(zx)^k}{(\alpha)_k} x^{\alpha-1} e^{-x}\,\mathrm dx\right)^2}. $$ The first definition I provided above comes from evaluating and simplifying this one.
Here are three ways to get $f(z)$:
(1):
Noting that $f(z)=\lim_{n\to\infty}f_n(z)$ we have from $(1)$ that $$ f(z)=\frac{\frac{1}{\Gamma(\alpha)}\int_0^\infty \left({_1F_1}(1;\alpha;zx)\right)^2 x^{\alpha-1} e^{-x}\,\mathrm dx}{\left(\frac{1}{\Gamma(\alpha)}\int_0^\infty {_1F_1}(1;\alpha;zx) x^{\alpha-1} e^{-x}\,\mathrm dx\right)^2}. $$ The numerator and denominator are directly evaluated with formulas $7.622.1$ and $7.621.5$ from Gradshteyn & Ryzhik (ed. 8), respectively. For the numerator, the corresponding formula requires $\alpha>0\land z<1/2$ while the formula for the denominator requires $\alpha>0\land z<1$. Evaluating yields $$ f(z)=\frac{(1-z)^{-2}{_2F_1}(1,1;\alpha;z^2(1-z)^{-2})}{\left((1-z)^{-1}\right)^2}={_2F_1}(1,1;\alpha;z^2(1-z)^{-2}). $$
(2):
Evaluating $(1)$ directly we find $$ \begin{aligned} f_n(z)% &=\frac{\frac{1}{\Gamma(\alpha)}\int_0^\infty \sum_{k,\ell=0}^n\frac{(zx)^{k+\ell}}{(\alpha)_k(\alpha)_\ell} x^{\alpha-1} e^{-x}\,\mathrm dx}{\left(\frac{1}{\Gamma(\alpha)}\int_0^\infty \sum_{k=0}^n\frac{(zx)^k}{(\alpha)_k} x^{\alpha-1} e^{-x}\,\mathrm dx\right)^2}\\ &=\frac{% \frac{1}{\Gamma(\alpha)}\sum_{k,\ell=0}^n\frac{z^{k+\ell}}{(\alpha)_k(\alpha)_\ell}% \int_0^\infty x^{\alpha+k+\ell-1} e^{-x}\,\mathrm dx% }{% \left(\frac{1}{\Gamma(\alpha)}\sum_{k=0}^n\frac{z^k}{(\alpha)_k}% \int_0^\infty x^{\alpha+k-1} e^{-x}\,\mathrm dx\right)^2% }\\ &=\frac{% \frac{1}{\Gamma(\alpha)}\sum_{k,\ell=0}^n\frac{z^{k+\ell}}{(\alpha)_k(\alpha)_\ell}% \Gamma(\alpha+k+\ell)% }{% \left(\frac{1}{\Gamma(\alpha)}\sum_{k=0}^n\frac{z^k}{(\alpha)_k}% \Gamma(\alpha+k)\right)^2% }\\ &=\frac{% \sum_{k,\ell=0}^n\frac{(\alpha)_{k+\ell}}{(\alpha)_k(\alpha)_\ell}z^{k+\ell}% }{% \left(\sum_{k=0}^n z^k\right)^2% } \end{aligned} $$ Simplifying yields $$ \tag{2} f_n(z)=\left(\frac{z-1}{z^{n+1}-1}\right)^2 \sum_{k,\ell=0}^n\frac{(\alpha)_{k+\ell}}{(\alpha)_k(\alpha)_\ell}z^{k+\ell}. $$ Now write $(2)$ as $$ f_n(z)=\left(\frac{z-1}{z^{n+1}-1}\right)^2 \sum_{k,\ell=0}^n\frac{(\alpha)_{k+\ell}(1)_k(1)_\ell}{(\alpha)_k(\alpha)_\ell k! \ell !}z^{k+\ell}. $$ Evaluating the limit $n\to\infty$ we find (assuming $z\in(0,1)$ $$ \begin{aligned} f(z)% &=(1-z)^2 \sum_{k,\ell=0}^\infty\frac{(\alpha)_{k+\ell}(1)_k(1)_\ell}{(\alpha)_k(\alpha)_\ell k! \ell !}z^{k+\ell}\\ &=(1-z)^2F_2(\alpha;1,1;\alpha,\alpha;z,z), \end{aligned} $$ where the double series converges if $|z|+|z|<1\implies 2z<1\implies z<1/2$. Using the reduction formulas of DLMF 16.16.3 and 16.16.1 we then find $$ f(z)={_2F_1}(1,1;\alpha;z^2(1-z)^{-2}). $$
(3):
Recalling equation $(2)$ we have $$ \begin{aligned} f_n(z)% &=\left(\frac{z-1}{z^{n+1}-1}\right)^2 \sum_{k,\ell=0}^n\frac{(\alpha)_{k+\ell}}{(\alpha)_k(\alpha)_\ell}z^{k+\ell}\\ &=\left(\frac{z-1}{z^{n+1}-1}\right)^2 \sum_{k,\ell=0}^n\frac{\Gamma(\alpha)\Gamma(\alpha+k+\ell)}{\Gamma(\alpha+k)\Gamma(\alpha+\ell)}z^{k+\ell}\\ &=\left(\frac{z-1}{z^{n+1}-1}\right)^2 \sum_{k,\ell=0}^n\frac{\Gamma(\alpha)\Gamma(\alpha-(-k)-(-\ell))}{\Gamma(\alpha-(-k))\Gamma(\alpha-(-\ell))}z^{k+\ell} \end{aligned} $$ According to Wolfram Functions Site if $\alpha>-(k+\ell)\implies \alpha>0$ (which it is), then $$ f_n(z)=\left(\frac{z-1}{z^{n+1}-1}\right)^2 \sum_{k,\ell=0}^n{_2F_1}(-k,-\ell;\alpha;1)z^{k+\ell}. $$ Evaluating the limit yields $$ \begin{aligned} f(z)% &=(1-z)^2 \sum_{k=0}^\infty z^k\sum_{\ell=0}^\infty{_2F_1}(-k,-\ell;\alpha;1)z^\ell\\ &=(1-z)^2 \sum_{k=0}^\infty z^k\sum_{\ell=0}^\infty\frac{(1)_\ell}{\ell!