Inspired by Kishlaya!

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

Now, $\displaystyle \frac{n!((n-1)!)}{(2n)!} = B(n+1,n)$

$\displaystyle \sum_{n=1}^{\infty}{\frac{B(n+1,n)}{(n-1)!}}$

$\displaystyle B(n+1,n) = \int_{0}^{1}{{t}^{n}{(1-t)}^{n-1}}$

$\displaystyle \sum_{n=1}^{\infty}{\frac{\int_{0}^{1}{{t}^{n}{(1-t)}^{n-1}}}{(n-1)!}}$

$\displaystyle \int_{0}^{1}{\sum_{n=1}^{\infty}{\frac{{t}^{n}{(1-t)}^{n-1}}{(n-1)!}}}$

$\displaystyle \int_{0}^{1}{\frac{1}{1-t}\sum_{n=1}^{\infty}{\frac{{t(1-t)}^{n}}{(n-1)!}}}$

$\displaystyle \sum_{n=1}^{\infty}{\frac{{t(1-t)}^{n}}{(n-1)!}} = t(1-t){e}^{t(1-t)}$

$\displaystyle \int_{0}^{1}{\frac{t(1-t){e}^{t(1-t)}}{1-t}}$

$\displaystyle \int_{0}^{1}{t{e}^{t(1-t)}}$

$\displaystyle \int_{0}^{1}{t{e}^{t-{t}^{2}-\frac{1}{4} + \frac{1}{4}}}$

$\displaystyle {e}^{\frac{1}{4}}(\int_{0}^{1}{t{e}^{{-(t-1/2)}^{2}}})$

Substituting $t-1/2 = x$ ,

$\displaystyle {e}^{1/4}(\int_{-1/2}^{1/2}{x{e}^{-{x}^{2}}} + \frac{1}{2}(\int_{-1/2}^{1/2}{{e}^{-{x}^{2}}}))$

$\displaystyle \int_{-1/2}^{1/2}{x{e}^{-{x}^{2}}} = -\frac{{e}^{-1/4}}{2} +\frac{{e}^{-1/4}}{2} = 0$

$\displaystyle \frac{1}{2}(\int_{-1/2}^{1/2}{{e}^{-{x}^{2}}}) = \frac{1}{2}2\int_{0}^{1/2}{{e}^{-{x}^{2}}}$

$\displaystyle = \frac{{\pi}^{1/2}}{2}*erf(1/2)$

And yes, we have ${e}^{1/4}$ in front as well, so, our answer becomes

$\displaystyle = \boxed{\frac{{e}^{1/4}{\pi}^{1/2}erf(1/2)}{2}}$

Therefore $\displaystyle \frac{1}{4} + \frac{1}{2} + \frac{1}{2} = \frac{5}{4}$ , hence $5 + 4 = \boxed{9}$