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Question: How do you prove this $\pi$ formula found by G.Bauer in $1859$$$\dfrac 2\pi=1-5\left(\dfrac {1}{2}\right)^3+9\left(\dfrac {1\cdot3}{2\cdot4}\right)^3-13\left(\dfrac {1\cdot3\cdot5}{2\cdot4\cdot6}\right)^3+\&\text{c}.$$

I'm not sure what to do in this case. I found it in a wikipedia article and would like to know how to prove it.

And do you think it's possible to use the expansion$$(1-x)^{-5}=1+5x+\dfrac {5\times6}{1\times2}x^2+\dfrac {5\times6\times7}{1\times2\times3}x^3+\dfrac {5\times6\times7\times8}{1\times2\times3\times4}x^5+\&\text{c}.\tag{2}$$

Thanks in advance!

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    What is the pattern for the coefficients in front of the term? Seems to be $4k+1$, but I want to confirm this – 2017-02-20
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    This is just a summation of double factorials cubed. Representing these in terms of Gamma function isn't hard from there... I would then try to represent it as a cubed value of Beta function – 2017-02-20
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    @BrevanEllefsen Yes, the coefficients obey the form $4k+1$. – 2017-02-20
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    OK. do you have a link to where you found the formula? – 2017-02-20
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    As usual in these matters, it's hypergeometric functions. See [this](http://www.maa.org/sites/default/files/pdf/pubs/amm_supplements/Monthly_Reference_5.pdf), § 10 for a summary and references. – 2017-02-20
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    @BrevanEllefsen The Ramanujan wikipedia article... – 2017-02-20
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    [Related question](http://math.stackexchange.com/questions/152577/what-and-where-in-the-notebooks-of-ramanujan-is-this-series). – 2017-02-21
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    See http://math.stackexchange.com/q/2095780/72031 – 2017-02-28

1 Answers 1

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This is what I have so far:

Start with the Hypergeometrical Series$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag1$$ And let $n=-x=-y=\frac 12$. Therefore, $(1)$ takes the form $$\begin{align*}_4F_3\left[\begin{array}{c c}\frac 54,\frac 12,\frac 12,\frac 12\\\frac 14,1,1\end{array};-1\right] & =\dfrac {1}{\Gamma\left(\frac 32\right)\Gamma\left(\frac 12\right)}\\ & =\dfrac 2\pi\tag{2}\end{align*}$$ And since the general Hypergeometrical series obeys$$_pF_q\left[\begin{array}{c c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_p\end{array};x\right]=\sum\limits_{k=0}^\infty\dfrac {(\alpha_1)_k(\alpha_2)_k\cdots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\cdots(\beta_p)_k}\dfrac {x^k}{k!}\tag{3}$$ We have the LHS as$$\begin{align*}_4F_3\left[\begin{array}{c c}\frac 54,\frac 12,\frac 12,\frac 12\\\frac 14,1,1\end{array};-1\right] & =\sum\limits_{k=0}^{\infty}\dfrac {\left(\frac 54\right)_k\left(\frac 12\right)_k\left(\frac 12\right)_k\left(\frac 21\right)_k(-1)^k}{\left(\frac 14\right)_k\left(1\right)_k\left(1\right)_kk!}\\ & =1-5\left(\dfrac 12\right)^3+9\left(\dfrac {1\cdot3}{2\cdot4}\right)^3-13\left(\dfrac {1\cdot3\cdot5}{2\cdot4\cdot6}\right)^3+\&\text c.\tag{4}\end{align*}$$ Therefore, the identity is established. However, now the question simplifies into

Question: How do we prove$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag{5}$$