Definition of the binomial coefficient $a_{n}=\binom{-\frac{1}{2}}{n}$ Applying the definition of the binomial coefficient I can't figure out how to simplify the following expression:
$$a_{n}=\binom{-\frac{1}{2}}{n}$$
I want to find:
$$a_{n}=(-1)^n\frac{(2n-1)!}{(n-1)!2^{2n}n!}$$
I'm stuck at the beginning. 
$$a_{n}=\frac{-\frac{1}{2}!}{n!\left ( n-\frac{1}{2} \right )!}$$
Thank you so much!

Here we use the following definition of the binomial coefficient 
\begin{align*}
  \binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}
\qquad\qquad\alpha\in\mathbb{C},\quad n\in \mathbb{N}\tag{1}
  \end{align*}
It is also convenient to use double factorials for a more compact notation
\begin{align*}
  (2n)!! &= (2n)(2n-2)\cdots 4\cdot2\qquad\qquad\qquad n\in\mathbb{N}\\
  (2n-1)!! &= (2n-1)(2n-3)\cdots 3\cdot1\\
  \end{align*}

We obtain
  \begin{align*}
\binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)
  \cdots\left(-\frac{1}{2}-n+1\right)}{n!}\tag{2}\\
  &=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)
  \cdots\left(-\frac{2n-1}{2}\right)}{n!}\\
  &=\frac{(-1)^n}{2^n}\cdot\frac{1\cdot 3\cdots \
  \left(2n-1\right)}{n!}\tag{3}\\
  &=\frac{(-1)^n}{2^n}\cdot\frac{(2n-1)!!}{n!}\tag{4}\\
  &=\frac{(-1)^n}{2^n}\cdot\frac{(2n)!}{(2n)!!n!}\tag{5}\\
  &=\frac{(-1)^n}{2^n}\cdot\frac{(2n)!}{2^nn!n!}\tag{6}\\
  &=\frac{(-1)^n(2n)!}{2^{2n}n!n!}\tag{7}\\
  &=\frac{(-1)^n(2n-1)!}{2^{2n-1}(n-1)!n!}\tag{8}\\
\end{align*}

Comment:


*

*In (2) we use the definition of the binomial coefficient with $\alpha=-\frac{1}{2}$

*In (3) we factor out $-\frac{1}{2}$ from each of the $n$ factors giving $(-1)^n\frac{1}{2^n}$

*In (4) we use double factorials $(2n-1)!!$

*In (5) we apply $$(2n)!=(2n)!!(2n-1)!!$$

*In (6) we use the formula from (5) and note that
$$(2n)!!=(2n)(2n-2)\cdots4\cdot 2=2^nn!$$


*

*In (7) we collect terms.

*In (8) we devide numerator and denominotor by $2n$.
Note there is a factor $2^{2n-1}$ in the denominator of $a_n$ which gives for $n=1$
\begin{align*}
\binom{-\frac{1}{2}}{1}=-\frac{1}{2}
\end{align*}
