<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] 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

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

</s>