Convergence of hypergeometric 2F1 with z=-1

Question

Encountering the hypergeometric function $_2F_1(4n+3,\ m+1;\ m+3;\ -1)$ where $n\in\textbf{N}$ and $m\in\{2,4,6,\ldots,4n-2\}$ I'm a bit confused about its convergence.

According to Erdélyi's "Higher Transcendental Functions Vol. 1," $_2F_1(a,b;c;z)$ diverges if $|z|=1$ and $\text{Re}(a+b-c)\geq1$ (also confirmed by Wikipedia), and this is clearly my case. However in Mathematica it seems to converge; I get well-defined results for all values of $n$ and $m$ that I try, and plotting the function with $z\in\textbf{R}$ it seems fine around $z=-1$.

Does Mathematica perform some sort of analytic continuation or am I missing something here?

Answer 1

As noted by O.L. in the comments Mathematica does indeed use an analytic continuation.

Answer 2

Note that, the radius of convergence is one. $z=-1$ is a boundary point and you need to handle it separately. To do this you need to look at Abel's theorem.