TITLE: Partial fraction expansion of $\frac{1}{x(x+1)(x+2)\cdots(x+n)}$
QUESTION [9 upvotes]: I try to find a partial fraction expansion of $\dfrac{1}{\prod_{k=0}^n (x+k)}$ (to calculate its integral).
After checking some values of $n$, I noticed that it seems to be true that $\dfrac{n!}{\prod_{k=0}^n (x+k)}=\sum_{k=0}^n\dfrac{(-1)^k{n \choose k}}{x+k}$. However, I can't think of a way to prove it. Can somebody please help me?

REPLY [15 votes]: Since every $k,\; k=-n,\ldots, 0$ is a simple pole of the given fraction then its decomposition take the form
$$\frac{1}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{a_k}{x+k}$$
and we have
$$a_k=\lim_{x \to -k}\sum_{i=0}^n\frac{a_i(x+k)}{x+i} = \lim_{x \to -k} (x+k)\sum_{i=0}^n\frac{a_i}{x+i}$$
$$= \lim_{x \to -k} \frac{x+k}{x(x+1)(x+2)...(x+n)}=\frac{1}{-k(-k+1)(-k+2)...(-k+n)}=\frac{(-1)^k}{k!(n-k)!}$$
so yes it's true that
$$\frac{n!}{x(x+1)(x+2)...(x+n)}=\sum_{k=0}^n\frac{(-1)^k{n\choose k}}{x+k}$$