# 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.

• To help sell the pattern, you should replace the $1$ on top with $2/2$. :)
– Blue
Dec 8, 2021 at 9:55

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 :)

• The summation starts at $n=1$. So, you result is correct with a $\color{red}{-1}$ in front of it Dec 8, 2021 at 10:46
• @ClaudeLeibovici thanks, i updated my answer accordingly Dec 8, 2021 at 10:49