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Few days ago I thought about proof of :$$\frac{1}{3}+\frac{1}{3\cdot 5} + \dots = \sqrt{\frac{e\pi}{2}}$$. I tried to represent my sum as : $$\sum\frac{2n!!}{(2n+1)!}$$, so after I was thinking about Wallis product, but that doesn't give me anything.

May anyone give me a hint what is the best way to start my proof?

openspace
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    May help: http://oeis.org/wiki/A_remarkable_formula_of_Ramanujan – Jean-Claude Arbaut May 10 '16 at 18:37
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    Note that holds $$\sum_{n\geq0}\frac{x^{2n+1}}{(2n+1)!!}=\sqrt{\frac{\pi}{2}}\textrm{erf}(\frac{x}{\sqrt{2}})e^{x^{2}/2}.$$ – Marco Cantarini May 10 '16 at 18:40
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    I thought the whole point about the R expression for $\sqrt{\frac{\pi e}{2}}$ was that it had two parts: the sum and the continued fraction. Why have you left out the continued fraction? – almagest May 10 '16 at 18:44
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    Possible duplicate of [An infinite series plus a continued fraction by Ramanujan](http://math.stackexchange.com/questions/833920/an-infinite-series-plus-a-continued-fraction-by-ramanujan) – Paramanand Singh May 11 '16 at 12:34
  • @almagest: see http://math.stackexchange.com/q/833920/72031 for continued fraction. – Paramanand Singh May 11 '16 at 12:36

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$$\begin{eqnarray*}\sum_{n\geq 1}\frac{(2n)!!}{(2n+1)!}=\sum_{n\geq 1}\frac{2^n \Gamma(n+1)}{\Gamma(2n+2)}&=&\sum_{n\geq 1}\frac{2^n}{n!}\,B(n+1,n+1)\\&=&\int_{0}^{1}\sum_{n\geq 1}\frac{2^n x^n(1-x)^n}{n!}\,dx\\&=&-1+\int_{0}^{1}\exp\left(2x(1-x)\right)\,dx\\&=&-1+\int_{-1/2}^{1/2}\exp\left(\frac{1}{2}-2z^2\right)\,dz\\&=&\color{red}{-1+\sqrt{\frac{e\pi}{2}}\,\text{Erf}\left(\frac{1}{\sqrt{2}}\right)}.\end{eqnarray*} $$

Jack D'Aurizio
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