Is there a way to simplify anyhow (also taking some limits of big/small parameters) the following hypergeometric function $_2F_1(1,-a,-a-b,1)$ for integer values of both $a$ and $b$ ?

  • $\begingroup$ If $\{a\in \mathbb{Z},b\in \mathbb{Z},a>0,b>0\}$ is: $\, _2F_1(1,-a;-a-b;1)=1+\frac{a}{1+b}$ $\endgroup$ – Mariusz Iwaniuk Mar 7 '18 at 20:35
  • $\begingroup$ how can I prove that ? $\endgroup$ – Ninja Warrior Mar 7 '18 at 20:48
  • $\begingroup$ Mathematica and Maple says that. $\endgroup$ – Mariusz Iwaniuk Mar 7 '18 at 21:28

Using the Chu-Vandermonde identity (a special case of Gauss's theorem) in the case of $a$ a positive integer, we have

$$\begin{align*} {}_2 F_1\left({{-a,1}\atop{-a-b}}\middle|1\right)&=\frac{(-a-b-1)_a}{(-a-b)_a}\\ &=\frac{\frac{(-1)^a(a+b+1)!}{(b+1)!}}{\frac{(-1)^a(a+b)!}{b!}}=\frac{a+b+1}{b+1} \end{align*}$$


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