How can I simplify, or transform, the following $_2F_1$ expression? Or evaluate its convergence? The $z$ term is problematic.
Expression: $$_2F_1\left(x+y,x;x+1;\dfrac{a+b+d}{b-a}\right)$$
where $x = 1,2,3,\ldots$ and $y = 0,1,2,\ldots$ and $a,b,d>0$ with $a I've already looked over Abramowitz and Stegun, DLMF (Digital Library of Mathematical Functions), and the Wolfram Functions Database in detail !
Given $\left(m,n\right)\in\mathbb{Z}_{\ge0}\times\mathbb{N}\land\left(a,b,c\right)\in\left(0,\infty\right)^{3}\land a $$f_{m,n}{\left(a,b,c\right)}:={_2F_1}{\left(m+n,n;n+1;\frac{a+b+c}{b-a}\right)},\tag{1}$$ where ${_2F_1}{\left(\alpha,\beta;\gamma;z\right)}$ is the Gauss hypergeometric function, which is primarily defined via the infinite series $${_2F_1}{\left(\alpha,\beta;\gamma;z\right)}:=\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{k}\,\left(\beta\right)_{k}\,z^{k}}{\left(\gamma\right)_{k}\,k!};~~~\small{\left|z\right|<1\lor\left[\left|z\right|=1\land\Re{\left(\gamma-\alpha-\beta\right)}>0\right]}.\tag{2}$$ Note that the argument is $z=\frac{a+b+c}{b-a}$ is greater than unity, provided that $0 $$z=\frac{a+b+c}{b-a}=1+\frac{2a+c}{b-a}>1.$$ This prevents us from defining our Gauss hypergeometric function by either its primary infinite series definition or Euler integral definition. Also complicating matters is the integer nature of the parameters. The complete definition for the Gauss hypergeometric function includes the following two cases: $$\begin{align}
{_2F_1}{\left(\alpha,\beta;\gamma;z\right)}
&:=\frac{\operatorname{B}{\left(\beta-\alpha,\gamma-\beta\right)}\,\left(-z\right)^{-\alpha}}{\operatorname{B}{\left(\beta,\gamma-\beta\right)}}\sum_{k=0}^{\infty}\frac{\left(\alpha\right)_{k}\,\left(\alpha-\gamma+1\right)_{k}\,z^{-k}}{\left(\alpha-\beta+1\right)_{k}\,k!}\\
&~~~~~+\frac{\operatorname{B}{\left(\beta,\alpha-\beta\right)}\,\left(-z\right)^{-\beta}}{\operatorname{B}{\left(\beta,\gamma-\beta\right)}}\sum_{k=0}^{\infty}\frac{\left(\beta\right)_{k}\,\left(\beta-\gamma+1\right)_{k}\,z^{-k}}{\left(\beta-\alpha+1\right)_{k}\,k!};~~~\small{\left|z\right|>1\land\alpha-\beta\notin\mathbb{Z}},\tag{3a}\\
\end{align}$$ and $${_2F_1}{\left(\alpha,\beta;\gamma;z\right)}:=\lim_{\epsilon\to0}{_2F_1}{\left(\alpha,\beta+\epsilon;\gamma;z\right)};~~~\small{\left|z\right|>1\land\alpha-\beta\in\mathbb{Z}}.\tag{3b}$$ Letting $\left(m,n\right)\in\mathbb{Z}_{\ge0}\times\mathbb{N}\land z\in\left(1,\infty\right)$, $$\begin{align}
{_2F_1}{\left(m+n,n;n+1;z\right)}
&=\lim_{\epsilon\to0}{_2F_1}{\left(m+n,n+\epsilon;n+1;z\right)}\\
&=\lim_{\epsilon\to0}\bigg{[}\frac{\operatorname{B}{\left(\epsilon-m,1-\epsilon\right)}\,\left(-z\right)^{-m-n}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\sum_{k=0}^{\infty}\frac{\left(m+n\right)_{k}\,\left(m\right)_{k}\,z^{-k}}{\left(m-\epsilon+1\right)_{k}\,k!}\\
&~~~~~+\frac{\operatorname{B}{\left(n+\epsilon,m-\epsilon\right)}\,\left(-z\right)^{-n-\epsilon}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\sum_{k=0}^{\infty}\frac{\left(n+\epsilon\right)_{k}\,\left(\epsilon\right)_{k}\,z^{-k}}{\left(\epsilon-m+1\right)_{k}\,k!}\bigg{]}\\
&=\lim_{\epsilon\to0}\bigg{[}\frac{\operatorname{B}{\left(\epsilon-m,1-\epsilon\right)}\,\left(-z\right)^{-m-n}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\,{_2F_1}{\left(m+n,m;m-\epsilon+1;\frac{1}{z}\right)}\\
