Investigate the convergence of the following series of complex numbers $$\sum_{n=1}^{\infty} \frac{a(a+1)...(a+n-1)b(b+1)...(b+n-1)}{n!c(c+1)...(c+n-1)}$$, when $Re(a+b-c)>0$.
I tried using D'Alembert ratio test, but since $a, b$ and $c$ are complex numbers and this works for real variables, it didn't help. Then I noticed that this expression looks like a general term of some Taylor series, but couldn't use this idea. Then I tried writing $\frac{a(a+1)...(a+n-1)b(b+1)...(b+n-1)}{n!c(c+1)...(c+n-1)}$ as ${e^{ln(\frac{a(a+1)...(a+n-1)b(b+1)...(b+n-1)}{n!c(c+1)...(c+n-1)})}}$, but this didn't get me far either.
Could you please help me, because this series has been haunting me for the last week and even my professor couldn't get this right.
 A: Sketch of a direct proof for the divergence of the series in the non-trivial case where $a,b,c$ are not non-positive integers as otherwise either the series is finite or is not well defined so nothing to prove.
Then we have the well-known limit  (see the Gauss formula down the page):
$$\Gamma (z) = \lim_{n \to \infty}\frac{n!n^z}{z(z+1)...z(z+n)}, z \ne 0,-1,...$$
In particular this means that one has $$\frac{a(a+1)...(a+n-1)\Gamma(a)}{(n-1)!(n-1)^a}=1+Q(n,a), Q(n,a) \to 0,  n \to \infty$$ and the analog results for $b,c$ hence for one has $$z_n(a,b,c)=\frac{a...(a+n-1)b...(b+n-1)}{n!c...(c+n-1)}=$$ $$=\frac{(1+Q(a,n))(1+Q(b,n))}{(1+Q(c,n))}\frac{(n-1)^{a+b-c}}{n}$$
Hence $z_n(a,b,c)=(1+Q(n))\frac{(n-1)^{a+b-c}}{n}, Q(n) \to 0, n \to \infty$
If $\Re (a+b-c-1) \ge 0$ then $\frac{(n-1)^{a+b-c}}{n}$ is not convergent to $0$, hence $z_n(a,b,c)$ is not convergent to zero either so $\sum z_n(a,b,c)$ diverges
If $0>\Re (a+b-c-1) >-1$ then $\frac{(n-1)^{a+b-c}}{n}=(n-1)^{a+b-c-1}+$ absolutely convergent term so $$z_{n+1}(a,b,c)=(1+Q(n))n^{-q}+y_n$$ where $q=1-a-b+c, 0<\Re q<1, \sum |y_n|<\infty, Q(n) \to 0$
But since $0<\Re q<1$ we know that $|\sum_{n=1}^m n^{-q}| \ge c(q)m^{1-\Re q}$ for $m$ large enough while $\sum_{n=1}^k |n^{-q}| \le c_1(q)k^{1-\Re q}$ so in particular $|\sum_{n=k}^m n^{-q}| \ge c_2(q)m^{1-\Re q}$ for $m \ge k^2$ and $k$ large say, hence choosing $k$ large enough st $|Q(n)|c_1(q)<c_2(q)/2, n \ge k$ one gets  $$|\sum_{n=k}^{m} z_{n+1}(a,b,c)| \ge |\sum_{n=k}^m n^{-q}|-\max_{k \le n \le m}|Q(n)|c_1(q)m^{1-\Re q} \ge \frac{c_2(q)m^{1-\Re q}}{2}$$ so $\sum z_n(a,b,c)$ cannot converge
Note that easy counterxamples like $x_n=\frac{1}{n}+(-1)^n\frac{2i}{\sqrt n}, n \ge 1$ and $Q(n)=i/\sqrt n$ for $n$ even and $Q(n)=0$ for $n$ odd so $Q(n) \to 0$, $\sum x_n$ diverges but $z_n=(1+Q(n))x_n=(-1)^{n+1}/n+(-1)^n\frac{2i}{\sqrt n}+\frac{(1+(-1)^n) i}{2n^{3/2}}$ shows that the relation $z_n=(1+Q(n))x_n, Q(n) \to 0$ and $\sum x_n$ divergent doesn't actually imply $\sum z_n$ divergent so one has to use the properties of the actual sequences here
