\[\sum_{k=0}^{m}\frac{{\left(\mathbf{a}\right)_{k}}}{{\left(\mathbf{b}\right)_{k%
}}}\frac{z^{k}}{k!}=\frac{{\left(\mathbf{a}\right)_{m}}z^{m}}{{\left(\mathbf{b%
}\right)_{m}}m!}{{}_{q+2}F_{p}}\left({-m,1,1-m-\mathbf{b}\atop 1-m-\mathbf{a}}%
;\frac{(-1)^{p+q+1}}{z}\right).\]