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!
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!