How can I prove this formula? $\cdots=\sqrt{\frac{e\pi}2}$ [duplicate]

I found in the book Escapades Arithmétiques written by Frédéric Laroche this formula:

$$1+\frac 1{1\cdot 3}+\frac 1{1\cdot 3\cdot 5}+\cdots+\frac 1{1+\frac 1{1+\frac 2{1+\frac 3{1+\cdots}}}}=\sqrt{\frac{e\pi}2}.$$

Perhaps in a more explicit way, the first part of this formula is:

$$\sum_{k=0}^\infty \left(\prod_{j=0}^k (2j+1)\right)^{-1}.$$

What I do not like is that this formula (beautiful in my opinion) is written without any proof nor reference.

• Do you have an idea on how to prove such a result?

• Do you know a book that gives the proof of this formula?

marked as duplicate by Bumblebee, egreg, J. M. is a poor mathematician, Hans Lundmark, Community♦Dec 27 '16 at 22:40

$$S= 1+\frac{1}{1\cdot 3}+\frac{1}{1\cdot 3\cdot 5}+\ldots = \sum_{n\geq 0}\frac{1}{(2n+1)!!}=\sum_{n\geq 0}\frac{2^n n!}{(2n+1)!}$$ can be written (through Euler's Beta function) as $$S = \sum_{n\geq 0}\frac{2^n}{n!}B(n+1,n+1) = \int_{0}^{1}\sum_{n\geq 0}\frac{(2x(1-x))^n}{n!}\,dx$$ or as: $$S = \int_{-1/2}^{1/2}\exp\left[\frac{1}{2}-2x^2\right]\,dx.$$ The remaining part follows from the well-known continued fraction expansion for the (complementary) error function: $$\frac{1}{\sqrt{2\pi}}\int_{z}^{+\infty}e^{-x^2/2}\,dx = \frac{e^{-z^2/2}}{\sqrt{2\pi}}\cdot \frac{1}{z+\frac{1}{z+\frac{2}{z+\frac{3}{\ldots}}}}$$