<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] We can rewrite the sum as $$\sum_{k=0}^n \binom{n}{k}(-1)^k\int_0^1x^{2k}dx = \int_0^1(1-x^2)^ndx.$$
Denote the last integral as $I(n)$. We integrate by parts:
$$I(n) = x(1-x^2)\big|_{x=0}^{x=1} - \int_0^1 x\cdot (-2nx)\cdot (1-x^2)^{n-1}dx = 2n \int_0^1 x^2(1-x^2)^{n-1}dx = 2n (I(n-1)-I(n)),$$
therefore
$$I(n) = \frac{2n}{2n+1}I(n-1).$$
Given that $I(1) = \frac{2}{3}$, a reasoning by recurrence allows us to conclude.
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