Mdthbs

I am trying to show that $forall Ninmathbb{N}$,

$$sumlimits_{n=0}^{N}sumlimits_{k=0}^{N}frac{left(-1right)^{n+k}}{n+k+1}{Nchoose n}{Nchoose k}{N+nchoose n}{N+kchoose k}=frac{1}{2N+1}$$

It's backed by numerical verifications, but I can't come up with a proof.

So far, I tried using the generating function of $left(frac{1}{2N+1}right)_{Ninmathbb{N}}$, which is $frac{arctanleft(sqrt{x}right)}{sqrt{x}}$, 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 $$int_{0}^{1}left(sumlimits_{k=0}^{N}(-1)^k{Nchoose k}{N+kchoose k}x^kright)^2dx$$

And we recognize here the shifted Legendre Polynomials $widetilde{P_N}(x)=displaystylesumlimits_{k=0}^{N}(-1)^k{Nchoose k}{N+kchoose k}x^k$.

And we know that the shifted Legendre Polynomials form a family of orthogonal polynomials with respect to the inner product $<f|g>=displaystyleint_{0}^{1}f(x)g(x)dx$, and that their squared norm with respect to this product is $<widetilde{P_n}|widetilde{P_n}>=frac{1}{2n+1}$;

So this basically provides the desired result immediatly.

We seek to verify that

$$sum_{n=0}^N sum_{k=0}^N
frac{(-1)^{n+k}}{n+k+1}
{Nchoose n} {Nchoose k}
{N+nchoose n} {N+kchoose k}
= frac{1}{2N+1}.$$

Now we have

$${Nchoose k} {N+nchoose n}
= frac{(N+n)!}{(N-k)! times k! times n!}
= {N+nchoose n+k} {n+kchoose k}.$$

We get for the LHS

$$sum_{n=0}^N sum_{k=0}^N
frac{(-1)^{n+k}}{n+k+1} {N+nchoose n+k}
{Nchoose n} {N+kchoose k} {n+kchoose k}
\ = sum_{n=0}^N frac{1}{N+n+1} sum_{k=0}^N
(-1)^{n+k} {N+n+1choose n+k+1}
{Nchoose n} {N+kchoose k} {n+kchoose k}
\ = sum_{n=0}^N frac{1}{N+n+1} {Nchoose n}
sum_{k=0}^N
(-1)^{n+k} {N+n+1choose N-k}
{N+kchoose k} {n+kchoose k}
\ = sum_{n=0}^N frac{1}{N+n+1}
[z^N] (1+z)^{N+n+1}
{Nchoose n}
sum_{k=0}^N
(-1)^{n+k} z^k
{N+kchoose N} {n+kchoose n}.$$

Now the coefficient extractor controls the range and we continue with

$$sum_{n=0}^N frac{1}{N+n+1}
[z^N] (1+z)^{N+n+1}
{Nchoose n}
\ times sum_{kge 0}
(-1)^{n+k} z^k
{N+kchoose N}
mathrm{Res}_{w=0} frac{1}{w^{n+1}} frac{1}{(1-w)^{k+1}}
\ = sum_{n=0}^N frac{1}{N+n+1}
[z^N] (1+z)^{N+n+1}
{Nchoose n}
mathrm{Res}_{w=0} frac{1}{w^{n+1}} frac{1}{1-w}
\ times sum_{kge 0}
(-1)^{n+k} z^k
{N+kchoose N} frac{1}{(1-w)^{k}}
\ = sum_{n=0}^N frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{N+n+1}
{Nchoose n}
\ times mathrm{Res}_{w=0} frac{1}{w^{n+1}} frac{1}{1-w}
frac{1}{(1+z/(1-w))^{N+1}}
\ = sum_{n=0}^N frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{N+n+1}
{Nchoose n}
\ times mathrm{Res}_{w=0} frac{1}{w^{n+1}}
frac{(1-w)^N}{(1-w+z)^{N+1}}
\ = sum_{n=0}^N frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{n}
{Nchoose n}
\ times mathrm{Res}_{w=0} frac{1}{w^{n+1}}
frac{(1-w)^N}{(1-w/(1+z))^{N+1}}
\ = sum_{n=0}^N frac{(-1)^n}{N+n+1}
[z^N] (1+z)^{n}
{Nchoose n}
\ times sum_{k=0}^n (-1)^k {Nchoose k}
{n-k+Nchoose N} frac{1}{(1+z)^{n-k}}
\ = sum_{n=0}^N frac{(-1)^n}{N+n+1}
{Nchoose n}
\ times sum_{k=0}^n (-1)^k {Nchoose k}
{n-k+Nchoose N} [z^N] (1+z)^k.$$

Now for the coefficient extractor to be non-zero we must have $kge N$
which happens just once, namely when $n=N$ and $k=N.$ We get

$$frac{(-1)^N}{2N+1}
{Nchoose N}
(-1)^N {Nchoose N}
{N-N+Nchoose N}.$$

This expression does indeed simplify to

$$bbox[5px,border:2px solid #00A000]{
frac{1}{2N+1}}$$

as claimed.