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.
 A: 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: $$S=\frac{1}{2n+1}.$$
A: Note that $$\frac{1}{i+j+1} = \int_0^1x^{i+j}dx$$ thus $$\sum_{i,j=0}^{n}\binom{n}{i}\binom{n}{j}\frac{(-2)^{i+j}}{i+j+1}=\sum_{i,j=0}^n{n\choose i}{n\choose j} (-2)^{i+j}\int_0^1x^{i+j}dx =\int_0^1\left(\sum_{i=0}^{n}{n\choose i}(-1)^i(2x)^i\right)\sum_{j=0}^n(-1)^j(2x)^j dx=\int_0^1(1-2x)^{2n}dx=\frac{(-1)^{2n}+1}{2(2n+1)}=\frac{1}{2n+1}$$
A: Without calculus:
$$\begin{align*}
\sum_{i=0}^n\sum_{j=0}^n\binom{n}i\binom{n}j\frac{(-2)^{i+j}}{i+j+1}&=\sum_{k=0}^{2n}\sum_{i=0}^n\binom{n}i\binom{n}{k-i}\frac{(-2)^k}{k+1}\\
&=\sum_{k=0}^{2n}\frac{(-2)^k}{k+1}\sum_{i=0}^n\binom{n}i\binom{n}{k-i}\\
&=\sum_{k=0}^{2n}\frac{(-2)^k}{k+1}\binom{2n}k\\
&=\sum_{k=0}^{2n}\frac{(-2)^k}{2n+1}\binom{2n+1}{k+1}\\
&=\sum_{k=0}^{2n}\frac{(-2)^{2n-k}}{2n+1}\binom{2n+1}{2n-k}\\
&=\frac1{2n+1}\sum_{k=1}^{2n+1}(-2)^{2n+1-k}\binom{2n+1}{2n+1-k}\\
&=\frac1{2n+1}\left(\sum_{k=0}^{2n+1}(-2)^{2n+1-k}\binom{2n+1}{2n+1-k}+2\right)\\
&=\frac1{2n+1}\left((1-2)^{2n+1}+2\right)\\
&=\frac1{2n+1}
\end{align*}$$
