Digamma function ratio limit

Question

I'd like to know how to go about proving the result

$$\lim_{x\to-n}\dfrac{\psi(x)}{\Gamma(x)} = (-1)^{n-1}n!$$

as it appears on p.g. 2 here http://www.math.usm.edu/lambers/mat415/lecture16.pdf.

Answer 1

Recall that the analytic continuation of the Gamma function can be obtained by using repeatedly the functional relationship $\Gamma(x+1)=x\Gamma(x)$. Then, for $x+m+1>0$, we can write

$$\Gamma (x)=\frac{\Gamma(x+m+1)}{x(x+1)(x+2)\cdots(x+m-1)(x+m)} \tag 1$$

Using $(1)$ reveals that for $x+m+1>0$, the digamma function can be expressed as

$$\psi(x)=\frac{d\log \Gamma (x)}{dx}=\psi(x+m+1)-\sum_{k=0}^m\frac{1}{x+k} \tag 2$$

Therefore, using $(1)$ and $(2)$, the ratio $\frac{\psi(x)}{\Gamma(x)}$ is given by

$$\frac{\psi(x)}{\Gamma(x)}=\left(\psi(x+m+1)-\sum_{k=0}^m\frac{1}{x+k}\right)\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\right) \tag 3$$

For $0<n\le m$, we note that for $x=-n$, $\Gamma(x+m+1)=(m-n)!$ and $\psi(x+m+1)=H_{m+1-n}-\gamma$. Therefore, we have

$$\begin{align} \lim_{x\to -n}\frac{\psi(x)}{\Gamma(x)}&=\lim_{x\to -n}\left(\frac{\psi(x+m+1)}{\Gamma(x+m+1)}\left(x(x+1)(x+2)\cdots(x+m-1)(x+m)\right)\right)\\\\ &-\lim_{x\to -n}\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\sum_{k=0}^{m}\frac{1}{x+k}\right)\\\\ &=-\lim_{x\to -n}\left(\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{\Gamma(x+m+1)}\sum_{k=0}^{m}\frac{1}{x+k}\right)\\\\ &=-\frac{1}{(m-n)!}\lim_{x\to -n}\sum_{k=0}^m\frac{x(x+1)(x+2)\cdots(x+m-1)(x+m)}{x+k}\\\\ &=-\frac{1}{(m-n)!}\left((-n)(-n+1)(-n+2)\cdots (-2)(-1)(1)(2)\cdots(m-n-1)(m-n)\right)\\\\ &=-(1)^{n+1}n! \end{align}$$

as was to be shown!