&~~~~~+\frac{\operatorname{B}{\left(n+\epsilon,m-\epsilon\right)}\,\left(-z\right)^{-n-\epsilon}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\,{_2F_1}{\left(n+\epsilon,\epsilon;\epsilon-m+1;\frac{1}{z}\right)}\bigg{]}\\
&=\lim_{\epsilon\to0}\bigg{[}\frac{\operatorname{B}{\left(\epsilon-m,1-\epsilon\right)}\,\left(-z\right)^{-m-n}\left(1-\frac{1}{z}\right)^{-m-n}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\,{_2F_1}{\left(n+m,1-\epsilon;1-\epsilon+m;\frac{1}{1-z}\right)}\\
&~~~~~+\frac{\operatorname{B}{\left(n+\epsilon,m-\epsilon\right)}\,\left(-z\right)^{-n-\epsilon}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\,{_2F_1}{\left(n+\epsilon,\epsilon;\epsilon-m+1;\frac{1}{z}\right)}\bigg{]}.\\
\end{align}$$ At this point it will be convenient to consider the $m=0$ and $m>0$ cases separately. Consider first the $m>0$ case, where $\left(m,n\right)\in\mathbb{N}^{2}\land z\in\left(1,\infty\right)$. $$\begin{align}
{_2F_1}{\left(m+n,n;n+1;z\right)}
&=\lim_{\epsilon\to0}\bigg{[}\frac{\operatorname{B}{\left(\epsilon-m,1-\epsilon\right)}\,\left(-z\right)^{-m-n}\left(1-\frac{1}{z}\right)^{-m-n}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\\
&~~~~~\times\,{_2F_1}{\left(n+m,1-\epsilon;1-\epsilon+m;\frac{1}{1-z}\right)}\\
&~~~~~+\frac{\operatorname{B}{\left(n+\epsilon,m-\epsilon\right)}\,\left(-z\right)^{-n-\epsilon}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\,{_2F_1}{\left(n+\epsilon,\epsilon;\epsilon-m+1;\frac{1}{z}\right)}\bigg{]}\\
&=\lim_{\epsilon\to0}\bigg{[}\frac{1}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\cdot\frac{\sin{\left(m\pi\right)}\Gamma{\left(1-\epsilon\right)}}{\left(m-\epsilon\right)\sin{\left(\pi\left(\epsilon-m\right)\right)}}\cdot\frac{\Gamma{\left(m\right)}}{\Gamma{\left(m-\epsilon\right)}}\\
&~~~~~\times\left(1-z\right)^{-m-n}\,{_2F_1}{\left(n+m,1-\epsilon;1-\epsilon+m;\frac{1}{1-z}\right)}\\
&~~~~~+\frac{\operatorname{B}{\left(n+\epsilon,m-\epsilon\right)}\,\left(-z\right)^{-n-\epsilon}}{\operatorname{B}{\left(n+\epsilon,1-\epsilon\right)}}\,{_2F_1}{\left(n+\epsilon,\epsilon;\epsilon-m+1;\frac{1}{z}\right)}\bigg{]}\\
&=-\frac{n}{m}\left(1-z\right)^{-m-n}\,{_2F_1}{\left(n+m,1;1+m;\frac{1}{1-z}\right)}\\
&~~~~~+\frac{n\,\operatorname{B}{\left(m,n\right)}}{\left(-z\right)^{n}}\lim_{\epsilon\to0}{_2F_1}{\left(n+\epsilon,\epsilon;\epsilon-m+1;\frac{1}{z}\right)}\\
&=\frac{n\operatorname{B}{\left(m,n\right)}}{\left(-z\right)^{n}}-\frac{n}{m}\left(1-z\right)^{-m-n}\,{_2F_1}{\left(n+m,1;1+m;\frac{1}{1-z}\right)}\\
&=n\left(-z\right)^{-n}\operatorname{B}{\left(m,n\right)}-n\left(-z\right)^{-n}\left(1-z\right)^{-m}\sum_{k=0}^{n-1}\frac{\left(1-n\right)_{k}\,\left(-1\right)^{k}}{\left(k+m\right)\,k!\,\left(z-1\right)^{k}}\\
&=n\left(-z\right)^{-n}\operatorname{B}{\left(m,n\right)}-n\left(-z\right)^{-n}\left(1-z\right)^{-m}\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}}{\left(k+m\right)\,\left(z-1\right)^{k}}\\
&=n\left(-z\right)^{-n}\left[\operatorname{B}{\left(m,n\right)}-\left(-1\right)^{m}\sum_{k=0}^{n-1}\frac{\binom{n-1}{k}}{\left(m+k\right)\,\left(z-1\right)^{m+k}}\right].\\
\end{align}$$ It remains to consider the $m=0$ special case, but for the time being I prefer to leave this as an exercise to the reader...
I found Chapter 15 of [Olver, Lozier, Boisvert, Clark (Eds.) (2010). NIST Handbook of Mathematical Functions. New York, NY: Cambridge University Press.] very clear. As far as I know, $\,_{2}F_{1}(a,b,c;z)$ with $a,b,c>0$ converges for $|z|<1$ and diverges otherwise (see page 384).