A nice Combinatorial Identity I am trying to show that $\forall N\in\mathbb{N}$,
$$\sum\limits_{n=0}^{N}\sum\limits_{k=0}^{N}\frac{\left(-1\right)^{n+k}}{n+k+1}{N\choose n}{N\choose k}{N+n\choose n}{N+k\choose k}=\frac{1}{2N+1}$$
It's backed by numerical verifications, but I can't come up with a proof.
So far, I tried using the generating function of $\left(\frac{1}{2N+1}\right)_{N\in\mathbb{N}}$, which is $\frac{\arctan\left(\sqrt{x}\right)}{\sqrt{x}}$, by showing that the LHS has the same generating function, but this calculation doesn't seem to lead me anywhere...
Any suggestion ?
Edit: the comment of bof (below this question) actually leads to a very simple proof.
Indeed, from bof's comment we have that the LHS is equal to $$\int_{0}^{1}\left(\sum\limits_{k=0}^{N}(-1)^k{N\choose k}{N+k\choose k}x^k\right)^2dx$$
And we recognize here the shifted Legendre Polynomials $\widetilde{P_N}(x)=\displaystyle\sum\limits_{k=0}^{N}(-1)^k{N\choose k}{N+k\choose k}x^k$.
And we know that the shifted Legendre Polynomials form a family of orthogonal polynomials with respect to the inner product $\langle f|g\rangle=\displaystyle\int_{0}^{1}f(x)g(x)dx$, and that their squared norm with respect to this product is $\langle\widetilde{P_n}|\widetilde{P_n}\rangle=\frac{1}{2n+1}$;
so this basically provides the desired result immediately.
 A: We seek to verify that
$$\sum_{n=0}^N \sum_{k=0}^N
\frac{(-1)^{n+k}}{n+k+1}
{N\choose n} {N\choose k}
{N+n\choose n} {N+k\choose k}
= \frac{1}{2N+1}.$$
Now we have
$${N\choose k} {N+n\choose n}
= \frac{(N+n)!}{(N-k)! \times k! \times n!}
= {N+n\choose n+k} {n+k\choose k}.$$
We get for the LHS
$$\sum_{n=0}^N \sum_{k=0}^N
\frac{(-1)^{n+k}}{n+k+1} {N+n\choose n+k}
{N\choose n} {N+k\choose k}  {n+k\choose k}
\\ = \sum_{n=0}^N \frac{1}{N+n+1} \sum_{k=0}^N
(-1)^{n+k} {N+n+1\choose n+k+1}
{N\choose n} {N+k\choose k}  {n+k\choose k}
\\ = \sum_{n=0}^N \frac{1}{N+n+1} {N\choose n} 
\sum_{k=0}^N
(-1)^{n+k} {N+n+1\choose N-k}
{N+k\choose k}  {n+k\choose k}
\\ = \sum_{n=0}^N \frac{1}{N+n+1} 
[z^N] (1+z)^{N+n+1}
{N\choose n} 
\sum_{k=0}^N
(-1)^{n+k} z^k
{N+k\choose N}  {n+k\choose n}.$$
Now the coefficient extractor controls the range and we continue with
$$\sum_{n=0}^N \frac{1}{N+n+1} 
[z^N] (1+z)^{N+n+1}
{N\choose n} 
\\ \times \sum_{k\ge 0}
(-1)^{n+k} z^k
{N+k\choose N}  
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1}{(1-w)^{k+1}}
\\ = \sum_{n=0}^N \frac{1}{N+n+1} 
[z^N] (1+z)^{N+n+1}
{N\choose n} 
\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1}{1-w}
\\ \times \sum_{k\ge 0}
(-1)^{n+k} z^k
{N+k\choose N} \frac{1}{(1-w)^{k}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1} 
[z^N] (1+z)^{N+n+1}
{N\choose n} 
\\ \times \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{(1+z/(1-w))^{N+1}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1} 
[z^N] (1+z)^{N+n+1}
{N\choose n} 
\\ \times \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} 
\frac{(1-w)^N}{(1-w+z)^{N+1}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1} 
[z^N] (1+z)^{n}
{N\choose n} 
\\ \times \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} 
\frac{(1-w)^N}{(1-w/(1+z))^{N+1}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1} 
[z^N] (1+z)^{n}
{N\choose n} 
\\ \times \sum_{k=0}^n (-1)^k {N\choose k}
{n-k+N\choose N} \frac{1}{(1+z)^{n-k}}
\\ = \sum_{n=0}^N \frac{(-1)^n}{N+n+1} 
{N\choose n} 
\\ \times \sum_{k=0}^n (-1)^k {N\choose k}
{n-k+N\choose N} [z^N] (1+z)^k.$$
Now for the coefficient extractor to be non-zero we must have $k\ge N$
which happens just once, namely when $n=N$ and $k=N.$ We get
$$\frac{(-1)^N}{2N+1} 
{N\choose N} 
(-1)^N {N\choose N}
{N-N+N\choose N}.$$
This expression does indeed simplify to
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2N+1}}$$
as claimed.
