<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] Binomial Theorem and Beta Functions
$$
\begin{align}
\sum_{k=0}^n(-1)^k\frac1{2k+1}\binom{n}{k}
&=\int_0^1\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}\,\mathrm{d}x\tag1\\
&=\int_0^1\left(1-x^2\right)^n\,\mathrm{d}x\tag2\\
&=\frac12\int_0^1\left(1-x\right)^nx^{-1/2}\,\mathrm{d}x\tag3\\
&=\frac12\frac{\color{#C00}{\Gamma(n+1)}\,\color{#090}{\Gamma(1/2)}}{\color{#090}{\Gamma(n+3/2)}}\tag4\\
&=\frac12\color{#C00}{n!}\color{#090}{\frac{2^{n+1}}{(2n+1)!!}}\tag5\\
&=\frac{2^n\,n!}{(2n+1)!!}\tag6
\end{align}
$$
Explanation:
$(1)$: $\int_0^1x^{2k}\,\mathrm{d}x=\frac1{2k+1}$
$(2)$: Binomial Theorem
$(3)$: substitute $x\mapsto x^{1/2}$
$(4)$: apply the Beta Function integral
$(5)$: apply $\Gamma(x+1)=x\Gamma(x)$
$(6)$: cancel

Repeated Differences
Define a double factorial version of the binomial coefficients:
$$\newcommand{\dinom}[2]{\left\langle{#1\atop#2}\right\rangle}
\dinom{n}{k}=\frac{(2n+1)!!}{(2n-2k+1)!!(2k-2)!!}\tag7
$$
Note that $\dinom{n}{1}=2n+1$. Furthermore,
$$
\begin{align}
\frac1{\dinom{n}{k}}-\frac1{\dinom{n-1}{k}}
&=(2k-2)!!\left(\frac{(2n-2k+1)!!}{(2n+1)!!}-\frac{(2n-2k-1)!!}{(2n-1)!!}\right)\tag8\\
&=(2k-2)!!\frac{(2n-2k-1)!!}{(2n+1)!!}((2n-2k+1)-(2n+1))\tag9\\[3pt]
&=-(2k)!!\frac{(2n-2k-1)!!}{(2n+1)!!}\tag{10}\\[3pt]
&=-\frac1{\dinom{n}{k+1}}\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(8)$: apply $(7)$ while extracting a common factor
$\phantom{1}(9)$: extract more common factors
$(10)$: simplify the right hand term and incorporate into the left hand term
$(11)$: apply $(7)$
Apply $(11)$ $n$ times to get
$$
\begin{align}
\overbrace{\sum_{j=0}^n(-1)^j\binom{n}{j}\frac1{\dinom{n-j}{1}}}^\text{$n^\text{th}$ order backward difference}
&=\frac{(-1)^n}{\dinom{n}{n+1}}\tag{12}\\
&=\frac{(-1)^n(-1)!!(2n)!!}{(2n+1)!!}\tag{13}\\[3pt]
&=(-1)^n\frac{2^nn!}{(2n+1)!!}\tag{14}
\end{align}
$$
Explanation:
$(12)$: apply $(11)$ $n$ times
$(13)$: apply $(7)$
$(14)$: $(-1)!!=1$ and $(2n)!!=2^nn!$
Multiply $(14)$ by $(-1)^n$ and substitute $j\mapsto n-j$:
$$
\sum_{j=0}^n(-1)^j\binom{n}{j}\frac1{2j+1}=\frac{2^nn!}{(2n+1)!!}\tag{15}
$$
</s>