Algebraic Addendum

Here's the series again.

$$ \frac{1}{\pi}= \frac{1}{53360\sqrt{640320}} \sum_{n=0}^{\infty} (-1)^n \frac{(6n)!}{(n!)^3(3n)!} \frac{545140134n+13591409}{640320^{3n}} $$

We define some convenience constants:

$$ A=13591409, \qquad B=545140134 $$

The series is then:

$$ \frac{1}{\pi}= \frac{1}{53360\sqrt{640320}} \sum_{n=0}^{\infty} (A+Bn)u_n $$

where

$$ u_n= (-1)^n \frac{(6n)!}{(n!)^3(3n)!} \frac{1}{640320^{3n}} $$

The series is a linear term multiplied by a hypergeometric part. The goal is to express the hypergeometric part as a product:

$$ u_n = \prod_{k=1}^{n} \frac{p_k}{q_k} $$

so that the whole sum has the binary-splitting form

$$ \sum_{n=0}^{\infty} a_n \prod_{k=1}^{n} \frac{p_k}{q_k} $$

We're gonna do this by finding the ration of two consecutive terms of $u_n$, simplifying it, then reconstructing $u_n$ as the product of the ratios.

Write down $u_n$ and $u_{n-1}$:

$$ u_n= (-1)^n \frac{(6n)!}{(n!)^3(3n)!} \frac{1}{640320^{3n}} $$

and

$$ u_{n-1}= (-1)^{n-1} \frac{(6n-6)!}{((n-1)!)^3(3n-3)!} \frac{1}{640320^{3n-3}} $$

Deduce the ratio, grouping together like-terms in fractions:

$$ \frac{u_n}{u_{n-1}}= -\frac{(6n)!}{(6n-6)!} \frac{((n-1)!)^3}{(n!)^3} \frac{(3n-3)!}{(3n)!} \frac{1}{640320^3} $$

Expand the three factorial fractions:

$$ \frac{(6n)!}{(6n-6)!}= (6n-5)(6n-4)(6n-3)(6n-2)(6n-1)(6n) \\ \frac{((n-1)!)^3}{(n!)^3}= \frac{1}{n^3}, \qquad \frac{(3n-3)!}{(3n)!}= \frac{1}{(3n-2)(3n-1)(3n)} $$

Put it all together:

$$ \frac{u_n}{u_{n-1}}= -\frac{ (6n-5)(6n-4)(6n-3)(6n-2)(6n-1)(6n) }{ n^3(3n-2)(3n-1)(3n)\,640320^3 } $$

Now, simplify with the identities:

$$ 6n-4=2(3n-2), \qquad 6n-3=3(2n-1), \qquad 6n-2=2(3n-1), \qquad 6n=2(3n) $$

After cancelling the common factors and tallying up the constant factor:

$$ \frac{u_n}{u_{n-1}}= -\frac{ 24(6n-5)(2n-1)(6n-1) }{ n^3\,640320^3 } $$

Equivalently,

$$ \frac{u_n}{u_{n-1}}= -\frac{ (6n-5)(2n-1)(6n-1) }{ n^3\left(640320^3/24\right) } $$

Define the Chudnovsky constant for convenience

$$ C= \frac{640320^3}{24} = 10939058860032000 $$

Then

$$ \frac{u_n}{u_{n-1}}= -\frac{ (6n-5)(2n-1)(6n-1) }{ n^3 C } $$

Now, we can choose:

$$ p_n= -(6n-5)(2n-1)(6n-1) $$

and

$$ q_n=n^3 C= n^3\frac{640320^3}{24} $$

Finally giving:

$$ \frac{u_n}{u_{n-1}}= \frac{p_n}{q_n} $$

Now we only need to turn this recurrence into a product. The zeroth hypergeometric term is 1, so we can directly deduce:

$$ u_n= \prod_{k=1}^{n} \frac{p_k}{q_k} $$

We have arrived at our desired form:

$$ \sum_{n=0}^{\infty} a_n \prod_{k=1}^{n} \frac{p_k}{q_k} $$

with

$$ a_n= 13591409+545140134n $$

$$ p_k= -(6k-5)(2k-1)(6k-1) $$

$$ q_k= k^3\frac{640320^3}{24} $$