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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).

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2 Answers 2

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Wolfram 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) 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.

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Maple does this in terms of the Meijer G function: $$ \sum _{k=n}^{\infty }{\frac { \left( k! \right) ^{2}}{ \left( k+x \right) \left( k-n \right) !\, \left( k+n+1 \right) !}} =G^{3, 1}_{3, 3}\left(-1\, \Big\vert\,^{1, 2+2\,n, 1+n+x}_{n+x, n+1, n+1}\right) $$ which probably just means it doesn't know a closed form.

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