Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$

You could consider the integral $$ \int_{0}^{1} (1-x^2)^n dx .$$

Sums of the form $$\sum_{k=0}^n(-1)^k{n\choose k}f(k)$$ can often be attacked via the Calculus of finite differences.

$$S_n=\sum\limits_{k=0}^n \dfrac{(-1)^k \binom{n}{k}}{2k+1}$$

We have:

$$\begin{align}S_n&=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1} \dfrac{(-1)^k \binom{n}{k}}{2k+1}\\ &=\dfrac{(-1)^n}{2n+1}+\sum\limits_{k=0}^{n-1}\left[\dfrac{n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{n(2n+1)}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk+2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2n^2-2nk}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\&+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1}\left[\dfrac{2nk+n}{n-k}.\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}\right]\\ &=\dfrac{(-1)^n}{2n+1}+\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \dfrac{n(-1)^k \binom{n-1}{k}}{n-k}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^{n-1} \left[(-1)^k \binom{n}{k}\right]+\dfrac{(-1)^n}{2n+1}\\ &=\dfrac{2n}{2n+1}\sum\limits_{k=0}^{n-1}\dfrac{(-1)^k \binom{n-1}{k}}{(2k+1)}+\dfrac{1}{2n+1}\sum\limits_{k=0}^n \left[(-1)^k \binom{n}{k}\right]\end{align}$$ Therefore, $$S_n=\dfrac{2n}{2n+1}S_{n-1}+0 \Rightarrow S_{n-1}=\dfrac{2n-2}{2n-1}S_{n-2} ... \Rightarrow S_1=\dfrac{2}{3}S_0$$ and $S_0=1$

Hence, $$S_n=\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\dfrac{(2n)!!}{(2n+1)!!}$$

Your identity is a special case of the Chu-Vandermonde identity.

${}_2 F_1(-n,b;c;1)=\frac{(c-b)_n}{(c)_n}$

with $b=\frac12$ and $c=\frac32$. More info on it is in A=B, as mentioned already by John.