Split $\prod_{i=0}^n (z+i)^{-1}$ into $\sum_{i=0}^n a_i (z+i)^{-1}$

Question

I'm trying to find a general formula to split

$$\prod_{i=0}^n (z+i)^{-1} = \sum_{i=0}^n a_i (z+i)^{-1}$$

If $n=1$, we have

$$ \frac{1}{z(z+1)} = \frac{1}{z}+\frac{-1}{z+1} $$

If $n=2$,

$$ \frac{1}{z(z+1)(z+2)} = \frac{1/2}{z}+\frac{-1}{z+1}+\frac{1/2}{z+2} $$

If $n=3$,

$$ \frac{1}{z(z+1)(z+2)(z+3)} = \frac{1/6}{z}+\frac{-1/2}{z+1}+\frac{1/2}{z+2} + \frac{-1/6}{z+3} $$

Motivation. I was calculating residues at $z=-m$ of $f(z)=\sum_{n=0}^\infty\prod_{i=0}^n (z+i)^{-1} $

Answer 1

The Heaviside method is $$\begin{align}\lim_{z\rightarrow-k}(z+k)\prod_{i=0}^n(z+i)^{-1}&=\lim_{z\rightarrow-k}(z+k)\sum_{i=0}^na_i(z+i)^{-1}\\ \frac{(-1)^k}{k!(n-k)!}=a_k\end{align}$$ So how this happened on the left was that $$\begin{align}\lim_{z\rightarrow-k}(z+k)\prod_{i=0}^n(z+i)^{-1}&=\lim_{z\rightarrow-k}\left[\prod_{i=0}^{k-1}(z+i)^{-1}\right]\frac{(z+k)}{(z+k)}\left[\prod_{i=k+1}^{n}(z+i)^{-1}\right]\\ &=\lim_{z\rightarrow-k}\left[\prod_{i=1}^{k}(z+k-i)^{-1}\right]\frac{(z+k)}{(z+k)}\left[\prod_{i=1}^{n-k}(z+k+i)^{-1}\right]\\ &=\left[\prod_{i=1}^{k}(-1)(i)^{-1}\right](1)\left[\prod_{i=1}^{n-k}(i)^{-1}\right]\\ &=\frac{(-1)^k}{k!(n-k)!}\end{align}$$ And on the right we have $$\lim_{z\rightarrow-k}(z+k)\sum_{i=0}^na_i(z+i)^{-1}=\lim_{z\rightarrow-k}\sum_{i=0}^na_i\frac{(z+k)}{(z+i)}=\sum_{k=0}^na_i\delta_{ik}=a_k$$

Answer 2

Hint: Induction using \begin{align} & \frac{1}{(z+a)(z+a+1)\cdots(z+a+n)} \\ = & \frac{1/n}{(z+a)\cdots(z+a+n-1)}-\frac{1/n}{(z+a+1)\cdots(z+a+n)}. \end{align}