TITLE: How to prove $\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{42n+5}{2^{12n+4}}=\frac1\pi$?
QUESTION [13 upvotes]: 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?

REPLY [2 votes]: 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.