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*}
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*}