A common problem in Combinatorial Analysis Please help me prove the fact below:
$$
\sum_{n=1}^N \frac{N!}{n!(N-n)!}\frac{(-1)^{n-1}n}{n+x} = \frac{N!}{\prod_{n=1}^N (n+x)}.
$$
I think this problem is common, but it is really hard for me to solve.
 A: It is a straightforward consequence of the residue theorem. The rational function:
$$ f(x) = \frac{1}{\prod_{n=1}^{N}(x+n)} $$
has simple poles in $x=-1,-2,\ldots,-N$, hence:
$$ f(x) = \sum_{j=1}^{N} \frac{c_j}{x+j} $$
where:
$$ c_j = \operatorname{Res}\left(f(z),z=-j\right).$$
A: Alternatively to Jack's answer, multiply both sides by $\dfrac{\prod_{n=1}^N (x+n)}{N!}$ and you get an equality of polynomials of degree at most $N$, and show that the equality is true for $x=0,-1,-2,\dots,-N$. 
The cases $x=-1,\dots,-N$, the left hand side is zero for all but one term. which you can check easily.
The case $x=0$ reduces to the binomial theorem for $(1+(-1))^N$, specifically:
$$1-\left(1+(-1)\right)^N = 1$$
A possible inductive proof would write the left hand side as $f_N(x)$ and then you'd get that you want to show that:
$$f_N(x) = \frac{N}{N+x} f_{N-1}(x)$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{n\ =\ 1}^{N}{N! \over n!\pars{N - n}!}\,
     {\pars{-1}^{n - 1}\,n \over n + x}
     ={N! \over \prod_{n\ =\ 1}^{N}\pars{n + x}}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 1}^{N}{N! \over n!\pars{N - n}!}\,{\pars{-1}^{n - 1}\,n \over n + x}}
=-\sum_{n\ =\ 0}^{N}{N \choose n}\,{\pars{-1}^{n}}
+x\sum_{n\ =\ 0}^{N}{N \choose n}\,{\pars{-1}^{n} \over n + x}
\\[5mm]&=-\bracks{1 + \pars{-1}}^{N}
+x\sum_{n\ =\ 0}^{N}{N \choose n}\pars{-1}^{n}\int_{0}^{1}t^{n + x - 1}\,\dd t
=x\int_{0}^{1}t^{x - 1}\sum_{n\ =\ 0}^{N}{N \choose n}\pars{-t}^{n}\,\dd t
\\[5mm]&=x\int_{0}^{1}t^{x - 1}\pars{1 - t}^{N}\,\dd t
=x\,{\Gamma\pars{x}\Gamma\pars{N + 1} \over \Gamma\pars{x + N + 1}}
={N! \over \Gamma\pars{x + 1 + N}/\Gamma\pars{x + 1}}
={N! \over \pars{x + 1}^{\rm\pars{N}}}
\\[5mm]&=\color{#66f}{\large{N! \over \prod_{n\ =\ 1}^{N}\pars{n + x}}}
\end{align}

$\ds{\pars{a}^{\rm\pars{m}}}$ is one of the Pochhammer Symbols: The raising factorial.

