$\displaystyle \left(\lim_{N \to \infty} a_N\right)-a_m = \sum_{n=m}^\infty(a_{n+1}-a_n)$を利用すると、様々な公式を導くことができます。
例
\begin{align}
&\frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)} =
1+\left(\frac{ab}{cd}-1\right)\hygeo{F}{3}{2}{a,b,1}{c+1,d+1}{1} \ [a+b=c+d] \\
&\frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)}
\left(1-\frac{\sin(\pi c)\sin(\pi d)}{\sin(\pi a)\sin(\pi b)}\right) =
\left(\frac{ab}{cd}-1\right)\hygeo{H}{2}{2}{a,b}{c+1,d+1}{1} \ [a+b=c+d] \\
&\frac{(a;q)_\infty}{(b;q)_\infty} =
1+\frac{b-a}{1-b}\hygeo{\phi}{2}{1}{a,q}{bq}{q,q} \\
&\frac{(a,b;q)_\infty}{(c,d;q)_\infty} - \frac{\left(\frac{q}{c},\frac{q}{d};q\right)_\infty}{\left(\frac{q}{a},\frac{q}{b};q\right)_\infty} =
\frac{c+d-a-b}{(1-c)(1-d)}\hygeo{\psi}{2}{2}{a,b}{cq,dq}{q,q} \ [ab=cd]
\end{align}
Wikipedia:
${}_rF_s$
,
${}_rH_r$
$(a_1,\cdots,a_j)_n := (a_1)_n\cdots(a_j)_n$と略記します。
\begin{align} \Gamma(z) = \lim_{n\to\infty} \frac{n^zn!}{(z)_{n+1}}, \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)} \end{align}より、
$a_1+\cdots+a_j=b_1+\cdots+b_j$のとき、
\begin{alignat}2
&\lim_{n\to\infty} \frac{(a_1,\cdots,a_j)_n}{(b_1,\cdots,b_j)_n} &&=
\frac{\Gamma(b_1)\cdots\Gamma(b_j)}{\Gamma(a_1)\cdots\Gamma(a_j)} \\
&\lim_{n\to-\infty} \frac{(a_1,\cdots,a_j)_n}{(b_1,\cdots,b_j)_n} &&=
\frac{\Gamma(1-a_1)\cdots\Gamma(1-a_j)}{\Gamma(1-b_1)\cdots\Gamma(1-b_j)} =
\frac{\Gamma(b_1)\cdots\Gamma(b_j)}{\Gamma(a_1)\cdots\Gamma(a_j)}
\frac{\sin(\pi b_1)\cdots\sin(\pi b_j)}{\sin(\pi a_1)\cdots\sin(\pi a_j)}
\end{alignat}
$a+b=c+d$のとき、
\begin{align}
&\frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)} =
1+\left(\frac{ab}{cd}-1\right)\hygeo{F}{3}{2}{a,b,1}{c+1,d+1}{1} \\
&\frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)}
\left(1-\frac{\sin(\pi c)\sin(\pi d)}{\sin(\pi a)\sin(\pi b)}\right) =
\left(\frac{ab}{cd}-1\right)\hygeo{H}{2}{2}{a,b}{c+1,d+1}{1}
\end{align}
\begin{align} \frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)} - \frac{(a,b)_m}{(c,d)_m} &= \sum_{n=m}^\infty \left(\frac{(a,b)_{n+1}}{(c,d)_{n+1}} - \frac{(a,b)_n}{(c,d)_n} \right) \\&= \sum_{n=m}^\infty \frac{(a,b)_n}{(c,d)_{n+1}}\lbrace(a+n)(b+n)-(c+n)(d+n)\rbrace \\&= \frac{ab-cd}{cd}\sum_{n=m}^\infty \frac{(a,b)_n}{(c+1,d+1)_n} \\&= \left(\frac{ab}{cd}-1\right)\sum_{n=m}^\infty \frac{(a,b)_n}{(c+1,d+1)_n} \end{align}
