<s>[INST] 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.
 [/INST] 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.
</s>