}{_2F_1}(-\ell,-k;\alpha;1)\left(1-\left(\frac{z-1}{z}\right)\right)^{-\ell}\\ \end{aligned} $$ According to DLMF $15.15.1$, the series over $\ell$ converges for $z<1$ yielding $$ \begin{aligned} f(z)% &=(1-z)\sum_{k=0}^\infty{_2F_1}(-k,1;\alpha;z(z-1)^{-1})z^k\\ &=(1-z)\sum_{k=0}^\infty\frac{(1)_k}{k!}{_2F_1}(-k,1;\alpha;z(z-1)^{-1})\left(1-\left(\frac{z-1}{z}\right)^2\left(\frac{z-1}{z}\right)^{-1}\right)^{-k}. \end{aligned} $$ Again calling on DLMF $15.15.1$, the series over $k$ converges for $z<1/2$ yielding $$ f(z)={_2F_1}(1,1;\alpha;z^2(1-z)^{-2}). $$
Remarks
In all three examples we required $\alpha>0$ and $z<1/2$. However, if we also impose $\alpha>2$ then using Wolfram Functions Site $$ f(1/2)=\frac{\alpha-1}{\alpha-2},\quad \alpha>2 $$ which is finite. If we further impose $\alpha>3$ and again call on Wolfram Functions Site then $$ \lim_{z\to 1/2^-} f^\prime(z)=8\frac{\alpha-1}{(\alpha-2)(\alpha-3)},\quad \alpha>3 $$ which again is finite. Also keep in mind that $f_n(z)$ is real, continuous, differentiable, and strictly increasing in $z$ for $z\in(0,1)$ and $n=1,2,3\dots$. Given these results it seems unlikely that $f(z)$ cannot be continued for $z>1/2$. Maybe my intuition is misguiding me here. I am open to being convinced otherwise.
This is not an answer, but some information in attempt to clear the air.
Using the first definition for the limit function, we have:
$$f(z)=(1-z)^2F_2(\alpha;1,1;\alpha,\alpha;z,z)$$
Using the integral definition for this Appell function we obtain:
$$f(z)=(1-z)^2(\alpha-1)^2 \int_0^1 \int_0^1 \frac{(1-u)^{\alpha-2}(1-v)^{\alpha-2} du dv}{(1-z (u+v))^\alpha}$$
Just as the series only converges for $z \leq \frac{1}{2}$ (if we are looking for $z>0$ at least), it's clear that the integral only exists for $z \leq \frac{1}{2}$, because otherwise it blows up for $\alpha >1$. And we can't have $\alpha <1$ either, because of the terms in the numerator.
We can define the function for $z>1/2$ as the Cauchy principal value of the integral, but that's all I can offer.
The fact that $$f_n(z)=\left(\frac{1-z}{1-z^{n+1}}\right)^2\sum_{k,\ell=0}^n\frac{(\alpha)_{k+\ell}(1)_k(1)_\ell}{(\alpha)_k(\alpha)_\ell k! \ell !}z^{k+\ell}$$
all exist and are real for $z>1/2$ doesn't mean anything, because what we are looking for is the limit $f(z) = \lim_{n \to \infty} f_n(z)$.
My claim is as follows. For $z>1/2$ neither the "limit of the partial sums definition" nor the "double integral definition" would work as stated.
Some further thoughts. The integrand is symmetric, so we can redefine:
$$g(z)=\frac{f(z)}{(1-z)^2(\alpha-1)^2}= 2\int_0^1 \int_0^u \frac{(1-u)^{\alpha-2}(1-v)^{\alpha-2} dv du}{(1-z (u+v))^\alpha}$$
$$v=u t$$
$$g(z)=2 \int_0^1 \int_0^1 \frac{(1-u)^{\alpha-2}(1-t u)^{\alpha-2} u dt du}{(1-z (1+t) u)^\alpha}$$
$$g(z)=2 \int_0^1 \int_0^1 \frac{(1-(1+t)u+t u^2)^{\alpha-2} u dt du}{(1-z (1+t) u)^\alpha}$$
Not sure if this could help us, but just in case.
Update:
Using the other definition:
$$f(z)=\frac{\int_0^\infty \left({_1F_1}(1,\alpha,z x)\right)^2 x^{\alpha-1} e^{-x}\,\mathrm dx}{\left(\int_0^\infty {_1F_1}(1,\alpha,z x) x^{\alpha-1} e^{-x}\,\mathrm dx\right)^2}+1$$
It's clear that the integral in the numerator doesn't converge for $z>1/2$, because the confluent hypergeometric function behaves like the exponential.
Meanwhile, the integral in the denominator does converge for $1/2<z<1$.
Unfortunately, I don't know how to provide the analytic continuation using any of those definitions.