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Is it possible to show that \begin{align}\frac{\left(a-n-1\right)!}{\left(a-1\right)!}\stackrel{?}{=}\frac{\left(-a\right)!\left(-1\right)^n}{\left(-a+n\right)!}\tag{1},\end{align} or, more conventionally (from the start of the problem, not the work I've done above), that \begin{align} \color{blue}{\frac{\Gamma\left(a-n\right)}{\Gamma\left(a\right)}=\frac{\left(-1\right)^n}{\left(1-a\right)_n},}\tag{2} \end{align} where $\left(1-a\right)_n$ is an increasing Pochhammer symbol? Hence, \begin{align} \left(1-a\right)_n=\frac{\Gamma\left(1-a+n\right)}{\Gamma\left(1-a\right)}.\tag{3} \end{align} I've been working with it a bit now and $\left(1\right)$ is where I was able to get (with no success).

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  • $\begingroup$ $(1)$ makes little sense because we cannot have $-a\in\mathbb N_0$ and $a-1\in\mathbb N_0$ at the same time. So better stay with $\Gamma$ $\endgroup$ – Hagen von Eitzen Apr 12 '15 at 19:48
  • $\begingroup$ @HagenvonEitzen I agree entirely $\endgroup$ – jm324354 Apr 12 '15 at 19:49
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Plugging $(3)$ into $(2)$, we get $$ \frac{\Gamma(a-n)}{\Gamma(a)}\stackrel?=\frac{(-1)^n\Gamma(1-a)}{\Gamma(1-a+n)}$$ or equivalently $$\tag 4 \Gamma(a-n)\Gamma(1-a+n)\stackrel?=(-1)^n\Gamma(1-a)\Gamma(a)$$ From Euler, we know $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin\pi x}$ for all $x\in\mathbb C-\mathbb Z$, hence $(4)$ reduces to the valid (outside $\mathbb Z$) equality $$\frac{\pi}{\sin(\pi(a-n))}=(-1)^n\frac{\pi}{\sin(\pi a)}$$

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I have found this quick paper that gave an identity I was not aware of, viz. \begin{align} \Gamma\left(\alpha\right)&=\frac{\Gamma\left(\alpha+1\right)}{\alpha},\tag{1} \end{align} so it seems that if $\alpha=-3.1$, then \begin{align} \Gamma\left(-3.1\right)&=\frac{\Gamma\left(-2.1\right)}{-3.1}\\ &=\frac{\Gamma\left(-1.1\right)}{6.51}\\ &=\frac{\Gamma\left(-0.1\right)}{-7.161}\\ &=\frac{\Gamma\left(0.9\right)}{0.7161}\approx 1.0685929671002\cdots \end{align} But then $\alpha\in\mathbb{R}\setminus\mathbb{Z^{-}}$?

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