How to prove: $$1+\frac13 \left(1+\frac{1}{5}\left(1+\frac{1}{7}\left(1+\frac{1}{9}\left(1+\dots \right) \right) \right) \right)=\sqrt{\frac{\pi e}{2}} \text{erf} \left( \frac{1}{\sqrt{2}} \right)$$
Since we can represent the series for $e$ in a pretty form:
$$1+\frac12 \left(1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\frac{1}{5}\left(1+\dots \right) \right) \right) \right)=e-1$$
I've been trying to search closed forms for some similar expressions. For example, it's easy to see that:
$$1+\frac12 \left(1+\frac{1}{4}\left(1+\frac{1}{6}\left(1+\frac{1}{8}\left(1+\dots \right) \right) \right) \right)=\sqrt{e}$$
However, I don't know how to prove the expression in the title. Seems to be something for Ramanujan to consider.
P.S. the closed form was given by Wolfram Alpha.
See a general formula provided by Lucian in the comments.
WA also provides another general formula:
$$\sum_{n=0}^\infty \frac{1}{\prod_{k=0}^n bk+c}=e^{\frac{1}{b}} \left(\frac{1}{b} \right)^{1-\frac{c}{b}} \left( \Gamma\left(\frac{c}{b} \right) -\Gamma\left(\frac{c}{b},\frac{1}{b} \right) \right)$$
Still don't know how to prove it, but I've noticed the emergence of the hypergeometric functions in some cases, which makes it possible to transform the expression to hypergeometric series.
Using Euler's Beta function,
$$ S = \sum_{n\geq 0}\frac{1}{(2n+1)!!} = \sum_{n\geq 0}\frac{2^n n!}{(2n+1)!} = \sum_{n\geq 0} \frac{2^n}{n!} B(n+1,n+1) \tag{1}$$ leads to: $$ S = \int_{0}^{1}\sum_{n\geq 0}\frac{\left(2x(1-x)\right)^n}{n!}\,dx = \int_{0}^{1}\exp\left(2x(1-x)\right)\,dx\tag{2}$$ and the last integral is easy to convert into the $\text{Erf}$ format through the substitution $x=\frac{1}{2}+t$.