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.