Binary splitting

A factor of ${\textstyle 1/{640320^{3/2}}}$ can be taken out of the sum and simplified to

$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}\sum _{k=0}^{\infty }{\frac {(-1)^{k}(6k)!(545140134k+13591409)}{(3k)!(k!)^{3}(640320)^{3k}}}}$$

Let ${\displaystyle f(n)={\frac {(-1)^{n}(6n)!}{(3n)!(n!)^{3}(640320)^{3n}}}}$, and substitute that into the sum.

$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}\sum _{k=0}^{\infty }{f(k)\cdot (545140134k+13591409)}}$$

${\displaystyle {\frac {f(n)}{f(n-1)}}}$ can be simplified to ${\displaystyle {\frac {-(6n-1)(2n-1)(6n-5)}{10939058860032000n^{3}}}}$, so

$${\displaystyle f(n)=f(n-1)\cdot {\frac {-(6n-1)(2n-1)(6n-5)}{10939058860032000n^{3}}}}$$

${\displaystyle f(0)=1}$ from the original definition of ${\displaystyle f}$, so

$${\displaystyle f(n)=\prod _{j=1}^{n}{\frac {-(6j-1)(2j-1)(6j-5)}{10939058860032000j^{3}}}}$$

This definition of ${\displaystyle f}$ isn't defined for ${\displaystyle n=0}$, so compute the first term of the sum and use the new definition of ${\displaystyle f}$

$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}{\Bigg (}13591409+\sum _{k=1}^{\infty }{{\Bigg (}\prod _{j=1}^{k}{\frac {-(6j-1)(2j-1)(6j-5)}{10939058860032000j^{3}}}{\Bigg )}\cdot (545140134k+13591409)}{\Bigg )}}$$

Let ${\displaystyle P(a,b)=\prod _{j=a}^{b-1}{-(6j-1)(2j-1)(6j-5)}}$ and ${\displaystyle Q(a,b)=\prod _{j=a}^{b-1}{10939058860032000j^{3}}}$, so

$${\displaystyle {\frac {1}{\pi }}={\frac {1}{426880{\sqrt {10005}}}}{\Bigg (}13591409+\sum _{k=1}^{\infty }{{\frac {P(1,k+1)}{Q(1,k+1)}}\cdot (545140134k+13591409)}{\Bigg )}}$$

Let ${\displaystyle S(a,b)=\sum _{k=a}^{b-1}{{\frac {P(a,k+1)}{Q(a,k+1)}}\cdot (545140134k+13591409)}}$ and ${\displaystyle R(a,b)=Q(a,b)\cdot S(a,b)}$

$${\displaystyle \pi ={\frac {426880{\sqrt {10005}}}{13591409+S(1,\infty )}}}$$

${\displaystyle S(1,\infty )}$ can never be computed, so instead compute ${\displaystyle S(1,n)}$ and as ${\displaystyle n}$ approaches ${\displaystyle \infty }$, the ${\displaystyle \pi }$ approximation will get better.

$${\displaystyle \pi \approx {\frac {426880{\sqrt {10005}}}{13591409+S(1,n)}}}$$

From the original definition of ${\displaystyle R}$, ${\displaystyle S(a,b)={\frac {R(a,b)}{Q(a,b)}}}$

$${\displaystyle \pi \approx {\frac {426880{\sqrt {10005}}\cdot Q(1,n)}{13591409Q(1,n)+R(1,n)}}}$$

import decimal


def binary_split(a, b):
    if b == a + 1:
        Pab = -(6*a - 5)*(2*a - 1)*(6*a - 1)
        Qab = 10939058860032000 * a**3
        Rab = Pab * (545140134*a + 13591409)
    else:
        m = (a + b) // 2
        Pam, Qam, Ram = binary_split(a, m)
        Pmb, Qmb, Rmb = binary_split(m, b)
        
        Pab = Pam * Pmb
        Qab = Qam * Qmb
        Rab = Qmb * Ram + Pam * Rmb
    return Pab, Qab, Rab


def chudnovsky(n):
    P1n, Q1n, R1n = binary_split(1, n)
    return (426880 * decimal.Decimal(10005).sqrt() * Q1n) / (13591409*Q1n + R1n)


print(chudnovsky(2))  # 3.141592653589793238462643384

decimal.getcontext().prec = 100
for n in range(2,10):
    print(f"{n=} {chudnovsky(n)}")  # 3.14159265358979323846264338...