Expressing double factorial for odd integers

[JamesGoh]

Homework Statement

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

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

where 0 <= n <= infinity

Homework Equations

The double factorial for odd integers is

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

The Attempt at a Solution

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

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

performing the sign inverse, i got

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

 

[SammyS]

Homework Statement

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

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

where 0 <= n <= infinity
...

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 $\displaystyle \frac{2^nn!}{(2n+1)!}(-1)^{n}\,x^{2n+1}\,.$