Perhaps a rather strange question. I am doing an exercise and the last part ends: $I = \frac{n!}{(2n)!}$
Now I know that: $$I = \frac{n(n-1)(n-2)(n-3)\cdots}{2n(2n-1)(2n-2)(2n-3)\cdots }$$
Is it possible to cancel out some terms? Or to make it a little bit more 'friendly'?
On
Note that $$I = \frac{n(n-1)(n-2)(n-3) \cdots 2 \cdot 1}{2n(2n-1)(2n-2)(2n-3) \cdots 2 \cdot 1} $$$$= \frac{n(n-1)(n-2)(n-3) \cdots 2 \cdot 1}{2^n \times n(2n-1)(n-1)(2n-3)(n-2) \cdots 1 \cdot 1}$$ and after cancelling out we get that$$ I = \frac{1}{2^n (2n-1) (2n-3) \cdots 3 \cdot 1} = \frac{1}{2^n} \prod_{i=1}^n \frac{1}{2i -1} $$This is one possible expression, but I think yours is more concise.
On
Separating the even and odd terms, we have $$ (2n)! = [ (2n)(2n-2)(2n-4)\dotsm \cdot 2 ] [(2n-1)(2n-3)\dotsm 1 ] \\ = 2^n [n(n-1)(n-2) \dotsm 1] [(2n-1)(2n-3)\dotsm 1] = 2^n n! (2n-1)!!, $$ where the double factorial is defined by $k!! = k(k-2)(k-4)\dotsm 3 \cdot 1 $ if $k$ is odd, or $\dotsm 4 \cdot 2$ if $k$ is even.
Alternatively, $(2n)!/n! = (2n)(2n-1)\dotsm (n+1) = (n+1)_{n}$, where $(a)_k$ is the rising factorial $ a(a+1)\dotsm (a+k-1) $.
On
Using the Double Factorial, $$ (2n)!=(2n)!!(2n-1)!!=2^nn!(2n-1)!! $$ Therefore, $$ \frac{n!}{(2n)!}=\frac1{2^n(2n-1)!!} $$ which does not really look any simpler.
On
Like @jam said you can use Stirling's approximation This approximates to : $$\frac {2^{\frac{-4n-1}{2}}e^{n}}{n^n}$$
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {n! \over \pars{2n}!} & = {n\pars{n - 1}! \over \pars{2n}\pars{2n - 1}!} = {1 \over 2}\,{\pars{n - 1}! \over \pars{2n - 1}!} = {1 \over 2}\,{\Gamma\pars{n} \over \Gamma\pars{2n}}\qquad \pars{~\Gamma:\ Gamma\ Function~} \\[5mm] & = \require{cancel} {1 \over 2}\,{\cancel{\Gamma\pars{n}} \over \pars{2\pi}^{-1/2} \,2^{2n - 1/2}\,\cancel{\Gamma\pars{n}}\Gamma\pars{n + 1/2}}\qquad \pars{\begin{array}{l} \Gamma\mbox{-}\,Duplication\ Formula\ \mbox{in the} \\ \textsf{denominator}. \end{array}} \\[5mm] & = \bbx{{2^{-2n} \over \pars{n - 1/2}!}\,\root{\pi}} \\ & \end{align}
On
To complete the "panorama", consider also the expression in terms of Rising ($x^{\,\overline {\,n\,}}$) and Falling ($x^{\,\underline {\, \,n\,}}$) Factorials $$ {{n!} \over {\left( {2n} \right)!}} = {{1^{\,\overline {\,n\,} } } \over {1^{\,\overline {\,2n\,} } }} = {{1^{\,\overline {\,n\,} } } \over {1^{\,\overline {\,n\,} } \;\left( {n + 1} \right)^{\,\overline {\,n\,} } }} = {1 \over {\;\left( {n + 1} \right)^{\,\overline {\,n\,} } }} = n^{\,\underline {\, - \,n\,} } $$
I think your possible options could be summarised as:
Leaving it as $\frac{n!}{(2n)!}$, which is a perfectly simple expression as it is.
Canceling out the terms and expressing it as $\left[\prod_{i=1}^n(n+i)\right]^{-1}$, but this doesn't really achieve much.
Using the Gamma function, to get $\frac{n!}{(2n)!}=\frac{\sqrt{π} }{2^{2 n}Γ(n + 0.5)}$. In a way this is simpler since the only hard part to compute is $Γ(n + 0.5)$.
Assuming that $n$ is sufficiently large and using $\frac{n!}{(2n)!}\sim \frac1{\sqrt{2}}\left(\frac{e}{4n}\right)^n $, by Stirling's approximation. In fact this approximation is pretty accurate, as shown for the true curve (blue), approximation (red) and absolute error (orange) below.