I try to calculate the following series \begin{align*} S_{n,m}(z)=\sum_{k=0}^{\infty}\, (-1)^{k} \frac{\Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{2^{2k}k!\,\Gamma(k+\frac{m}{2})} \, z^{2k}. \qquad \qquad (*) \end{align*} where $\Gamma(z)$ is the Gamma function and where $n,m\in \mathbb Z^+$ (non-negative integers).
More precisely I would like to show that the series $(*)$ is an elementary function.
I used the Legendre's duplication formula: \begin{align*} \Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z), \end{align*} to simplify the expression of the series $(*)$, but without success.
Otherwise, I thought of the hypergeometric functions: $$\displaystyle {}_{2}F_{1}(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{z^{n}}{n!}.$$ Then, returning back to $(*)$, keeping in mind the expression of the hypergeometric functions $\displaystyle {}_{2}F_{1}(a,b;c;z)$, we obtain that \begin{align*} S_{n,m}(z)&=\sum_{k=0}^{\infty}\, (-1)^{k} \frac{\Gamma(k+\frac{n+m}{2})\,\Gamma(k+\frac{m-1}{2})}{k!\,\Gamma(k+\frac{m}{2})} \, \left(\frac{z}{2}\right)^{2k}\\ &= {}_{2}F_{1}\left(\frac{n+m}{2},\frac{m-1}{2};\frac{m}{2};z\right)\\ &= {}_{2}F_{1}\left(\frac{m}{2}+\frac{n}{2},\frac{m}{2}-\frac{1}{2};\frac{m}{2};-(z/2)^2\right)?? \end{align*} If that's right, how can I show that $S_{n,m}(z)$ is an elementary function?
Thank you in advance