Hypergeometric Function divergence Consider the Hypergeometric function
\begin{align}
{}_{2}{F}_{1}(a,b,c,z)=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(c)_n}\frac{z^n}{n!}
\end{align}
with $(q)_{n}$ the Pochhammer symbol
$(q)_{n}=
\begin{cases}
\begin{aligned}
1 \quad &\text{if } n=0\\
q\cdot(q+1)\cdot(q+2)\cdot \dots \cdot (q+n-1) \quad &\text{if } n>0
\end{aligned}
\end{cases}
$
In my case I have $a\in \mathbb{R}$ and $b=1-a$ and also $c=1$. Then we get
\begin{align}
{}_{2}{F}_{1}(a,1-a,1,z)=\sum_{n=0}^{\infty}\frac{(a)_{n}(1-a)_{n}}{(n!)^2}z^n
\end{align}
I would like to show that $\lim_{z \to 1}{}_{2}{F}_{1}(a,1-a,1,z)=\pm \infty$, or equivalently, that the sum $\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{(n!)^2}$ is divergent.
Sadly the quotient criterion is not applicable and other convergence criteria don't seem to work either.
 A: This series only diverges as $z\searrow 1$ for certain parameters. For example, suppose $a\in\Bbb N$ then
$$
{_2F_1}\left({a,1-a\atop 1};z\right)=\sum_{k=0}^{a-1}\frac{(a)_n(1-a)_n}{(1)_n}\frac{z^k}{k!},
$$
which is just a polynomial and clearly does not diverge at $z=1$.
Likewise, if $-a\in\Bbb N_0$ we again have a polynomial of the form
$$
{_2F_1}\left({a,1-a\atop 1};z\right)=\sum_{k=0}^{-a}\frac{(a)_n(1-a)_n}{(1)_n}\frac{z^k}{k!}.
$$

As such if $a\in\Bbb Z$ we always end up with a polynomial in $z$, which is finite in the limit $z\searrow 1$.

For $a\in\Bbb R\setminus\Bbb Z$ we say the hypergeometric function ${_2F_1}(1,1-a;1;z)$ is zero-balanced because the sum of the top parameters equals the bottom parameter, i.e. $(a)+(1-a)=1$. In such cases, our hypergeometric function has a logarmithmic singularity at $z\searrow 1$. According to DLMF 15.4.21:
$$
{_2F_1}\left({a,1-a\atop 1};z\right)\sim-\frac{1}{\Gamma(a)\Gamma(1-a)}\log(1-z),\quad z\searrow 1,
$$
and so we can use this fact to conclude our hypergeometric function diverges in the limit. For a proof of this fact I will point you towards the paper Inequalities for Zero-Balanced Hypergeometric Functions.
