Error in the Chudnovsky Formula

A digit is gained 'linearly' if the summand behaves as $10^{-k}.$ (In many contexts this would be considered exponential convergence.) As I understand the question, we want to show the summand behaves as $(10^{14})^{-k}.$ The constant factors and (-1)^k are immaterial, so we write, (with $a$ and $b$ easily pulled from above) $$ S:=\frac{(6k)!(a\,k+b)}{(3k)!k!^3} (640320)^{-3k} = \binom{6k}{3k}\binom{2k}{k}\binom{3k}{k}(ak+b) (640320)^{-3k}$$ where binomials have been inserted to make the asymptotic expansion a little simpler. Use $$ \binom{2k}{k} \sim \frac{2^{2k}}{\sqrt{\pi\,k}} \quad \text{ and } \quad \binom{3k}{k} \sim \frac{\sqrt{3}}{2} \,\frac{3^{3k}\,2^{-2k}}{\sqrt{\pi\,k}}$$ which are derivable by Stirling's formula. The $\binom{6k}{3k}$ can use the first asymptotic formula. Thus we have $$ S \sim \bigg\{ \frac{a+b/k}{2\pi\,\sqrt{\pi\,k}} \bigg\} \,\bigg(\frac{12}{640320}\bigg)^{3k}$$ The expression in curly brackets can be ignored as ratio of polynomials won't get you 'linear' convergence. Then it's a simple manner of seeing that $$ (53360)^{-3k} \sim (1.52\,\text{x}\,10^{14})^{ -k} .$$

While Stirling's formula yields an approximation of the error, Binet's Formula yields an error bound for the Chudnovsky Formula:

First we observe that the series is an alternating series whose absolute values are strictly decreasing to zero. Thus, the Alternating series test tells that the error you produce by summing only until $k=N$ is smaller than the next term's absolute value: $$\left|\frac{1}{\pi}-\frac{1}{\pi_N}\right| = \left|12\sum_{k=N+1}^\infty s_k\right| < 12\left| s_{N+1} \right|.$$

But since it holds $$\frac{1}{\pi}-\frac{1}{\pi_N}=\frac{\pi_N-\pi}{\pi~\pi_N},$$ we have proven $$\left|\pi_N-\pi\right| < \left|12\pi~\pi_N~s_{N+1} \right|.$$

Second we use Binet's Formula for $n!$, which says $$\exp\left(\frac{1}{12n}-\frac{1}{360n^3}\right) < \frac{n!}{\sqrt{2\pi n}\cdot(n/e)^n} < \exp\left(\frac{1}{12n}\right)$$ This yields $$\frac{(6k)!}{(3k)!(k!)^3} < \frac{1728^k}{2(\pi k)^{3/2}}.$$ Using $545140134k+13591409\leq 545140134k+13591409k=558731543k$ yields $$|s_k|<\frac{1728^k}{2(\pi k)^{3/2}}\cdot\frac{558731543k}{640320^{3k+3/2}}<\frac{1}{10~\sqrt{k}~53360^{3k}}$$ and thus the following error bound for the Chudnovsky Formula (where $N\geq 0$): $$\left|\pi_N-\pi\right| < \left|12\pi~\pi_N~s_{N+1} \right|<\frac{12}{\sqrt{N+1}~53360^{3(N+1)}}$$

Remark: For $N\geq 128$, using $13591409\leq 13591409k/128$, we obtain with the same method: $$\left|\pi_N-\pi\right| < \frac{11.317}{\sqrt{N+1}~53360^{3(N+1)}} < 53360^{-3(N+1)}$$ Thus every term of the summation reduces the error by a factor of $53360^{-3}=10^{-14.1816}$.