$m=0$のとき、
$\displaystyle
\frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)} - 1 =
\left(\frac{ab}{cd}-1\right)\hygeo{F}{3}{2}{a,b,1}{c+1,d+1}{1}
$
$m\to-\infty$のとき、
\begin{align}
\frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)} -
\frac{\Gamma(c)\Gamma(d)}{\Gamma(a)\Gamma(b)}
\frac{\sin(\pi c)\sin(\pi d)}{\sin(\pi a)\sin(\pi b)} =
\left(\frac{ab}{cd}-1\right)\hygeo{H}{2}{2}{a,b}{c+1,d+1}{1}
\end{align}
\begin{align}
A&:=abc-dfg, \\
B&:=ab+bc+ca-df-fg-gd
\end{align}
$a+b+c=d+f+g$のとき、
\begin{align}
&\frac{\Gamma(d)\Gamma(f)\Gamma(g)}{\Gamma(a)\Gamma(b)\Gamma(c)} =
1+\left(\frac{abc}{dfg}-1\right)
\hygeo{F}{5}{4}{a,b,c,\frac{A}{B}+1,1}{d+1,f+1,g+1,\frac{A}{B}}{1} \\
&\frac{\Gamma(d)\Gamma(f)\Gamma(g)}{\Gamma(a)\Gamma(b)\Gamma(c)}
\left(1-\frac{\sin(\pi d)\sin(\pi f)\sin(\pi g)}
{\sin(\pi a)\sin(\pi b)\sin(\pi c)}\right) =
\left(\frac{abc}{dfg}-1\right)\hygeo{H}{4}{4}{a,b,c,\frac{A}{B}+1}{d+1,f+1,g+1,\frac{A}{B}}{1}
\end{align}
\begin{align} \frac{\Gamma(d)\Gamma(f)\Gamma(g)}{\Gamma(a)\Gamma(b)\Gamma(c)} - \frac{(a,b,c)_m}{(d,f,g)_m} &= \sum_{n=m}^\infty \left(\frac{(a,b,c)_{n+1}}{(d,f,g)_{n+1}} - \frac{(a,b,c)_n}{(d,f,g)_n} \right) \\&= \sum_{n=m}^\infty \frac{(a,b,c)_n}{(d,f,g)_{n+1}}\lbrace(a+n)(b+n)(c+n)-(d+n)(f+n)(g+n)\rbrace \\&= \sum_{n=m}^\infty \frac{(a,b,c)_n}{(d,f,g)_{n+1}}(A+Bn) \\&= \frac{A}{dfg}\sum_{n=m}^\infty \frac{(a,b,c,\frac{A}{B}+1)_n}{(d+1,f+1,g+1,\frac{A}{B})_n} \\&= \left(\frac{abc}{dfg}-1\right)\sum_{n=m}^\infty \frac{(a,b,c,\frac{A}{B}+1)_n}{(d+1,f+1,g+1,\frac{A}{B})_n} \end{align}
$m=0$のとき、
$\displaystyle
\frac{\Gamma(d)\Gamma(f)\Gamma(g)}{\Gamma(a)\Gamma(b)\Gamma(c)} - 1 =
\left(\frac{abc}{dfg}-1\right)\hygeo{F}{5}{4}{a,b,c,\frac{A}{B}+1,1}{d+1,f+1,g+1,\frac{A}{B}}{1}
$
$m\to-\infty$のとき、
\begin{align}
\frac{\Gamma(d)\Gamma(f)\Gamma(g)}{\Gamma(a)\Gamma(b)\Gamma(c)} -
\frac{\Gamma(d)\Gamma(f)\Gamma(g)}{\Gamma(a)\Gamma(b)\Gamma(c)}
\frac{\sin(\pi d)\sin(\pi f)\sin(\pi g)}
