Is it possible to simplify this equation?
Not looking for something long (that would defeat the purpose of simplifying)...
I'm doing some partial sums of sequences and was just wondering if I could simplify the equation before substituting n for {$1,2,3,...$}.
I doubt there's anything that's highly satisfying. As Rahul mentioned in the comments: $$\frac{n!}{(2n)!} = \frac{n(n-1)(n-2)... 3*2}{2n(2n-1)...n(n-1)...3*2} = \frac{1}{(n+1)(n+2)...(2n-1)2n}$$
This is actually something called a falling factorial, defined as: $$(a)_n = a(a-1)...(a-n+1)$$ So, $$(2n)_n = 2n(2n-1)...(n+1)$$ Therefore, $$\frac{n!}{(2n)!} = \frac{1}{(2n)_n}$$
Moreover, $$\frac{1}{n!\binom{2n}{n}} =\frac{n!}{(2n)!} $$
You can also use the standard notation for Permutations to directly represent the quantity. However, all of these are really more of a way to condense the problem, not how to actually simplify it.