= Chudnovsky series
{c}
{title2=$\pi^{-1}=\frac{12}{640320^{3/2}}\sum_{k\geq0}\frac{(-1)^k(6k)!(13591409+545140134k)}{(3k)!(k!)^3\,640320^{3k}}$}
The <Chudnovsky series> is the rapidly converging identity
$$
\frac1\pi=\frac{12}{640320^{3/2}}\sum_{k=0}^{\infty}\frac{(-1)^k(6k)!(13591409+545140134k)}{(3k)!(k!)^3\,640320^{3k}}.
$$
The absolute ratio of successive summands tends to $1728/640320^3$, approximately $6.58\times10^{-15}$, giving about fourteen new decimal digits per term asymptotically. Keeping only $k=0$ gives $\pi\approx640320^{3/2}/(12\cdot13591409)$. The factorial ratio, not a direct geometric approximation of a <circle>, controls the numerical <convergence>.<Chudnovsky’s formula for 1 π revisited by Yue Zhao>[§1, equation (1.4), printed p. 1]{ref}
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