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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 $e-2$ 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 $e-2$ to find closed form for continued fraction given above. Any hint is welcomed.

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    $\begingroup$ To help sell the pattern, you should replace the $1$ on top with $2/2$. :) $\endgroup$
    – Blue
    Dec 8, 2021 at 9:55
  • $\begingroup$ @Pedja I changed your expression $2-e$ to the correct $e-2$, since the second cfrac is clearly positive. $\endgroup$ Oct 21, 2023 at 7:06
  • $\begingroup$ See the series in math.stackexchange.com/q/833920/72031 $\endgroup$
    – Paramanand Singh
    Oct 21, 2023 at 10:56
  • $\begingroup$ @ParamanandSingh Thanks to Pedja's insight, I realized that $\sqrt{\frac{\pi\,e}{2}}$ you discussed in the link can be expressed as the sum of TWO nice continued fractions. Kindly see math.stackexchange.com/q/4791520/4781 $\endgroup$ Oct 22, 2023 at 7:22

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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 :)

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

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