Is there a closed form expression for
$$ \sum\_{k=n}^\infty\frac{k!^2}{(k+x)(k-n)!(k+n+1)!} $$
where $0<x<1$ ?


(For $n=0$, I know that
$$\sum\_{k=0}^\infty\frac{1}{(k+x)(k+1)}=\frac{\psi(x)+\gamma}{x-1}$$
where $\psi(x)$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant).



Factorial Series
[Wolfram Programming Lab](http://www.wolfram.com/programming-lab/) gives the slightly "simpler"
$$
\sum\_{k=n}^{\infty} \frac{(k!)^2}{(k+x)(k-n)!(k+n+1)!} \\ = \frac{(n!)^2}{(n+x)(2 n+1)!}\ {}\_3F\_2(n+1,n+1,n+x; 2n+2,n+x+1;1),
$$
with a generalized hypergeometric function instead of Meijer's G-function.


**Edit:** In case $x$ is a negative integer and $x \le -n$ ($n$ is assumed to be a non-negative integer) the hypergeometric function is Saalschützian (see [MathWorld](http://mathworld.wolfram.com/SaalschuetzsTheorem.html)) and thus we get:
$$
{}\_3F\_2(n+1,n+1,n+x; 2n+2,n+x+1;1) =(-1)^{n+x} \Gamma(1-(n+x)) \frac{((n+1)\_{|n+x|})^2}{(2 n + 2)\_{|n+x|}}
$$
with $(a)\_k$ the [Pochhammer symbol](http://mathworld.wolfram.com/PochhammerSymbol.html).


