Is there an evaluation of this sum (possibly involving gamma functions)? $k$ and $n$ are natural numbers and $x$ is real with $0<x<1$.
$$ \sum\_{\substack{k=0\\n-k\text{ even}}}^n \frac{(-1)^{(n-k)/2}}{k+x} \frac{(n+k)!}{k!(\frac{n+k}{2})!(\frac{n-k}{2})!}$$
Any help is much appreciated.


EDIT: I know the result for a similar sum:
$$ \sum\_{k=0}^n\frac{(-1)^{n-k}}{k+x}\frac{(n+k)!}{k!^2(n-k)!}=\frac{1}{x}\prod\_{k=1}^n\frac{x-k}{x+k} $$



Evaluation of sum of factorials
Plugging the sum into MAPLE gives
$$
g1:=(-1)\text{^}((n-k)/2)\*1/(k+x)\*(n+k)!/k!/((n+k)/2)!/((n-k)/2)!;
$$
Plug in $n=2m$ even:
$$
g2:=\text{subs}(k=n-2\*l,n=2\*m,g1);
$$
Then perform summation:
$$
Seven:=\text{SumTools[DefiniteSummation]}(g2,l=0..m);
$$
Result (modulo typos):
$$
Seven=\frac{(-1)^m4^m\Gamma(1+\frac x2)\Gamma(m-\frac x2+\frac12)}{x\Gamma(1/2-\frac x2)\Gamma(m+\frac x2+1)}
$$
Example for $n=10$:
$$
simplify(subs(m=5,Seven));
$$
yields
$$
\frac{1024(-1+x)(-3+x)(-5+x)(-7+x)(-9+x)}{(x(x+2)(x+4)(6+x)(8+x)(10+x)}
$$
For odd $n=2m+1$ one gets something similar:
$$
Sodd=\frac{2(-1)^m4^m\Gamma(\frac32+\frac x2)\Gamma(m-\frac x2+1)}{(1+x)\Gamma(-\frac x2+1)\Gamma(m+\frac32+\frac x2)}
$$
Example $n=11$
$$
simplify(subs(m=5,Sodd));
$$
yields
$$
2048\,{\frac { \left( -2+x \right) \left( -4+x \right) \left( -6+x
 \right) \left( -8+x \right) \left( -10+x \right) }{ \left( 1+x
 \right) \left( x+3 \right) \left( x+5 \right) \left( 7+x \right) 
 \left( 9+x \right) \left( 11+x \right) }}
$$


So the result is
$$
2^n\prod\_{i=0}^{n\_0}(x+(-1)^i(n-i))^{(-1)^{i-1}}
$$
where $n\_0=n$ if $n$ is even and $n\_0=n-1$ if $n$ is odd


Remark: Maple uses the Wilf-Zeilberger algorithm. It is even possible generate a "human readable" proof of the formula (a so-called certificate) using Maple.


