I recently found a proof for the following sum \begin{align*} S_n & =\sum_{k=0}^n\mathcal S_n^{(k)}(\Phi(z,-k,\omega)-z^\nu\Phi(z,-k,\omega+\nu))\\ & =\frac{1}{n+1}(\omega+\nu)^{(n+1)}z^\nu{_2F_1}(1,\omega+\nu+1;n+2;1-z), \end{align*} where $\mathcal S_n^{(k)}$ are Stirling numbers of the 1st-kind, $\Phi(\cdot)$ is the Lerch transcendent, $(s)^{(n)}=\Gamma(s+1)/\Gamma(s-n+1)$ is the falling factorial, and ${_2F_1}(\cdot)$ is the hypergeometric function.
My question is about finding an expression for $S=\lim_{n\to\infty} S_n$. If we restrict $z\in(0,1)$ each of the special functions listed may be represented as a convergent power series which may aid in this calculation. The r.h.s. then becomes $$ S=z^\nu\Gamma(\omega+\nu+1)\lim_{n\to\infty}\frac{1}{n+1}\frac{1}{\Gamma(\omega+\nu-n)}\sum_{k=0}^\infty\frac{(\omega+\nu+1)_k}{(n+2)_k}(1-z)^k. $$ Its not immediately obvious to me what happens with the $\Gamma(\omega+\nu-n)$ term as $n$ becomes large and this is where I got stuck. How to proceed?
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.