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)} $. I have to prove that
$$
\forall n \in \mathbb{N}_{0}, ~ \forall x \in \mathbb{R} \setminus \mathbb{N}_{0}:
\qquad
  \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k}
= \frac{n!}{(x + 0) (x + 1) \cdots (x + n)}.
$$
However, I was unable to find a proof. I have tried to use the binomial expansion of $ (1 + x)^{n} $ to get the l.h.s., by performing a suitable multiplication followed by integration, but I was unable to obtain the required form. Please help me out with the proof. Thanks in advance.
 A: \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*}
A: 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.
A: 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.
A: I thought it might be instructive to present a proof by induction.  
First, we establish a base case.  For $n=0$, it is straightforward to show that the expression holds.  
Second, we assume that for some $n\geq 0$, we have
$$\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{x+k}=n!\prod_{k=0}^n\frac{1}{x+k}$$
Third, we analyze the sum $\sum_{k=0}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{x+k}$.  Clearly, we can write
$$\begin{align}
\sum_{k=0}^{n+1} \binom{n+1}{k}\frac{(-1)^k}{x+k}&=\frac1x+\sum_{k=1}^{n} \binom{n+1}{k}\frac{(-1)^k}{x+k}+\frac{(-1)^{n+1}}{x+n+1}\\\\
&=\color{green}{\frac1x}+\sum_{k=1}^{n} \left(\color{green}{\binom{n}{k}}+\color{blue}{\binom{n}{k-1}}\right)\frac{(-1)^k}{x+k}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\
&=\color{green}{n!\prod_{k=0}^n\frac{1}{x+k}}+\color{blue}{\sum_{k=0}^{n-1} \binom{n}{k}\frac{(-1)^{k+1}}{(x+1)+k}}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\
&=\color{green}{n!\prod_{k=0}^n\frac{1}{x+k}}+\color{blue}{-n!\prod_{k=0}^n\frac{1}{x+1+k}-\frac{(-1)^{n+1}}{x+n+1}}+\color{red}{\frac{(-1)^{n+1}}{x+n+1}}\\\\
&=(n+1)!\prod_{k=0}^{n+1}\frac{1}{x+k}
\end{align}$$ 
And we are done!
A: Remark. What follows  is an exact duplicate of  an earlier post of
mine  which  I am  unable  to  locate  despite making  a  considerable
effort. 
We seek to verify that
$$\sum_{k=0}^n \frac{1}{x+k} (-1)^k {n\choose k}
= n! \times \prod_{q=0}^n \frac{1}{x+q}.$$
Consider the function
$$f(z) = n! \frac{1}{x+z} \prod_{q=0}^n \frac{1}{z-q}$$
where clearly  $x$ is not equal  to $0,-1,-2,\ldots -n$ or  the LHS of
the target formula becomes singular.
We compute  the residues of $f(z).$  We get for the  poles at $p\in
[0,n]$ the residue
$$\mathrm{Res}_{z=p} f(z) =
n!\frac{1}{x+p} \prod_{q=0}^{p-1} \frac{1}{p-q}
\prod_{q=p+1}^n \frac{1}{p-q}
\\ = n! \frac{1}{x+p} \frac{1}{p!} \frac{(-1)^{n-p}}{(n-p)!}
= \frac{1}{x+p} (-1)^{n-p} {n\choose p}.$$
The residue at $z=-x$ yields
$$\mathrm{Res}_{z=-x} f(z) =
n! \prod_{q=0}^n \frac{1}{-x-q}
= n! (-1)^{n+1} \times
\prod_{q=0}^n \frac{1}{x+q}.$$
The residue at infinity is
$$\mathrm{Res}_{z=\infty} f(z) =
- \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
n! \frac{1}{x+1/z} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = - \mathrm{Res}_{z=0} \frac{1}{z^2}
n! \frac{z}{zx+1} \prod_{q=0}^n \frac{z}{1-qz}
\\ = - \mathrm{Res}_{z=0} \frac{z^{n+2}}{z^2}
n! \frac{1}{zx+1} \prod_{q=0}^n \frac{1}{1-qz}
\\ = - \mathrm{Res}_{z=0} z^n
n! \frac{1}{zx+1} \prod_{q=0}^n \frac{1}{1-qz}
= 0.$$
To  conclude observe  that the  residues  sum to  zero and  collecting
everything we obtain
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{n-p} {n\choose p}
+ n! (-1)^{n+1} \times
\prod_{q=0}^n \frac{1}{x+q} = 0.$$
This is
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{p} {n\choose p}
- n! \times
\prod_{q=0}^n \frac{1}{x+q} = 0.$$
or
$$\sum_{p=0}^n\frac{1}{x+p} (-1)^{p} {n\choose p}
= n! \times
\prod_{q=0}^n \frac{1}{x+q}$$
which is the claim.
