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}={2n\choose n}$$\dfrac{(2n)!}{n!^2}=$
$=\displaystyle{2n\choose n}$, can be re-writtenrewritten as $\displaystyle\sum_{n=2}^\infty\frac{2^n}{2n\choose n}=\frac\pi2$$\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}}{{2n\choose n}n^2}=2\arcsin^2x$$\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2}=2\arcsin^2x$, itself ultimately a consequence of integrating
the Cauchy product between the binomial series expansion of $\arcsin'x=\dfrac1{\sqrt{1-x^2}}$ and
that of its primitive. Hope this helps.