Identity involving the rising factorial

Question

I am reading a book about hypergeometric functions and in a proof of a transformation they use the supposedly obvious fact $$ \displaystyle\frac{(c-a-b)_{n-r}}{(n-r)!} = \frac{(c-a-b)_{n} (-n)_{r}}{n!\ (1+a+b-c-n)_{r} } $$

I don't see how the two expression are equivalent.

Answer 1

Well this is maybe not obvious but doable:

• First note that $$\frac{(c-a-b)_{n-r}}{(c-a-b)_n}=\frac{\Gamma(c-a-b+n-r)}{\Gamma(c-a-b+n)}$$

• Next use the reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$ to write $$(1+a+b-c-n)_r\times\frac{\Gamma(c-a-b+n-r)}{\Gamma(c-a-b+n)}=\frac{\sin\pi(c-a-b+n)}{\sin\pi(c-a-b+n-r)}=(-1)^r.$$

• Now it remains to prove that $$\frac{(-n)_r(n-r)!}{n!}=(-1)^r,$$ which is more or less straightforward.