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I apoligize if this is a stupid/obvious question, but last night I was wondering how we can compute limits for factorials of negative integers, for instance, how do we evaluate:


Neither $x!$, nor $(2x)!$ are defined for $x\in\mathbb{Z}^{-}$, and indeed, both are singularities according to the graph of $\Gamma(x+1)$.

The book I am reading calculates this using a previously shown identity that:

$$F\left(\left.{1-c-2n,-2n \atop c}\right|-1\right)=(-1)^{n}\frac{(2n)!}{n!}\frac{(c-1)!}{(c+n-1)!},\space\forall n\in\mathbb{Z}^{*}$$

And then, the more general Kummer's Formula:

$$F\left(\left.{a,b \atop 1+b-a}\right|-1\right)=\frac{(b/2)!}{b!}(b-a)^{\underline{b/2}}$$

It then shows that they would only produce consistent results if:

$$(-1)^{n}\frac{(2n)!}{n!}=\lim_{b\to-2n}{\frac{(b/2)!}{b!}}=\lim_{x\to-n}{\frac{x!}{(2x)!}},\space n\in\mathbb{Z}^{*}$$

It then gives the example of $n=3$, proving that:


However, using Wolfram|Alpha, I can see that there are other such limits defined (such as $\lim_{x\to-3}{\frac{x!}{(8x)!}}=-103408066955539906560000$.

Without using the hypergeometric series, how could we evaluate limits such as these?

Again, sorry if this is a stupid question, thanks in advance!

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(Some of your limits have $n\to$ but $x$ rather than $n$ in the expression.) The Gamma function has simple poles at negative integers with easily-describable residues (alternating sign reciprocals of factorials); this could probably be exploited. – anon Jul 8 '12 at 12:22
@anon I've fixed the limits. So would it be legal to do something like: $$\lim_{x\to-3}{\frac{x!}{(4x)!}}=\frac{(-1)^{3}}{3!}\div\frac{(-1)^{12}}{12!}=-‌​79833600$$ Which Wolfram|Alpha says is the correct answer? – Shaktal Jul 8 '12 at 12:30
@Shaktal: Would it be legal? Not without additional explantion. – GEdgar Jul 8 '12 at 12:36
up vote 3 down vote accepted

You want to compute $\displaystyle \lim_{x\to -n} \frac {\Pi(x)}{\Pi(mx)}$ when $x$ is near a negative integer.
$\Pi$ is the 'natural' extension of the factorial : $\Pi(n)=n!$ and $\Pi(z)=\Gamma(z+1)$ (see Wikipedia)

In this form the "Euler's reflection formula" becomes simply (for $\operatorname{sinc}(z)=\frac{\sin(\pi z)}{\pi z}$) : $$\Pi(-z)\Pi(z)=\frac 1{\operatorname{sinc}(z)}$$

$$ \lim_{x\to -n}\ \frac {\Pi(x)}{\Pi(mx)}=\lim_{x\to -n}\frac {\Pi(-mx)\operatorname{sinc}(-mx)}{\Pi(-x)\operatorname{sinc}(-x)}$$ $$ =\lim_{t\to n}\frac {\Pi(mt)\operatorname{sinc}(mt)}{\Pi(t)\operatorname{sinc}(t)}$$

It remains to prove that $\ \lim_{t\to n} \frac {\operatorname{sinc(mt)}}{\operatorname{sinc(t)}}=\frac {(-1)^{(m-1)n}}m$ (you may use l'Hôpital's rule for that) and to conclude!

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Actually, $\displaystyle\lim_{t\to n} \frac {\operatorname{sinc(mt)}}{\operatorname{sinc(t)}}=\frac{(-1)^{(m-1)t}}{m}$ – Generic Human Jul 8 '12 at 13:37
@GenericHuman: Oops you are right, thanks to notice! – Raymond Manzoni Jul 8 '12 at 13:44

Using anon's "pole" idea, with definition $x! := \Gamma(x+1)$ we have: $$\begin{align} x! &= \Gamma(x+1) = \frac{1}{2}\;\frac{1}{x+3}+O(1)\qquad\text{as $x \to -3$}, \\ (2x)! &= \Gamma(2x+1) = -\frac{1}{240}\;\frac{1}{x+3} + O(1)\qquad\text{as $x \to -3$}, \\ \frac{x!}{(2x)!} &= \frac{1/2}{-1/240}+O(x+3)\qquad\text{as $x \to -3$}, \\ \frac{x!}{(2x)!} &\to -120\qquad\text{as $x \to -3$}. \end{align}$$

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Let me add to the above, that while $\Gamma(-n)$ for a positive integer $n$ is undefined, let $m$ be such an integer as well, and then the ratio $\Gamma(-n)/\Gamma(-m)$ is well defined, and the Euler reflection formula above leads to its value being equal to $\Gamma(m+1)/\Gamma(n+1)(-1)^{n-m}$. This shows, by the way that the ratio mentioned at the beginning of the this sequence, effectively $(-3)!/(-6)!$ is $-60$, and not as suggested.

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