<s>[INST] Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$ Mathematica tells me that
$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$
Although I have not been able to come up with a proof.
Proofs, hints, or references are all welcome.
 [/INST] In trying to prove
$$S_n = \sum_{k=0}^n \frac{(-1)^k}{2k+1} {n\choose k}
= \frac{(2n)!!}{(2n+1)!!}$$
we introduce
$$f(z) = n! (-1)^n \frac{1}{2z+1} \prod_{q=0}^n \frac{1}{z-q}.$$
This has the property that for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z)
= n! (-1)^n \frac{1}{2k+1}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = n! (-1)^n \frac{1}{2k+1}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= \frac{(-1)^k}{2k+1} {n\choose k}.$$
Note that the residue of $f(z)$ at infinity is zero by inspection and
residues sum to zero so that we have
$$S_n = \sum_{k=0}^n \mathrm{Res}_{z=k} f(z)
= - \mathrm{Res}_{z=-1/2} f(z)
\\ = - \frac{1}{2} \mathrm{Res}_{z=-1/2} \; n! (-1)^n
\frac{1}{z+1/2} \prod_{q=0}^n \frac{1}{z-q}
= - \frac{1}{2} n! (-1)^n
\prod_{q=0}^n \frac{1}{-1/2-q}
\\ = \frac{1}{2} n! \prod_{q=0}^n \frac{1}{1/2+q}
= 2^n n! \prod_{q=0}^n \frac{1}{2q+1}
= \frac{(2n)!!}{(2n+1)!!}.$$
This is the claim.
</s>