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?

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.

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.

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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 $n$.