Simplify a factorial I have the problem to evaluate the following:
$$
(2n)!\over
2^n(n!)
$$
Does this reduce to anything in particular?
I stuck it into a computer and it's 
1: 1
2: 3
3: 15
4: 105
5: 945
6: 10395

No pattern immediately apparent.
 A: You won’t get a nice closed form, but there is another way to write it that is sometimes useful. Notice that
$$\begin{align*}2^nn!&=\underbrace{2\cdot2\cdot2\cdot\ldots\cdot2}_n\cdot1\cdot2\cdot3\cdot\ldots\cdot n\\&=(2\cdot1)(2\cdot2)(2\cdot3)\dots(2\cdot n)\\&=2\cdot4\cdot6\cdot\ldots\cdot 2n\;,\end{align*}$$
do some cancelling, and look at Qiaochu’s comment.
A: You can use the identity
$$ (2n)! = \Gamma(2n+1) = {\frac {{2}^{2n} \Gamma  \left( n + 1\right) \Gamma 
\left( n + \frac{1}{2} \right) }{\sqrt {\pi }}}\,.$$
This leads to the simplification
$$\frac{(2n)!}{2^n n!} = \frac{ 2^n \Gamma(n+\frac{1}{2})}{\sqrt{\pi}} = \frac{ 2^n (n-\frac{1}{2})!}{\sqrt{\pi}}\,.$$
A: Since $(2n)! = (2n) \times (2n-1) \times \cdots \times 2 \times 1$. Split the product into products of even factors and odd factors:
$$
   (2n)! = \prod_{m=1}^{n} (2m) \cdot \prod_{m=1}^{n} (2m-1) = 2^n \prod_{m=1}^n m \cdot \prod_{m=1}^{n} (2m-1) = 2^n n! \prod_{m=1}^{n} (2m-1) 
$$
Therefore:
$$
   \frac{(2n)!}{2^n n!} = \prod_{m=1}^n (2m-1) = (2n-1)!!
$$
where $m!!$ denotes double factorial.
A: As Brian said, you will not get a nice closed form expression. You can also do a little algebra and arrive at an expression with gamma functions instead of factorials.
http://en.wikipedia.org/wiki/Gamma_function
