Is the numerator of $\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+1}}$ always a power of $2$ in lowest terms? Is the numerator of
$$
\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+1}}
$$
always a power of $2$ in lowest terms, and if so, why? Is there a combinatorial or probabilistic proof of this?
 A: Consider the function
$$
F(q)=\sum_{k=0}^n (-1)^k {n\choose k} q^{2k}=(1-q^2)^n
$$
If we denote the sum you want to calculate by $\phi=\sum_{k=0}^n (-1)^k {n\choose k}\frac{1}{2k+1}$, then
$$
\phi = \int_0^1 F(q)dq = \int_0^1 (1-q^2)^n dq=\frac{(2n)!!}{(2n+1)!!}
$$
where $k!!=k(k-2)(k-4)\cdots$. This simplifies to
$$
\phi=\frac{2^{2n} (n!)^2}{(2n+1)!}
$$
Now note that
$$
{2n \choose n}=\frac{(2n)!}{(n!)^2}\in \mathbb{Z}
$$
so $(n!)^2$ divides $(2n)!$ which in turn means it divides $(2n+1)!$. So the numerator of $\phi$ is necessarily a power of $2$ in lowest terms.
A: Proved in edit at the end: this sum is equal to the fraction 
$${\prod_{i=1}^n 2i\over\prod_{i=1}^{n} 2i+1}$$
Now: $\prod_{i=1}^n 2i=2^n n!$ 
and $\prod_{i=1}^{n} {2i+1}={(2n+1)!\over \prod_{i=1}^n 2i}={(2n+1)!\over 2^n n!}$
Therefore your sum is equal to $${(2^n n!)^2\over (2n+1)!}={2^{2n}\over(2n+1){2n \choose n}}$$
Where the result is now clear.

Edit: combinatorial proof of the identity ${\prod_{i=1}^n 2i\over\prod_{i=1}^{n} 2i+1}$.
Define the sum $$S_{n,m}=\sum_{k=0}^{n}{(-1)^k\binom{n}{k}\frac{1}{2k+m}}$$
Claim: for any $n,m$, one has $$S_{n,m}={\prod_{i=1}^{n}2i\over \prod_{i=0}^{n}2i+m }$$  
The claim implies the desired identity when $m=1$.
Proof of the claim: 
The identity ${n+1\choose k}={n\choose k}+{n\choose k-1} \ $ implies immediately that $$S_{n+1,m}=S_{n,m}-S_{n,m+2}$$
Moreover, the claim holds when $n=0$, where $S_{0,m}=\frac1m$
The result follows now from an immediate induction on $n$.
