Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$

One may first observe that your initial sum may be rewritten as $$ \begin{align} \sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k} &=\sum_{k=0}^n (-1)^k\left(\int_0^1 x^{2k}\:dx\right)\binom{n}{k}\\\\ &=\int_0^1\sum_{k=0}^n (-1)^k\binom{n}{k}x^{2k}\:dx\\\\ &=\int_0^1(1-x^2)^ndx. \end{align} $$ Then, integrating by parts, we have $$ \begin{align} \int_0^1(1-x^2)^ndx &=\left. x(1-x^2)^n\right|_0^1+2n\int_0^1x^2(1-x^2)^{n-1}dx\\\\ &=0-2n\int_0^1(1-x^2-1)(1-x^2)^{n-1}dx\\\\ &=-2n\int_0^1(1-x^2)^ndx+2n\int_0^1(1-x^2)^{n-1}dx. \end{align} $$ which is equivalent to $$ \int_0^1(1-x^2)^ndx=\frac{2n}{2n+1}\int_0^1(1-x^2)^{n-1}dx $$ giving $$ \int_0^1(1-x^2)^ndx=\frac{2}{3}\frac{4}{5}\cdots\frac{2n}{2n+1}=\frac{2^{2n}(n!)^2}{(2n+1)(2n)!} $$ Finally

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

${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)!!} $

Since $(1+x)^n =\sum_{k=0}^n \binom{n}{k} x^k $, $(1-x^2)^n =\sum_{k=0}^n \binom{n}{k}(-1)^k x^{2k} $.

Integrating from $0$ to $1$, $\int_0^1 (1-x^2)^n\,dx =\sum_{k=0}^n \binom{n}{k}(-1)^k \int_0^1 x^{2k}\,dx =\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{2k+1} $.

$\begin{align*} \int_0^1 (1-x^2)^n\,dx &=\int_0^1 (1-x)^n (1+x)^n\,dx\\ &=\int_0^1 y^n (2-y)^n\,dy \qquad(y = 1-x)\\ &=2\int_0^\frac12 (2z)^n (2-2z)^n\,dz \qquad(y = 2z)\\ &=2^{2n+1}\int_0^\frac12 z^n (1-z)^n\,dz\\ &=2^{2n}\int_0^1 z^n (1-z)^n\,dz\\ &=2^{2n}B(n+1, n+1) \qquad\text{(Beta function)}\\ &=2^{2n}\frac{(n!)^2}{(2n+1)!}\\ \end{align*} $

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.