Product formula for $\sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}$

Question

How to prove the following identity: $$\sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}=\prod_{k=1}^n\frac{2k}{2k+1}$$

Answer 1

Expanding $(1-x^2)^n$ we obtain $$ (1-x^2)^n=\sum_{k=0}^n (-1)^{k}x^{2k}\binom{n}{k}. $$ Then integrate the right hand side in $[0,1]$: $$ \sum_{k=0}^n(-1)^{k}\binom{n}{k}\int_0^kx^{2k}\,dx=\sum_{k=0}^n(-1)^{k}\binom{n}{k}\frac{1}{2k+1}=\sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}. $$ Then, for the left hand side set $x=\sin t$, and this, together with integration by parts, provides \begin{align} I_{n}&=\int_0^1 (1-x^2)^n\,dx=\int_0^{\pi/2} \cos^{2n+1}t\,dt=\cos^{2n}t\,\sin t\big|_0^{\pi/2} +2n\int_0^{\pi/2} \cos^{2n-1}t\sin^2 t\,dt\\ &=2n\int_0^{\pi/2} \cos^{2n-1}t\,dt-2n\int_0^{\pi/2} \cos^{2n+1}t\,dt=2nI_{n-1}-2nI_n, \end{align} which in turn provides the recursion relation $$ I_{n}=\frac{2n}{2n+1}I_{n-1}=\frac{2n}{2n+1}\frac{2n-2}{2n-1}I_{n-2}=\prod_{k=0}^n \frac{2k}{2k+1}, $$ as $I_0=1$. Therefore

$$ \sum_{k=0}^n\frac{(-1)^k}{2k+1}\binom{n}{k}=\prod_{k=0}^n \frac{2k}{2k+1}. $$