Convergence of Hyper-Geometric Function at $x=1$ I know that the Hyper-Geometric function is given by
$$_2F_1(a,b,c,x)=\sum_{i=0}^{\infty}h_nx^n=\sum_{i=0}^{\infty}\frac{(a)_n(b)_n}{n!(c)_n}x^n \tag{1}$$
I want to know that under what conditions the Hyper-Geometric function will converge at $x=1$.
Here in wiki, it is said that according to Gauss theorem 
$$_2F_1(a,b,c,1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}, \qquad \Re(c) \gt \Re(a+b)\tag{2}$$ 
I used the ratio test to get something but the test is inconclusive.
$$\lim_{n \to \infty}\frac{h_{n+1}}{h_n}=\lim_{n \to \infty}\frac{(a+n)(b+n)}{(c+n)(1+n)}=1$$

Questions
$1$- How can I prove the relation $(2)$?
$2$- What happens if $\Re(c) \lt \Re(a+b)$?
 A: $$\phantom{}_2 F_1\left(a,b;c;1\right)=\sum_{n\geq 0}\frac{(a)_n (b)_n}{n! (c)_n} = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum_{n\geq 0}\frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(n+1)\Gamma(c+n)}$$
is a series whose convergence can be investigated through Gautschi's inequality, and by recalling that the singularities of the $\Gamma$ function are simple poles at $0,-1,-2,-3,-4,\ldots$ with residues $1,-1,\frac{1}{2},-\frac{1}{6},\ldots $. By the Weierstrass product, the $\Gamma$ function has no zeroes.
For instance, if $a,b,c>0$ and $\color{red}{c>a+b}$ the above series is clearly convergent. There might be issues if $a$ or $b$ is a non-positive integer, but in such a case the series is just a finite sum, by the vanishing of a Pochhammer symbol for any $n$ big enough.
A: The best paper I know on HyperGeometric functions, and specifically about their convergence, is
Higher Trascendental Functions, A. Erdelyi, California Institute of Technology, Volume 1, Chapter 2, Section 2.1.1, Page 57
where, for instance, it is explicitely stated that
Begin Citation
$$
\begin{align}
  \frac{{\left( a \right)_n \left( b \right)_n }}
{{\left( c \right)_n \left( 1 \right)_n }} = \frac{{a^{\,\overline {\,n\,} } b^{\,\overline {\,n\,} } }}
{{c^{\,\overline {\,n\,} } 1^{\,\overline {\,n\,} } }} &= \frac{{\Gamma (a + n)}}
{{\Gamma (a)}}\frac{{\Gamma (b + n)}}
{{\Gamma (b)}}\frac{{\Gamma (c)}}
{{\Gamma (c + n)}}\frac{{\Gamma (1)}}
{{\Gamma (1 + n)}}   \hfill \\
   &= \frac{{\Gamma (c)\Gamma (1)}}
{{\Gamma (a)\Gamma (b)}}n^{\,a + b - c - 1} \left[ {1 + O(n^{\, - 1} )} \right] \hfill \\ 
\end{align} 
$$
we see by Raabe's test (see e.g. Bromwich 1947, pp. 39, 241) that for $\left| z \right| = 1$ we have
$$\begin{array}{rcr}
\text{Absolute Convergence if} &  & \text{Re}(a+b-c)<0 \\
\text{Conditional Convergence if} & z \ne 1,& 0 \le \text{Re}(a+b-c) \lt 1 \\
\text{Divergence if} & &\text{Re}(a+b-c) \ge 1
\end{array}$$
End Citation
and some pages later it also provides the Gauss Theorem.
But it is to be noted that, speaking of convergence, we shall discriminate between the series above and its analytical continuation.
