Evaluating $\lim_{n\to\infty}\frac{1}{n+1}(\omega+\nu)^{(n+1)}z^\nu{_2F_1}(1,\omega+\nu+1;n+2;1-z)$

By (15.12.3) and (5.11.13), we have \begin{align*} S_n & = \frac{1}{{n + 1}}(\omega + \nu )^{(n + 1)} z^\nu \frac{{\Gamma (n + 2)}}{{\Gamma (n - \omega - \nu + 1)}}\frac{1}{{n^{\omega + \nu + 1} }}\left( {1 + \mathcal{O} \! \left( {\frac{1}{n}} \right)} \right) \\ & = (\omega + \nu )^{(n + 1)} z^\nu \frac{{\Gamma (n + 2)}}{{\Gamma (n - \omega - \nu + 1)}}\frac{1}{{n^{\omega + \nu + 2} }}\left( {1 + \mathcal{O} \! \left( {\frac{1}{n}} \right)} \right) \\ & = z^\nu (\omega + \nu )^{(n + 1)} \frac{1}{n}\left( {1 +\mathcal{O} \! \left( {\frac{1}{n}} \right)} \right) \\ & = ( - 1)^{n+1} \frac{{z^\nu }}{{\Gamma ( - \omega - \nu )}}\frac{{\Gamma (n - \omega - \nu )}}{n}\left( {1 + \mathcal{O} \! \left( {\frac{1}{n}} \right)} \right). \end{align*} If $\omega+\nu$ is not a non-negative integer, $S_n$ will tend to infinity in absolute value and will oscillate in sign.