Closed form expression for continued fraction

While I was trying to determine the value of the following infinite series: $$\displaystyle\sum_{n=1}^{\infty}\displaystyle\prod_{k=1}^n\frac{1}{2k+1}$$ I realized that it is equal to the value of the following continued fraction: $$\cfrac{1}{2+\cfrac{3/2}{3+\cfrac{4/2}{4+\cdots}}}$$ I know that the value of $2-e$ is given by: $$\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cdots}}}$$ So, there is similarity between these two continued fractions but I don't know if I can use the continued fraction for $2-e$ to find closed form for continued fraction given above. Any hint is welcomed.

Solution 1:

i think it is easier to go for the sum, i will show a sketch why:

The product gives: $$ P_n=\prod_{k=1}^n (2k+1)^{-1} =\frac{2^n n!}{(2n+1)!} $$

using $n! = \int_0^{\infty} t^n e^{-t}$ and the series expansion for $\sinh(x)$ we get (i don't jusitfy exchange of integral and series, but it should be fine!) for the sum:

$$ S=\sum_{n \geq 0} P_n=\int_0^{\infty}dte^{-t}\frac{\sinh{\sqrt{2t}}}{\sqrt{2t}} $$

writing $x^2=t$ we get some integrals which can be expressed as Error functions yielding

$$ S= \frac{\sqrt{e \pi}}{\sqrt{2}}\text{erf}\left(\frac{1}{\sqrt{2}}\right) $$

@Claude rightfully pointed out that the sum in question (call it $s$) starts at $1$ so we have

$$ s=S-1 $$

which is also a very unexpected form for your partial fraction :)