This is equivalent to showing that $\displaystyle\sum_{n=2}^\infty\frac{n!}{(2n-1)!!}=\frac\pi2$ , which, after multiplying both the 

numerator and the denominator with $(2n)!!=2^n\,n!$, and taking into account that $\dfrac{(2n)!}{n!^2}=$ 

$=\displaystyle{2n\choose n}$, can be rewritten as $\displaystyle\sum_{n=2}^\infty\frac{2^n}{\displaystyle{2n\choose n}}=\frac\pi2$ , which can ultimately be deduced from the 

wider identity $\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2}=2\arcsin^2x$, itself ultimately a consequence of integrating <br> the [Cauchy product](http://en.wikipedia.org/wiki/Cauchy_product) between the [binomial series expansion](http://en.wikipedia.org/wiki/Binomial_series) of $\arcsin'x=\dfrac1{\sqrt{1-x^2}}$ and 

that of its primitive. Hope this helps.