Limit of ratio of two Gamma functions with negative integer arguments When using the hypergeometric representation for a Legendre polynomial, I encounter, for integer n and l, the following ratio: $$\frac{\Gamma(n-l)}{\Gamma(-l)}$$
Where $n \leq l$ (the quantity is definitely zero for $n > l$, as it should be in the definition of a Legendre polynomial). I am unsure as to how to evaluate this ratio; as it stands, it is indeterminate. My original idea was to use: $$\Gamma(k+1) = (k+1)\Gamma(k)$$
Multiple times to reduce $\Gamma(n-l)$ to: $$(n-l)(n-l-1)(...)(-l+1) \ \Gamma(-l)$$
Then the $\Gamma(-l)$ terms would cancel and I'd be left only with some sensible terms. Unfortunately, I do not think that this approach is valid, as it does not yield the correct representation for the Legendre polynomial. Secondly, the above can be written as: $$(-1)^n\frac{(l-1)!}{(l-n-1)!} \ \Gamma(-l)$$
Which now no longer permits us to set $n=l$ as is necessary to obtain a polynomial of order $l$. I'm trying to remain brief on the references to Legendre polynomials as it is specifically the ratio of the Gamma functions listed at the start of this post that I am interested in evaluating.
 A: I am not sure where you obtain your equation but the correct recursive formula for gamma function should be $$\Gamma(k+1)=k\Gamma(k)$$
If I proceed in a similar way as yours, then $\Gamma(n-l)$ can easily be reduced to
$$\begin{aligned}\Gamma(n-l)&=(n-l-1)(n-l-2)\cdots (-l)\Gamma(-l)\\
&=(-1)^{n}\frac{l!}{(l-n)!}\Gamma(-l)\end{aligned}$$
The Legendre polynomial defined in terms of hypergeometric function can be written as
$$\begin{aligned}
P_l(z)&={}_2F_1(-l,l+1;1;\frac{1-z}{2})\\
&=\sum^{\infty}_{n=0}\frac{(-l)_n(l+1)_n}{(1)_n}\frac{1}{n!}\left(\frac{1-z}{2}\right)^2
\end{aligned}$$
Using $(a)_n\equiv \frac{\Gamma(a+n)}{\Gamma(a)}$,
$$\begin{aligned}
P_l(z)&=\sum^{\infty}_{n=0}\frac{\Gamma(n-l)}{\Gamma(-l)}\frac{\Gamma(l+n+1)}{\Gamma(l+1)\Gamma(n+1)n!}\left(\frac{1-z}{2}\right)^2\\
&=\sum^{\infty}_{n=0}\frac{(-1)^{n}l!}{n!(l-n)!}\frac{\Gamma(l+n+1)}{\Gamma(l+1)\Gamma(n+1)}\left(\frac{1-z}{2}\right)^2
\end{aligned}$$
(Btw, I believe the $\Gamma(1-n)$ in the denominator inside the summation series of the your comment is a mistake) The binomial coefficient for complex argument is related to gamma function via
$$
\begin{pmatrix}x \\y\end{pmatrix}=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma(x-y+1)}
$$
and using the identity
$$
\begin{pmatrix}x \\y\end{pmatrix}=(-1)^y\begin{pmatrix}y-x-1 \\y\end{pmatrix}
$$, we can establish the relation
$$\begin{aligned}
\begin{pmatrix}-l-1 \\n\end{pmatrix}&=(-1)^n\begin{pmatrix}l+n \\n\end{pmatrix}\\
&=(-1)^n\frac{\Gamma(l+n+1)}{\Gamma(n+1)\Gamma(l+1)}
\end{aligned}$$
Finally, we can substitute the above relation into the Legendre polynomial equation to obtain
$$P_l(z)=\sum^{l}_{n=0}\begin{pmatrix}l \\n\end{pmatrix}\begin{pmatrix}-l-1 \\n\end{pmatrix}\left(\frac{1-z}{2}\right)^2$$
where the series terminates after $n=l$. This is exactly the other form of Legendre polynomial expression, therefore the ratio I showed here
$$\frac{\Gamma(n-l)}{\Gamma(-l)}=(-1)^{n}\frac{l!}{(l-n)!}$$
is consistent with the formulas even if the arguments are negative integers.
A: We have for $k\in\mathbf{N}$ and $x\rightarrow 0$
$$\Gamma(-k+x) \sim \frac{1}{k!x} + O(1)$$
and therefore 
$$\lim_{x\rightarrow 0}\frac{\Gamma(n-l+x)}{\Gamma(-l+x)} = \frac{1}{(l-n)!}/\frac{1}{l!} = \frac{l!}{(l-n)!}$$
Of course this is only a very special limit. It is finite for $n=l$ but I do not know what you expect for the value. If the above does reproduce it, you will at least have a simple arugument.
