A proof of the identity $ \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k} = \frac{n!}{(x + 0) (x + 1) \cdots (x + n)} $.

Solution 1:

Write down the partial fraction decomposition

$$\frac{1}{x(x+1)\cdots(x+n)}=\sum_{k=0}^n \frac{c_k}{x+k}. \tag{$\circ$}$$

To determine the coefficient $c_k$, multiply by $x+k$ then compute the limit $x\to -k$, getting

$$c_k=\frac{1}{(-k)(1-k)(2-k)\cdots (-2)\cdot (-1)\cdot 1\cdot2\cdots(n-k)}=\frac{(-1)^k}{k!(n-k)!}$$

Substitute this into $(\circ)$, muliply both sides by $n!$, then use $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ to conclude.

Solution 2:

\begin{align*} (1-t)^n = \sum_{k=0}^n (-1)^k \binom{n}{k} t^k \end{align*} Multiplying by $t^{x-1}$ and integrating between 0 and 1, we get \begin{align*} \int_0^1 t^{x-1}(1-t)^n dt = \sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{x+k} \end{align*} and the left hand side is \begin{align*} \beta(x,n+1) &= \frac{\Gamma(x)\Gamma(n+1)}{\Gamma(x+n+1)} \end{align*} This simplifies to \begin{align*} \frac{n!}{x(x+1)(x+2)\ldots(x+n)} \end{align*}

Solution 3:

An induction on $n$ will work. Let

$$f(x,n)=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\;,$$

and for the induction step suppose that

$$f(x,n)=\frac{n!}{x(x+1)\ldots(x+n)}\;.$$

Then

$$\begin{align*} \frac{(n+1)!}{x(x+1)\ldots(x+n+1)}&=\frac{n+1}{x+n+1}f(x,n)\\ &=\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n}k\\ &=(n+1)\sum_{k=0}^n(-1)^k\binom{n}k\frac1{(x+k)(x+n+1)}\\ &=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1-k}\binom{n}k\left(\frac1{x+k}-\frac1{x+n+1}\right)\\ &=(n+1)\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\frac1{x+k}\\ &\qquad-\frac{n+1}{x+n+1}\sum_{k=0}^n\frac{(-1)^k}{n+1}\binom{n+1}k\\ &=\sum_{k=0}^n\frac{(-1)^k}{x+k}\binom{n+1}k-\frac1{x+n+1}\color{red}{\sum_{k=0}^n(-1)^k\binom{n+1}k}\\ &=f(x,n+1)-\frac{(-1)^{n+1}}{x+n+1}-\frac1{x+n+1}\color{red}{\left(0-(-1)^{n+1}\right)}\\ &=f(x,n+1)\;, \end{align*}$$

as desired.