Convergence or divergence of the gaussian hypergeometric function $_{2}F_1(a,b;c;z)$ where $a,b$ and $c$ are complex. I'm studying the convergence of the hypergeometric function and I have some questions. First, I define the hypergeometrical functions as
\begin{equation}
_{2}F_{1} (a,b;c;z)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(a)_n(b)_n}{(c)_n}\dfrac{1}{n!}z^n,
\end{equation}
where $(w)_n$ is the rising factorial and it is defined by
\begin{equation}
(w)_n=\left\lbrace \begin{array}{cc}
1 & \text{for}~n=0\\
w(w+1)\cdots (w-n+1) & \text{for}~n\neq 0
\end{array}\right.
\end{equation}
and the parameters $a$, $b$ and $c$ are complex.
Then, I know that:

*

*For $|z|=1,z\neq 1$, we have that it is not absolutely convergent, but convergent for $0\leq \Re(a+b-c)<1$. But I don't know how to prove that it is convergent there. Moreover, in this case, if $\Re(a+b-c)>1$ we obtain that it is divergent and if $\Re(a+b-c)=1$ and $\Re(a+b)\leq\Re(ab)$ it is also divergent. But I don't know again how to see it.

Thanks a lot.
 A: Let's first assume that $a$, $b$, and $c$ are real numbers.
Saying that $|z|=1$, $z \ne1$, is equivalent to saying that $z^n= e^{i n \theta} = \cos(n \theta) + i \sin(n\theta)$ for some $\theta$ between $0$ and $2 \pi$.
And the partial sums of both $\cos(n \theta)$ and $\sin (n \theta)$ are bounded.
Also, for large $n$,  $$ \begin{align}\frac{(a)_n(b)_n}{(c)_n}\frac{1}{n!} &= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \frac{\Gamma(a+n) \Gamma(b+n)}{\Gamma(c+n)} \frac{1}{\Gamma(1+n)} \frac{\Gamma(n)}{\Gamma(n)} \\ &= \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} \frac{\Gamma(a+n)}{\Gamma(n)} \frac{\Gamma(b+n)}{\Gamma(n)} \frac{\Gamma(n)}{\Gamma(c+n)} \frac{1}{n} \\ & \sim  \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} n^{a}n^{b}n^{-c}n^{-1} \tag{1} \\&=  \frac{\Gamma(c)}{\Gamma(a) \Gamma(b)} n^{a+b-c-1}, \end{align}$$ which decreases monotonically to zero if $a+b-c-1 <0$.
So when $|z|=1, z \ne 1$, we can use Dirichlet's test to conclude that the series converges if $a+b-c<1$.

I wasn't sure how to extend this result to complex-valued parameters.  But looking at the proof of Dirichlet's test, it looks like we need to show that $$ \sum_{n=m}^{\infty} \left| n^{a+b+c-1} - (n+1)^{a+b+c-1}\right| \tag{2}$$ converges for $\Re(a+b-c) <  1$ for some positive integer $m$.
To show that it converges, let $\alpha = a+b-c$.
Then for large $n$, $$ \begin{align} n^{\alpha-1}-(n+1)^{\alpha-1} &= n^{\alpha-1}-n^{\alpha-1} \left(1+ \frac{1}{n} \right)^{\alpha-1} \\ &\sim n^{\alpha-1} -n^{\alpha-1} \left(1+ \frac{\alpha}{n} + \mathcal{O}(n^{-2})  \right) \\ &= - \alpha n^{\alpha-2} + \mathcal{O}(n^{\alpha-3}). \end{align}$$
Therefore, $$|n^{\alpha-1} - (n+1)^{\alpha-1}| \sim |\alpha|n^{\Re(\alpha)-2},$$ which, by the p-series test, shows that $(2)$ converges for $\Re(\alpha) < 1$.

$(1)$ $\lim_{n \to \infty} \frac{\Gamma(\beta+n)}{\Gamma(n)\beta^{n}} =1$, which can be proven using Stirling's approximation
