TITLE: Cardinality summation
QUESTION [5 upvotes]: How do I prove that, for set $X$,
$$\sum_{S\subseteq X, S\neq \emptyset}\frac{(-1)^{|S|}}{|X|+|S|} = \frac{|X|!(|X|-1)!}{(2|X|)!}$$
I have been around this exercise all day and would much appreciate your help.

REPLY [7 votes]: Let us denote $n = |X|$. Then
\begin{align*}
\sum_{S\subset X} \frac{(-1)^{|S|}}{|X|+|S|}
&= \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^{k}}{n+k}
 = \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{0}^{1} x^{n+k-1} \, dx \\
&= \int_{0}^{1} \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} x^{n+k-1} \, dx
 = \int_{0}^{1} x^{n-1} (1 - x)^{n} \, dx \\
&= \beta(n, n+1) = \frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)} = \frac{(n-1)!n!}{(2n)!}
\end{align*}
as desired.