If $a,b,c$ are positive integers with $c\leq a+b,$ can I conclude that $_2F_1(a,b;c;1)$ diverges?

Question

Recall that the hypegeometric series is defined by $$_2F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}$$ where $z\in \mathbb{C}$ with $|z|<1$ and $(a)_n = a(a+1)...(a+n-1)$ if $n\geq 1$ and $(a)_0 = 1.$

Wiki states the following Gauss's Theorem.

(Gauss's Theorem) $$_2F_1(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$$ if $Re(c)> Re(a+b)$

where $\Gamma(a)$ is the Gamma function.

Question: If $a,b,c$ are positive integers with $c\leq a+b,$ can I conclude that $_2F_1(a,b;c;1)$ diverges?

Answer 1

No. Note that if $a$ or $b$ is a negative integer (and $c$ is not) the series has only finitely many nonzero terms.

EDIT: If $a,b,c$ are positive integers, $ \dfrac{(a)_n (b)_n}{(c)_n n!}$ is a rational function of $n$ with degree of denominator - degree of numerator $c+1-a-b$. Thus the series converges at $z=1$ iff $c+1-a-b > 1$, i.e. $c > a+b$.

Answer 2

In fact, on $\{\lvert z\rvert=1,z\neq 1\}$, $\sideset{_2}{_1}F(a,b;c;z)$ converges only conditionally if $0\geq\operatorname{Re}(c-a-b)>-1$ and diverges if $\operatorname{Re}(c-a-b)\leq -1$.