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.

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    $\begingroup$ You can look here math.stackexchange.com/questions/833920/… $\endgroup$ – E. Joseph Jan 9 '17 at 12:54
  • $\begingroup$ In general, $$\sum_{n=0}^\infty~2^n\cdot\frac{x^{2n+1}}{(2n+1)!!} ~=~ \frac{\sqrt\pi}2\cdot e^{x^2}\cdot\text{erf}(x).$$ $\endgroup$ – Lucian Jan 9 '17 at 12:58
  • $\begingroup$ Start by writing the well-known Taylor series expansion for exponential function, then integrate it term by term to obtain that of the error function, and lastly evaluate their Cauchy product. $\endgroup$ – Lucian Jan 9 '17 at 14:10
  • $\begingroup$ @Lucian, thanks for the advice, but this looks complicated. Maybe using hypergeometric functions will be easier for me $\endgroup$ – Yuriy S Jan 9 '17 at 14:12
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    $\begingroup$ The link from @E.Joseph has your answer. $\endgroup$ – Paramanand Singh Jan 9 '17 at 14:21

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 $x=\frac{1}{2}+t$.

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    $\begingroup$ Simplest solution. +1 $\endgroup$ – Paramanand Singh Jan 9 '17 at 18:35

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