<s>[INST] Split $\prod_{i=0}^n (z+i)^{-1}$ into $\sum_{i=0}^n a_i (z+i)^{-1}$ 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} $
 [/INST] 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}
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