Infinite factorial summation

This series can be written as

$S = \dfrac{1}{2}\left(\dfrac{2}{3!} + \dfrac{4}{5!} + \dfrac{6}{7!} + \dfrac{8}{9!} + ....\right) =$

$\dfrac{1}{2}\left(\dfrac{3 - 1}{3!} + \dfrac{5 - 1}{5!} + \dfrac{7 - 1}{7!} + \dfrac{9 - 1}{9!} + ....\right) =$

$\dfrac{1}{2}\left(\displaystyle\sum_{n=1}^{\infty} \dfrac{2n + 1}{(2n + 1)!} - \sum_{n=1}^{\infty} \dfrac{1}{(2n + 1)!}\right) =$

$\dfrac{1}{2}\left(\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{(2n)!} - \sum_{n=1}^{\infty} \dfrac{1}{(2n + 1)!}\right) =$

$\dfrac{1}{2}\displaystyle\sum_{n=2}^{\infty} \dfrac{(-1)^{n}}{n!} = \dfrac{1}{2}\left(\left(\sum_{n=0}^{\infty} \dfrac{(-1)^{n}}{n!}\right) - \dfrac{1}{0!} + \dfrac{1}{1!}\right) = \dfrac{1}{2}(e^{-1} - 1 + 1) = \dfrac{1}{2e}.$

Thus $\lfloor 1000S \rfloor = \lfloor \dfrac{500}{e} \rfloor = \lfloor 183.9397... \rfloor = \boxed{183}.$

Note that $e^{x} = \displaystyle\sum_{n=0}^{\infty} \dfrac{x^{n}}{n!}.$

We recall that the Maclaurin series for $\displaystyle \sinh x = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$ . Thus,

$\sum_{n=0}^{\infty} \frac{2nx^{2n-1}}{(2n+1)!} = \frac{d}{dx} \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n+1)!} = \frac{d}{dx} \frac{\sinh x}{x} = \frac{x\cosh x - \sinh x}{x^{2}}$

from the product rule and that $\frac{d}{dx} \sinh x = \cosh x$ .

Since our desired sum matches half our obtained series at $x = 1$ , we thus get

$S = \sum_{n=0}^{\infty} \frac{n}{(2n+1)!} = \left. \frac{x\cosh x - \sinh x}{2x^{2}} \right|_{x=1} = \frac{\cosh 1 - \sinh 1}{2} = \boxed{\frac{1}{2e}}$