<s>[INST] Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$ 
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1$).

How to prove this equation?
Is it possible to use induction?
$${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1};$$
$$(2n+1)!!=\frac{(2n)!(2n+1)}{2^nn!}\Rightarrow \frac{n!2^n}{(2n+1)!!}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)};$$
$$\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)}$$
What now?
 [/INST] Permit me  to contribute  an algebraic  proof that  does not  use Beta
functions.
Suppose we seek to verify that 
$$S_n = \sum_{k=0}^n \frac{(-1)^k}{2k+1} {n\choose k} = 
\frac{2^{2n} (n!)^2}{(2n+1) (2n)!}.$$
We have by inspection that
$$S_n = \sum_{k=0}^n \mathrm{Res}(f(z); z=k)$$
where $$f(z) =
(-1)^n n! \frac{1}{2z+1} \prod_{q=0}^n \frac{1}{z-q}.$$
This is because
$$\mathrm{Res}(f(z); z=k) =
(-1)^n \frac{n!}{2k+1} \prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\= (-1)^n \frac{n!}{2k+1} \frac{1}{k!} (-1)^{n-k} \frac{1}{(n-k)!}
= \frac{(-1)^k}{2k+1} {n\choose k}.$$
Now with $f(z)$ being rational we must have
$$S_n = - \mathrm{Res}(f(z); z=-1/2)
- \mathrm{Res}(f(z); z=\infty).$$
From $z=-1/2$ we get including the sign
$$- (-1)^n \frac{n!}{2} \prod_{q=0}^n \frac{1}{-1/2-q}
= - (-1)^n n! 2^n \prod_{q=0}^n \frac{1}{-1-2q}
= n! 2^n \prod_{q=0}^n \frac{1}{2q+1}
\\ = n! 2^n \times \frac{2^n n! }{(2n+1)(2n)!}
= \frac{2^{2n} (n!)^2}{(2n+1) (2n)!}.$$
For the residue at infinity we get including the sign
$$\mathrm{Res}_{z=0} \frac{1}{z^2}
(-1)^n n! \frac{1}{2/z+1} \prod_{q=0}^n \frac{1}{1/z-q}
\\ = \mathrm{Res}_{z=0} \frac{1}{z}
(-1)^n n! \frac{1}{z+2} \prod_{q=0}^n \frac{z}{1-qz} = 0.$$
Collecting the two contributions we obtain
$$\frac{2^{2n} (n!)^2}{(2n+1)!}$$
QED.
</s>