How to prove $\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{42n+5}{2^{12n+4}}=\frac1\pi$?

Question

In an article about $\pi$ in a popular science magazine I found this equation printed in light grey in the background of the main body of the article: $$ \color{black}{ \sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{42n+5}{2^{12n+4}}=\frac1\pi } $$ It's true, I checked it at Wolfram, who gives a even more cryptic answer at first glance, but finally confirms the result.

The appearance of $42$ makes me confident that there is someone out there in this universe, who can help to prove that?

Answer 1

This is a famous identity of Ramanujan in "Modular equations and approximations of $\pi$".
There is a proof by the Borweins in "Pi and the AGM" (no preview) p. $177$ to $188$ (this proof and others are rather long!).

UPDATE: "Ramanujan’s Series for 1/π: A Survey" provides the history of the subject with all the technical details.
The brothers Borwein proposed a derivation in $1987$ in "Ramanujan's rational and algebraic series for $\dfrac 1{\pi}$".
Guillera proposed different "Kind of proofs of Ramanujan-like series" in $2012$.

A proof 'by computer' using the WZ algorithm may be found in the paper of Ekhad and Zeilberger "A WZ proof of Ramanujan's formula for $\pi$".
Aycock proposes to compute many similar series using hypergeometric identities like (page $6$ and $28$) : $$_3F_2\left(\frac12,\frac12,\frac12,1,1,x\right)=(1-x)^{-1/2}\;_3F_2\left(\frac14,\frac34,\frac12,1,1,-\frac{4x}{(1-x)^2}\right)$$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\large\tt Hint:$ $\ds{% {1 \over \pi} = \sum_{n=0}^{\infty}\bracks{\pars{2n}! \over \pars{n!}^{2}}^{3} {42n + 5 \over 2^{12n + 4}} = {21 \over 8}\sum_{n=0}^{\infty}{2n \choose n}^{3}n\pars{2^{-12}}^{n} + {5 \over 16}\sum_{n=0}^{\infty}{2n \choose n}^{3}\pars{2^{-12}}^{n}\,, \qquad{\large ?}}$ Let's consider the function $\ds{{\cal F}\pars{x} \equiv \sum_{n=0}^{\infty}{2n \choose n}^{3}x^{n}}$ and we have to evaluate $\ds{\braces{\bracks{{21 \over 8}\,x\,\partiald{}{x} + {5 \over 16}}{\cal F}\pars{x}}_{x = 2^{-12}}}$ $\ds{\pars{~\mbox{this expression returns the value}\ {1 \over \pi}~}}$:

\begin{align} {\cal F}\pars{x} &\equiv \sum_{n=0}^{\infty}x^{n}\int_{\verts{z_{1}} = 1} {\dd z_{1} \over 2\pi\ic}\,{\pars{1 + z_{1}}^{2n} \over z_{1}^{n + 1}} \int_{\verts{z_{2}} = 1} {\dd z_{1} \over 2\pi\ic}\,{\pars{1 + z_{2}}^{2n} \over z_{2}^{n + 1}}\int_{\verts{z_{1}} = 1} {\dd z_{1} \over 2\pi\ic}\,{\pars{1 + z_{3}}^{2n} \over z_{3}^{n + 1}} \\[3mm]&= \prod_{i = 1}^{3}\pars{\int_{\verts{z_{i}} = 1} {\dd z_{i} \over 2\pi\ic}\,{1 \over z_{i}}}\sum_{n = 0}^{\infty}\bracks{% x\pars{1 + z_{1}}^{2}\pars{1 + z_{2}}^{2}\pars{1 + z_{3}}^{2} \over z_{1}z_{2}z_{3}}^{n} \\[3mm]&= \prod_{i = 1}^{3}\int_{\verts{z_{i}} = 1} {\dd z_{i} \over 2\pi\ic}\, {1 \over z_{1}z_{2}z_{3} - x\pars{1 + z_{1}}^{2}\pars{1 + z_{2}}^{2}\pars{1 + z_{3}}^{2}} \end{align}

Answer 3

Ah, 42 and The Hitchhiker's Guide to the Galaxy. Would you like to know how this is connected to the 24-dimensional Leech lattice? :)

Given the Ramanujan-type formula,

$$\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{An+B}{(C)^{n+1/2}}=\frac1\pi$$

there are relatively simple expressions for $A,C$. Define,

$$A(\tau) = \sqrt{d\big(C(\tau)-64\big)}$$

and the 24th power of the Weber modular function as,

$$C(\tau) = \mathfrak{f}^{24}(2\tau)=\left(\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\right)^{24}$$

with the Dedekind eta function,

$$\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})$$

The connection between the appearance of 24 in $\eta(\tau)$ and the Leech lattice is quite well-known.

Example: Let $\tau = \frac{1}{2}\sqrt{-d},\; d = 7$, then,

$$A(\tau) =4\cdot42$$

$$C(\tau) = 2^{12}$$

$$e^{\pi\sqrt{7}} \approx 2^{12} - 24.06\dots$$

$$\frac{4(42n+B')}{(2^{12})^{n+1/2}}$$

so,

$$\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{42n+B'}{2^{12n+4}}=\frac1\pi$$

hence why, using a 24th power of an eta quotient, the number 42 appears.