How to show $\lim_{x \to -N} (x+N) \Gamma(x) $ =$(-1)^N/N!$?

Question

I am trying to evaluate $\lim_{x \to -N} (x+N) \Gamma(x) $ =$(-1)^N/N!$, for N = 0, 1, 2, ... but I keep getting zero.

I tried using the definition of $\Gamma(x) = \Gamma(x+1)/x $ with no luck.

Is there any identity for gamma I could use to get the desired result?

Answer 1

Assuming you are exposed to the Euler's reflection formula: $$ \Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)} \tag{1} $$ we get: $$ \lim_{x \to -n} (x+n)\Gamma(x) = \lim_{x\to-n} \frac{\pi (x+n)}{\sin(\pi x)}\frac{1}{\Gamma(1-x)} \stackrel{y=x+n}{=} \lim_{y \to 0} \frac{\pi y}{\sin(\pi y - \pi n)} \frac{1}{\Gamma(1 + n -y )} $$ Using identity $\sin(\pi y - \pi n) =(-1)^n \sin(\pi y)$, valid for integer $n$, we conclude $$ \lim_{x \to -n} (x+n)\Gamma(x) = \frac{(-1)^n}{\Gamma(1+n)} \, \underbrace{ \lim_{y \to 0} \frac{\pi y}{\sin(\pi y)} }_{ \lim_{z\to 0} \frac{z}{\sin(z)}=1}= \frac{(-1)^n}{n!} $$

Answer 2

Let $L$ denote the desired limit. Write $(x+N)\Gamma(x)=\dfrac{\Gamma(x)}{(x+N)^{-1}}$.

Recall that $\Gamma(x)$ has simple poles at the nonpositive integers, with residue at $x=-N$ equal to $l=\frac{(-1)^N}{N!}.$ Thus, near $x=-N,$ we can express $\Gamma(x)=\frac{l}{x+N}+h(x)$ with $h(x)$ holomorphic. We obtain

$$L=\lim_{x\to-N}\dfrac{\Gamma(x)}{(x+N)^{-1}}=\lim_{x\to-N}\dfrac{\frac{l}{x+N}+h(x)}{(x+N)^{-1}}=l+0=l,$$

as desired. Of course, the real crux of the question is to prove that the residue of the gamma function at $x=-N$ is $l.$ I'll leave this to you, because it really is a nice exercise on the idea of analytic continuation.

Answer 3

Recalling the rising and falling factorials

$$ x^{(N)}=\frac{\Gamma(x+N)}{\Gamma(x)}\,,\quad (x)_N=\frac{\Gamma(x+1)}{\Gamma(x-N+1)}\,, \quad {(-a)}^{(N)} = {(-1)}^N {(a)}_{{N}}\,,$$

where $ \Gamma(x+1)= x! \,.$

$$ (x+N)\Gamma(x)= \frac{\Gamma(x)(x+N)\Gamma(x+N)}{\Gamma(x+N)}=\frac{\Gamma(x)}{\Gamma(x+N)}(x+N)!=\frac{(x+N)!}{x^{(N)}}\,,$$

$$\Rightarrow (x+N)\Gamma(x) = \frac{(x+N)!}{x^{(N)}} $$

Taking the limit of the above equation gives

$$ \lim_{n\to \infty} (x+N)\Gamma(x) = \frac{0!}{(-N)^{(N)}} = \frac{1}{(-1)^N(N)_N}= \frac{1}{(-1)^N N!}\,.$$

Note that, $(N)_N = N!\,,$ by the second identity of the falling factorial.