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?
 A: ${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*}
$
A: 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.
A: 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 $j\mapsto n-j$:
$$
\sum_{j=0}^n(-1)^j\binom{n}{j}\frac1{2j+1}=\frac{2^nn!}{(2n+1)!!}\tag{15}
$$
A: 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.
A: 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

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

