Calculate the sum $\sum_{n=0}^{\infty }\frac{1}{n!!}$ \begin{align*}
S =\sum_{n=0}^{\infty }\frac{1}{n!!}=\sum_{n=0}^{\infty }\left ( \frac{1}{(2n)!!}+\frac{1}{(2n+1)!!} \right )
&= \sum_{n=0}^{\infty }\left ( \frac{1}{(2n)!!}+\frac{(2n)!!}{(2n+1)!} \right )=\sum_{n=0}^{\infty }\left ( \frac{1}{2^nn!}+\frac{2^nn!}{(2n+1)!} \right )\\
& =e^{1/2}+\sum_{n=0}^{\infty }\frac{2^nn!}{(2n+1)!}; \; \sum_{n=0}^{\infty }\frac{n!x^{2n+1}}{(2n+1)!}=F(x)\\
&\Rightarrow \frac{\mathrm{d} F}{\mathrm{d} x}=\sum_{n=0}^{\infty }\frac{n!x^{2n}}{(2n)!}\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x}\left [ F(x)\exp \left \{ -\frac{x^2}{4} \right \} \right ]\\
&=\left ( \frac{\mathrm{d} F}{\mathrm{d} x}-\frac{1}{2}xF(x) \right )\exp \left \{ -\frac{x^2}{4} \right \}\\
& =\frac{1}{2}\left ( 2-x^2+\sum_{n=1}^{\infty }\left ( \frac{(n-1)!x^{2n}}{(2n-1)!}-\frac{n!x^{2n+2}}{(2n+1)!} \right ) \right )\exp \left \{ -\frac{x^2}{4} \right \}\\
& =\exp\left \{ -\frac{x^2}{4} \right \}\Rightarrow F(x)=\exp \left \{ \frac{x^2}{4} \right \}\int\limits_{0}^{x}\exp \left \{ -\frac{\xi ^2}{4} \right \}d\xi
\end{align*}
$$\left \{ \int\limits_{\mathbb{R}}e^{-x^2}dx \right \}^2=\iint\limits_{\mathbb{R}}e^{-x^2-y^2}dxdy=\int\limits_{0}^{2\pi}\int\limits_{0}^{\infty }e^{-r^2}rdrd\varphi =\frac{1}{2}\int\limits_{0}^{2\pi}\left ( 1-\lim_{r\rightarrow \infty } e^{-r^2}\right )d\varphi =\pi$$
\begin{align*}
\int\limits_{0}^{\infty }e^{-\xi ^2}d\xi =\frac{\sqrt{\pi}}{2}; \; \lim_{x\rightarrow \infty }\mathrm{erf}x=1
&\Rightarrow \int\limits_{0}^{x}\exp \left \{ -\frac{\xi ^2}{4} \right \}d\xi =\sqrt{\pi}\mathrm{erf}\frac{x}{2}\\
& \Rightarrow F(x)=\sqrt{\pi}\exp \left \{ \frac{x^2}{4} \right \}\mathrm{erf}\frac{x}{2}
\end{align*}
$$\sum_{n=0}^{\infty }\frac{1}{n!!}=\exp \left \{ \frac{1}{2} \right \}+\sum_{n=0}^{\infty }\frac{2^nn!}{(2n+1)!}=\sqrt{e}+\sqrt{\frac{\pi e}{2}}\mathrm{erf}\frac{1}{\sqrt{2}} $$
I showed my attempt at a solution above. This is the standard method of calculating the sum. The question is, is it possible to calculate this sum through a contour integral?

This is not an answer but it is too long for a comment.
Without integration, the partial sums can be computed (thanks to Mathematica).
$$S_m=\sum_{n=0}^{m }\frac{1}{n!!}$$
$$S_{2p}=\sqrt{\frac{e \pi }{2}} \,\frac{\Gamma
   \left(p+\frac{1}{2},\frac{1}{2}\right)}{\Gamma
   \left(p+\frac{1}{2}\right)}+\sqrt{e}\,\frac{ \Gamma
   \left(p+1,\frac{1}{2}\right)}{\Gamma (p+1)}-\sqrt{\frac{e \pi }{2}}
   \text{erfc}\left(\frac{1}{\sqrt{2}}\right)$$
$$S_{2p+1}=S_{2p}+\frac{\sqrt{\pi } }{2^{p+1}\,\Gamma \left(p+\frac{3}{2}\right)}$$ All of the above leads to
$$S_\infty=\sqrt e \left(1+\sqrt{\frac{\pi }{2}} \text{erf}\left(\frac{1}{\sqrt{2}}\right)\right)$$
