Proof of famous Ramanujan sum

Question

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?

May help: http://oeis.org/wiki/A_remarkable_formula_of_Ramanujan — Jean-Claude Arbaut, May 10 '16 at 18:37
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
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
Possible duplicate of 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

score 1 · Accepted Answer · answered May 10 '16 at 18:45

1

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

answered May 10 '16 at 18:45

Jack D'Aurizio

353,855

Yeah, I forgot about Gamma and Beta functions – openspace May 10 '16 at 20:01

Proof of famous Ramanujan sum

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