How to prove that $\sum_{i=0}^n \sum_{j=0}^{n}\binom{n}{i}\binom{n}{j}\frac{(-2)^{i+j}}{i+j+1}=\frac{1}{2n+1}, n \in \mathbb N$ I want help in proving that
$$\sum_{i=0}^n \sum_{j=0}^{n}\binom{n}{i}\binom{n}{j}\frac{(-2)^{i+j}}{i+j+1}=\frac{1}{2n+1},\quad  n \in \mathbb N.$$
the non-separable factor $\frac{1}{i+j+1}$ is causing the problem.

Let
$$S=\sum_{i=0}^n \sum_{j=0}^{n}(-2)^{i+j}\frac{{n \choose i} {n \choose j}}{i+j+1}.$$
By binomial theorem we can write
$$ \sum_{i=0}^n \sum_{j=0}^{n}x^{i+j} {n\choose i}{n\choose j}=(1+x)^{2n}.$$
Integrate both sides w.r.t. $x$ from $x=0$ to $x=-2$ to get
$$\sum_{i=0}^n \sum_{j=0}^{n}(-2)^{i+j+1}\frac{{n \choose i} {n \choose j}}{i+j+1}=\frac{(1-2)^{2n+1}-1}{2n+1}.$$
Theredore, we get: $$S=\frac{1}{2n+1}.$$
