Sum: $\sum\limits_{n=0}^\infty \frac{n!}{(2n)!}$ I'm struggling with the following sum:
$$\sum_{n=0}^\infty \frac{n!}{(2n)!}$$
I know that the final result will use the error function, but will not use any other non-elementary functions. I'm fairly sure that it doesn't telescope, and I'm not even sure how to get $\operatorname {erf}$ out of that.
Can somebody please give me a hint? No full answers, please.

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\begin{align}
\sum_{n = 0}^{\infty}{n! \over \pars{2n}!} & =
1 + \sum_{n = 1}^{\infty}{\Gamma\pars{n} \over \pars{n - 1}!}\,
{\Gamma\pars{n + 1} \over \Gamma\pars{2n + 1}}
\\[5mm] & =
1 + \sum_{n = 0}^{\infty}{1 \over n!}\,\ \overbrace{%
{\Gamma\pars{n + 1}\Gamma\pars{n + 2} \over \Gamma\pars{2n + 3}}}
^{\ds{\substack{\ds{=\ \mrm{B}\pars{n + 1,n + 2}.}\\[1mm] \ds{\mrm{B}:\ Beta\ Function}}}}
\\[5mm] & =
1 + \sum_{n = 0}^{\infty}{1 \over n!}\ \overbrace{%
\int_{0}^{1}t^{n}\pars{1 - t}^{n + 1}\,\dd t}^{\ds{\mrm{B}\pars{n + 1,n + 2}}}
\\[5mm] & =
1 + \int_{0}^{1}\pars{1 - t}\sum_{n = 0}^{\infty}
{\bracks{t\pars{1 - t}}^{\,n} \over n!}\,\dd t
\\[5mm] & =
1 + \int_{0}^{1}\pars{1 - t}\expo{t\,\pars{1 - t}}\,\dd t
\\[5mm] & =
1 +
\int_{-1/2}^{1/2}\pars{{1 \over 2} - t}\exp\pars{{1 \over 4} - t^{2}}\,\dd t 
\\[5mm] & =
1 + \expo{1/4}\ \overbrace{\int_{0}^{1/2}\expo{-t^{2}}\,\dd t}
^{\ds{{1 \over 2}\,\root{\pi}\,\mrm{erf}\pars{1 \over 2}}}\qquad
\pars{~\mrm{erf}:\ Error\ Function~}
\\[5mm] & =
\bbx{1 + {1 \over 2}\,\root{\pi}\expo{1/4}\,\mrm{erf}\pars{1 \over 2}}
\approx 1.5923
\end{align}
