# Expressing double factorial for odd integers

1. Sep 17, 2011

### JamesGoh

1. The problem statement, all variables and given/known data

Express $\frac{1}{(2n+1)!!}$ as the following

$\frac{(-2)^{n}n!x^{2n+1}}{(2n+1)!}$

where 0 <= n <= infinity

2. Relevant equations

The double factorial for odd integers is

(2n+1)!! = (2n+1)(2n-1)(2n-3)....1 where 0 <= n <= infinity

3. The attempt at a solution

Visited this website http://mathworld.wolfram.com/DoubleFactorial.html and scrolled
down to the expression for

$(-2n-1)!! = \frac{(-1)^{n}2^{n}n!}{2n!}$

$(2n+1)!! = \frac{(1)^{n}-2^{n}n!}{-2n!}$

2. Sep 17, 2011

### SammyS

Staff Emeritus
It's not at all clear what the x2n+1 is doing in there.

As for (2n+1)!!, start by working with (2n+1)! and see what needs to be canceled out.

$\displaystyle (2n+1)!=(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)\dots(5)(4)(3)(2)(1)$
$\displaystyle =\{(2n+1)(2n-1)(2n-3)\dots(5)(3)(1)\}\{(2n)(2n-2)\dots(4)(2)\}$

$\displaystyle =\{(2n+1)!!\}\{(2^n)(n!)\}$​

Solve for (2n+1)!! .

As for $\displaystyle \frac{(-2)^{n}n!x^{2n+1}}{(2n+1)!}\,,$ that's the same as $\displaystyle \frac{2^nn!}{(2n+1)!}(-1)^{n}\,x^{2n+1}\,.$