{\sin(\pi a)\sin(\pi b)\sin(\pi c)} =
\left(\frac{abc}{dfg}-1\right)\hygeo{H}{4}{4}{a,b,c,\frac{A}{B}+1}{d+1,f+1,g+1,\frac{A}{B}}{1}
\end{align}
$a_1 \cdots a_j=b_1 \cdots b_j$のとき、
\begin{align}
\lim_{n\to-\infty} \frac{(a_1,\cdots,a_j;q)_n}{(b_1,\cdots,b_j;q)_n} &=
\frac{\left(\frac{q}{b_1},\cdots,\frac{q}{b_j};q\right)_\infty}
{\left(\frac{q}{a_1},\cdots,\frac{q}{a_j};q\right)_\infty}
\end{align}
\begin{align} \frac{(a;q)_\infty}{(b;q)_\infty} &= 1+\frac{b-a}{1-b}\hygeo{\phi}{2}{1}{a,q}{bq}{q,q} \\ \end{align}
\begin{align}
\frac{(a;q)_\infty}{(b;q)_\infty} - \frac{(a;q)_m}{(b;q)_m} &=
\sum_{n=m}^\infty \left(\frac{(a;q)_{n+1}}{(b;q)_{n+1}} - \frac{(a;q)_n}{(b;q)_n} \right) \\&=
\sum_{n=m}^\infty \frac{(a;q)_n}{(b;q)_{n+1}}\lbrace(1-aq^n)-(1-bq^n)\rbrace \\&=
(b-a)\sum_{n=m}^\infty \frac{(a;q)_n}{(b;q)_{n+1}}q^n \\&=
\frac{b-a}{1-b}\sum_{n=m}^\infty \frac{(a;q)_n}{(bq;q)_n}q^n
\end{align}
$m=0$のとき、
\begin{align}
\frac{(a;q)_\infty}{(b;q)_\infty} - 1 &=
\frac{b-a}{1-b}\hygeo{\phi}{2}{1}{a,q}{bq}{q,q} \\
\end{align}
\begin{align}
A&:=ab-cd, \\
B&:=a+b-c-d
\end{align}
\begin{align}
\frac{(a,b;q)_\infty}{(c,d;q)_\infty} &=
1 + \frac1{(1-c)(1-d)}\left(-B\hygeo{\phi}{3}{2}{a,b,q}{cq,dq}{q,q} +
A\hygeo{\phi}{3}{2}{a,b,q}{cq,dq}{q,q^2}\right)
\end{align}
$A=0$のとき、
\begin{align}
\frac{(a,b;q)_\infty}{(c,d;q)_\infty} - \frac{\left(\frac{q}{c},\frac{q}{d};q\right)_\infty}{\left(\frac{q}{a},\frac{q}{b};q\right)_\infty} &=
-\frac{B}{(1-c)(1-d)}\hygeo{\psi}{2}{2}{a,b}{cq,dq}{q,q}
\end{align}
\begin{align} \frac{(a,b;q)_\infty}{(c,d;q)_\infty} - \frac{(a,b;q)_m}{(c,d;q)_m} &= \sum_{n=m}^\infty \left(\frac{(a,b;q)_{n+1}}{(c,d;q)_{n+1}} - \frac{(a,b;q)_n}{(c,d;q)_n} \right) \\&= \sum_{n=m}^\infty \frac{(a,b;q)_n}{(c,d;q)_{n+1}}\lbrace(1-aq^n)(1-bq^n)-(1-cq^n)(1-dq^n)\rbrace \\&= \sum_{n=m}^\infty \frac{(a,b;q)_n}{(c,d;q)_{n+1}}(-Bq^n+Aq^{2n}) \\&= \frac1{(1-c)(1-d)}\sum_{n=m}^\infty \frac{(a,b;q)_n}{(cq,dq;q)_n}(-Bq^n+Aq^{2n}) \end{align}
$m=0$のとき、
\begin{align}
\frac{(a,b;q)_\infty}{(c,d;q)_\infty} - 1 &=
\frac1{(1-c)(1-d)}\left(-B\hygeo{\phi}{3}{2}{a,b,q}{cq,dq}{q,q} +
A\hygeo{\phi}{3}{2}{a,b,q}{cq,dq}{q,q^2}\right)
