Originally Posted by

**Soroban**

Here's a stunning pattern in Partial Fractions.

I "discovered" it many years ago and a recent posting reminded me of it.

$\displaystyle \begin{array}{c}\;\;\dfrac{1}{n(n+1)} \;=\;\dfrac{1}{1!}\left(\dfrac{1}{n} - \dfrac{1}{n+1}\right) \qquad\;\; \\ \\[-3mm]

\dfrac{1}{n(n+1)(n+2)} \;=\; \dfrac{1}{2!}\left(\dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2}\right);\;\;\; \\ \\[-3mm]

\dfrac{1}{n(n+1)(n+2)(n+3)} \;=\;\dfrac{1}{3!}\left(\dfrac{1}{n} - \dfrac{3}{n+1} + \dfrac{3}{n+2} - \dfrac{1}{n+3}\right)\;\; \\ \\[-3mm]

\dfrac{1}{n(n+1)(n+2)(n+3)(n+4)} \;=\; \dfrac{1}{4!}\left(\dfrac{1}{n} - \dfrac{4}{n+1} + \dfrac{6}{n+2} - \dfrac{4}{n+3} + \dfrac{1}{n+4}\right)\\ \\[-3mm]

\vdots \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\q quad \vdots \qquad

\end{array}$

This is fairly well-known actually. Let

$\displaystyle \displaystyle P(x)=\sum_{k=1}^{m}(-1)^k{m\choose k}\prod_{\substack{j=0\\ j\ne k}}^m (x+j)$

It's evident that $\displaystyle \deg P\leqslant m$ yet it's easy to see that $\displaystyle P(0)=P(-1)=\cdots=P(-m)=m!$ from where it follows that $\displaystyle P(x)=m!$. Thus, dividing both sides by $\displaystyle \displaystyle m!\prod_{j=0}^{m}(x+j)$ gives $\displaystyle \displaystyle \left(\prod_{j=0}^{m}(x+j)\right)^{-1}=\frac{1}{m!}\sum_{k=1}^{m}{m\choose k}\frac{(-1)^k}{x+k}$ just as you noticed.