Help in simplifying an alternating factorial

Question

What is $\dfrac{(n-1)(n-3)(n-5)\cdots}{(n-2)(n-4)(n-6)\cdots}$ ?

Can it be simplified in terms of $n$? I could not convert it into a factorial form.

Plz solve

Answer 1

It can be written in terms of the Gamma function.

When $n$ is even, $$ {\frac { \left( n-1 \right) !!}{ \left( n-2 \right) !!}}={\frac { 2\;\Gamma \left( n/2+1/2 \right) }{\sqrt {\pi}\;\Gamma \left( n/2 \right) }} $$ and when $n$ is odd, $$ {\frac { \left( n-1 \right) !!}{ \left( n-2 \right) !!}}={\frac {\Gamma \left( n/2+1/2 \right) \sqrt {\pi}}{\Gamma \left( n/ 2 \right) }} $$

Answer 2

Suppose $n=2m$. We have $k=(n-2)(n-4)\dots2=2^{m-1}(m-1)!$ and so the given expression is $$\frac{(2m-1)!}{k^2}=\frac{2m-1}{2^{2m-2}}\ {2m-2\choose m-1}$$

If $n=2m+1$ we have $h=(n-1)(n-3)\dots2=2^mm!$ and so the given expression is $$\frac{h^2}{(2m)!}=\frac{2^{2m}}{{2m\choose m}}$$