5
$\begingroup$

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!

  • 0
    @Shaktal: Would it be legal? Not without additional explantion.2012-07-08

3 Answers 3

0

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.

4

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}$

4

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!

  • 0
    @GenericHuman: Oops you are right, thanks to notice!2012-07-08