Infinite sum of falling factorial quotients. Computing the infinite sum of falling factorial quotients in Mathematica I've got the result: 
$$
\sum_{k=0}^\infty\frac{z_1^\underline{k}}{(-z_2)^\underline{k}}=\frac{\Gamma(z_2)\Gamma(z_1+z_2-1)}{\Gamma(z_2-1)\Gamma(z_1+z_2)}.
$$
How to prove the result? What are the necessary and sufficient conditions for convergence of the series, assuming that both $z_1$ and $z_2$ are not integer? Any hint is appreciated.
 A: Evaluation
$$\newcommand{\Re}{\operatorname{Re}}
\begin{align}
\sum_{k=0}^\infty\frac{z_1^\underline{k}}{(-z_2)^\underline{k}}
&=\sum_{k=0}^\infty\frac{(k-z_1-1)^{\underline{k}}}{(k+z_2-1)^{\underline{k}}}\tag1\\
&=\sum_{k=0}^\infty\frac{\Gamma(k-z_1)}{\Gamma(-z_1)}\frac{\Gamma(z_2)}{\Gamma(k+z_2)}\tag2\\
&=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\sum_{k=0}^\infty\frac{\Gamma(k-z_1)\Gamma(z_1+z_2)}{\Gamma(k+z_2)}\tag3\\
&=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\sum_{k=0}^\infty\int_0^\infty\frac{t^{z_1+z_2-1}}{(1+t)^{k+z_2}}\,\mathrm{d}t\tag4\\
&=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\int_0^\infty\frac{t^{z_1+z_2-2}}{(1+t)^{z_2-1}}\,\mathrm{d}t\tag5\\[6pt]
&=\frac{\Gamma(z_2)}{\Gamma(-z_1)\Gamma(z_1+z_2)}\frac{\Gamma(z_1+z_2-1)\Gamma(-z_1)}{\Gamma(z_2-1)}\tag6\\[3pt]
&=\bbox[5px,border:2px solid #C0A000]{\frac{z_2-1}{z_1+z_2-1}}\tag7
\end{align}
$$
Explanation:
$(1)$: $z^{\underline{k}}=(-1)^k(k-z-1)^{\underline{k}}$
$(2)$: write the falling factorial using the Gamma Function
$(3)$: algebraic manipulation
$(4)$: apply the Beta Function integral
$(5)$: sum the geometric series
$(6)$: apply the Beta Function integral, which converges for $\Re(z_1)\lt0$ and $\Re(z_1+z_2)\gt1$
$(7)$: $\Gamma(z+1)=z\,\Gamma(z)$

Convergence
Note that as $k\to\infty$,
$$
\begin{align}
\frac{z_1^{\underline{k}}}{(-z_2)^{\underline{k}}}
&=\frac{(k-z_1-1)^{\underline{k}}}{(k+z_2-1)^{\underline{k}}}\\
&=\frac{\Gamma(k-z_1)}{\Gamma(-z_1)}\frac{\Gamma(z_2)}{\Gamma(k+z_2)}\\
&\sim\frac{\Gamma(z_2)}{\Gamma(-z_1)}k^{-z_1-z_2}\tag8
\end{align}
$$
Therefore, as long as $z_2$ is not a non-positive integer (so that $(-z_2)^{\underline{k}}$ never vanishes), the series will converge for $\Re(z_1+z_2)\gt1$. By analytic continuation, $(7)$ will hold for $\Re(z_1+z_2)\gt1$.
A: Another approach is through the hypergeometric function.
In fact the ratio can be written in terms of Rising Factorials and then as terms of the hypergeometric sum 
$$
{{z_{\,1} ^{\,\underline {\,k\,} } } \over {\left( { - z_{\,2} } \right)^{\,\underline {\,k\,} } }}
 = {{\left( { - 1} \right)^{\,k} \left( { - z_{\,1} } \right)^{\,\overline {\,k\,} } } \over {\left( { - 1} \right)^{\,k} \left( {z_{\,2} } \right)^{\,\overline {\,k\,} } }}
 = {{\left( { - z_{\,1} } \right)^{\,\overline {\,k\,} } 1^{\,\overline {\,k\,} } } \over {\left( {z_{\,2} } \right)^{\,\overline {\,k\,} } }}{1 \over {k!}}
$$
The infinite sum is therefore the hypergeometric function computed at the unitary value of its argument
$$
\sum\limits_{k = 0}^\infty  {{{z_{\,1} ^{\,\underline {\,k\,} } } \over {\left( { - z_{\,2} } \right)^{\,\underline {\,k\,} } }}}  = {}_2F_{\,1} \left( {\left. {\matrix{
   { - z_{\,1} ,1}  \cr 
   {z_{\,2} }  \cr 
 } \;} \right|\;1} \right)
$$
which in virtue of the Gauss theorem
 gives
$$
\eqalign{
  & \sum\limits_{k = 0}^\infty  {{{z_{\,1} ^{\,\underline {\,k\,} } } \over {\left( { - z_{\,2} } \right)^{\,\underline {\,k\,} } }}}  = {}_2F_{\,1} \left( {\left. {\matrix{
   { - z_{\,1} ,1}  \cr 
   {z_{\,2} }  \cr 
 } \;} \right|\;1} \right) = {{\Gamma (z_{\,2} )\Gamma (z_{\,2}  + z_{\,1}  - 1)} \over {\Gamma (z_{\,2}  + z_{\,1} )\Gamma (z_{\,2}  - 1)}} =   \cr 
  &  = {{\left( {z_{\,2}  - 1} \right)\Gamma (z_{\,2}  - 1)\Gamma (z_{\,2}  + z_{\,1}  - 1)} \over {\left( {z_{\,2}  + z_{\,1}  - 1} \right)\Gamma (z_{\,2}  + z_{\,1} )\Gamma (z_{\,2}  - 1)}}
 = {{\left( {z_{\,2}  - 1} \right)} \over {\left( {z_{\,2}  + z_{\,1}  - 1} \right)}}\quad \left| {\;1 < {\mathop{\rm Re}\nolimits} \left( {z_{\,1} } \right) + {\mathop{\rm Re}\nolimits} \left( {z_{\,2} } \right)} \right. \cr} 
$$
including the range of convergence.
