Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

$$\lim_{x\to-3}\frac{x!}{(2x)!}=-120$$

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:

$$\lim_{x\to-3}{\frac{x!}{(2x)!}}=-\frac{6!}{3!}=-120$$

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!

share|improve this question
1  
(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

3 Answers 3

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!

share|improve this answer
2  
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

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.