Playing around with this series for natural values of $a,b$, it appears that more generally for $c\in\mathbb N$, $$\sum_{k=0}^\infty \frac{ (a+k)! \ (b+k)!}{k!\ (a+b+c+ k+1)! }=\frac{a!\ b!\ (c-1)!}{(a+c)!(b+c)!}$$ and obviously the factorials should be extendable to Gamma functions for all $a,b,c>0$ or even $\Re(a),\Re(b),\Re(c)>0$.
Moreover when introducing a variable $z\in[-1,1]$, it seems that for rational $z$ $$\sum_{k=0}^\infty \frac{ (a+k)! \ (b+k)!}{k!\ (a+b+c+k+1)! }z^k=p\log(1-z)+q$$ with rational $p,q$ (depending of course not only on $a,b,c$, but also on $z$).

I don't think this kind of identities is new, as they seem too elementary for that. But I can't find anything related, though Abramowitz & Stegun or Ryzhik should have them. (?)

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    $\begingroup$ 1 is the Gauss' sum en.wikipedia.org/wiki/… . 2 you can obtain as a solution of hypergeometric equation $\endgroup$
    – Nemo
    Jan 1, 2018 at 12:10
  • $\begingroup$ @Nemo Yes of course it is a Generalized hypergeometric series. The question is about a reference. $\endgroup$
    – Wolfgang
    Jan 1, 2018 at 12:23
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    $\begingroup$ The reference is Whittaker Watson, or any book called Special functions, these books have a chapter on hypergeometric functions. $\endgroup$ Jan 1, 2018 at 14:29
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    $\begingroup$ The first series looks divergent $\endgroup$ Jan 1, 2018 at 15:25
  • $\begingroup$ @FedorPetrov sorry, that was a typo of course. $\endgroup$
    – Wolfgang
    Jan 1, 2018 at 16:53

1 Answer 1


The sum $$\sum_{k=0}^\infty \frac{(a+k)!\,(b+k)!}{k!\,(a+b+c+k+1)!}z^k.$$ is not only a generalized hypergeometric series; it's the original ungeneralized Gauss hypergeometric series, $$\frac{\Gamma(a+1)\,\Gamma(b+1)}{\Gamma(a+b+c+2)}{}_2F_1\left({a+1,b+1\atop a+b+c+2}\ \Big |\, z\right).$$ Chapter 15 of Abramowitz and Stegun is on hypergeometric functions. Gauss's theorem (your first formula) is 15.1.20. For your second formula, see formulas 15.3.11 to 15.3.14 of Abramowitz and Stegun.

  • $\begingroup$ Thank you for the answer and the link! Typo corrected. $\endgroup$
    – Wolfgang
    Jan 1, 2018 at 16:36
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    $\begingroup$ The formulae 15.3.11 to 15.3.14 still contain an infinite sum. The fact that for rational $z$, the coefficient of $\log(1-z)$ as well as the sum over the digamma terms are rational seems to indicate for me that there should be a closed form e.g. for $\sum_{n=0}^\infty \frac{(a+m)_n\,(b+m)_n}{n!\,(n+m)!}z^n$. Don't you think so? $\endgroup$
    – Wolfgang
    Jan 1, 2018 at 16:50
  • $\begingroup$ You're right—there is an infinite sum. You should be able to evaluate $_2F_1(a+1, b+1;a+b+c+2;z)$ when $a$, $b$, and $c$ are nonnegative integers by using the formulas of section 15.2 of Abramowitz and Stegun, starting with $_2F_1(1,1;2;z) = -\log(1-z)/z$. There may be an explicit formula somewhere in the literature, but I wasn't able to find it. $\endgroup$
    – Ira Gessel
    Jan 1, 2018 at 18:44
  • $\begingroup$ That looks like it might be much easier to come up with some recursion formula, reducing $_2F_1(a+1, b+1;a+b+c+2;z)$ in terms of $_2F_1(a+1, b;a+b+c+1;z)$. $\endgroup$
    – Wolfgang
    Jan 1, 2018 at 20:25

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