$\sum_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} z^{2k}$ is an elementary function 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?
 A: As Carlo noted, for $n$ an even integer, $S_{n,m}(z)$ is an elementary function of $z$.
What about $n$ odd?
When $n,m$ are both odd, I get something in terms of arcsinh, also elementary.

But for $n$ odd and $m$ even, Maple gets complete elliptic integrals, which are not elementary... Examples
$$
S_{1,2}(z) = \frac{1}{\sqrt{z^2+1}}\;E\left(\frac{z}{\sqrt{z^2+1}\;}\right)
$$
and
$$
S_{7,6}(z) =
3{\frac {24\,{z}^{10}+148\,{z}^{8}+398\,{z}^{6}+669\,{z}^{4}-280
\,{z}^{2}-35}{ 16\left( {z}^{2}+1 \right) ^{11/2}{z}^{4}}{E}
 \left( {\frac {z}{\sqrt {{z}^{2}+1}}} \right) }-{\frac {72\,{z}^{8}+
381\,{z}^{6}+864\,{z}^{4}-1575\,{z}^{2}-210}{32\, \left( {z}^{2}+1
 \right) ^{11/2}{z}^{4}}{K} \left( {\frac {z}{\sqrt {{z}^{
2}+1}}} \right) }
$$
A: The arguments of the term ${}_{2}F_{1}\left(\frac{m}{2}+\frac{n}{2},\frac{m}{2}-\frac{1}{2};\frac{m}{2};-z^2\right)$ are not special. So, it is very unlikely that this term can be expressed anyhow other than tautologically. Mathematica cannot do anything with it:

