I try to calculate the following series \begin{align*} S_{n,m}(z)=\sum_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} \, z^{2k}, \end{align*} where $\Gamma(z)$ is the Gamma function and where $n,m\in \mathbb Z^+$ (positive integers).

More precisely I would like to show that the series $S_{n,m}(z)$ 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 $S_{n,m}(z)$, 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 $S_{n,m}(z)$, keeping in mind the expression of the hypergeometric functions $\displaystyle {}_{2}F_{1}(a,b;c;z)$, we get \begin{align*} S_{n,m}(z)&=\sum_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} \, z^{2k}\\ &=\sum_{k=0}^{\infty} {(-1)^{k} \frac{\Gamma(k+\frac{n+m}{2})\,\Gamma(k+\frac{m-1}{2})}{\Gamma(k+\frac{m}{2})}} \, \frac{(z^{2})^k}{k!}\\ &=\frac{\Gamma(\frac{n+m}{2})\Gamma(\frac{m-1}{2})}{\Gamma(\frac{m}{2})}\, {}_{2}F_{1}\left(\frac{n+m}{2},\frac{m-1}{2};\frac{m}{2};-z^2\right)\\ &=\frac{\Gamma(\frac{n+m}{2})\Gamma(\frac{m-1}{2})}{\Gamma(\frac{m}{2})}\, {}_{2}F_{1}\left(\frac{m}{2}+\frac{n}{2},\frac{m}{2}-\frac{1}{2};\frac{m}{2};-z^2\right)? \end{align*} If that's right, how can I show that $S_{n,m}(z)$ is an elementary function?

elementary functionis a function of a single variable (typically real or complex) that is defined as taking sums, products, and compositions of finitely many polynomial, rational, trigonometric, hyperbolic, and exponential functions, $\endgroup$