\end{align}
$A=0 \land m\to-\infty$のとき、
\begin{align}
\frac{(a,b;q)_\infty}{(c,d;q)_\infty} - \frac{\left(\frac{q}{c},\frac{q}{d};q\right)_\infty}{\left(\frac{q}{a},\frac{q}{b};q\right)_\infty} &=
-\frac{B}{(1-c)(1-d)}\hygeo{\psi}{2}{2}{a,b}{cq,dq}{q,q}
\end{align}
\begin{align}
A&:=abc-dfg, \\
B&:=ab+bc+ca-df-fg-gd, \\
C&:=a+b+c-d-f-g
\end{align}
\begin{align}
\frac{(a,b,c;q)_\infty}{(d,f,g;q)_\infty} &=
1 + \frac1{(1-d)(1-f)(1-g)}\left(-C\hygeo{\phi}{4}{3}{a,b,c,q}{dq,fq,gq}{q,q} +
B\hygeo{\phi}{4}{3}{a,b,c,q}{dq,fq,gq}{q,q^2} -
A\hygeo{\phi}{4}{3}{a,b,c,q}{dq,fq,gq}{q,q^3}\right)
\end{align}
$A=0$のとき、
\begin{align}
\frac{(a,b,c;q)_\infty}{(d,f,g;q)_\infty} -
\frac{\left(\frac{q}{d},\frac{q}{f},\frac{q}{g};q\right)_\infty}
{\left(\frac{q}{a},\frac{q}{b},\frac{q}{c};q\right)_\infty} &=
\frac1{(1-d)(1-f)(1-g)}\left(-C\hygeo{\psi}{3}{3}{a,b,c}{dq,fq,gq}{q,q} +
B\hygeo{\psi}{3}{3}{a,b,c}{dq,fq,gq}{q,q^2}\right)
\end{align}
\begin{align} \frac{(a,b,c;q)_\infty}{(d,f,g;q)_\infty} - \frac{(a,b,c;q)_m}{(d,f,g;q)_m} &= \sum_{n=m}^\infty \left(\frac{(a,b,c;q)_{n+1}}{(d,f,g;q)_{n+1}} - \frac{(a,b,c;q)_n}{(d,f,g;q)_n} \right) \\&= \sum_{n=m}^\infty \frac{(a,b,c;q)_n}{(d,f,g;q)_{n+1}} \lbrace(1-aq^n)(1-bq^n)(1-cq^n)-(1-dq^n)(1-fq^n)(1-gq^n)\rbrace \\&= \sum_{n=m}^\infty \frac{(a,b,c;q)_n}{(d,f,g;q)_{n+1}}(-Cq^n+Bq^{2n}-Aq^{3n}) \\&= \frac1{(1-d)(1-f)(1-g)}\sum_{n=m}^\infty \frac{(a,b,c;q)_n}{(dq,fq,gq;q)_n}(-Cq^n+Bq^{2n}-Aq^{3n}) \end{align}
$m=0$のとき、
\begin{align}
\frac{(a,b,c;q)_\infty}{(d,f,g;q)_\infty} - 1 &=
\frac1{(1-d)(1-f)(1-g)}\left(-C\hygeo{\phi}{4}{3}{a,b,c,q}{dq,fq,gq}{q,q} +
B\hygeo{\phi}{4}{3}{a,b,c,q}{dq,fq,gq}{q,q^2} -
A\hygeo{\phi}{4}{3}{a,b,c,q}{dq,fq,gq}{q,q^3}\right)
\end{align}
$A=0 \land m\to-\infty$のとき、
\begin{align}
\frac{(a,b,c;q)_\infty}{(d,f,g;q)_\infty} -
\frac{\left(\frac{q}{d},\frac{q}{f},\frac{q}{g};q\right)_\infty}
{\left(\frac{q}{a},\frac{q}{b},\frac{q}{c};q\right)_\infty} &=
\frac1{(1-d)(1-f)(1-g)}\left(-C\hygeo{\psi}{3}{3}{a,b,c}{dq,fq,gq}{q,q} +
B\hygeo{\psi}{3}{3}{a,b,c}{dq,fq,gq}{q,q^2}\right)
\end{align}
同じ方法でいくらでも変数を増やすことができます。また、それぞれの式が変数が少ない式を特別な場合として含みます。