<s>[INST] Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$ 
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1$).

How to prove this equation?
Is it possible to use induction?
$${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1};$$
$$(2n+1)!!=\frac{(2n)!(2n+1)}{2^nn!}\Rightarrow \frac{n!2^n}{(2n+1)!!}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)};$$
$$\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)}$$
What now?
 [/INST] ${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*}
$
</s>