Hypergeometric Function divergence

Solution 1:

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.