The fact is,
\(\theta_0=\cfrac{\pi}{4}\)
\(S_{\theta_i}=\cfrac{\pi}{2^2}+\cfrac{\pi}{2^3}+...+\cfrac{\pi}{2^{i+1}}+\cfrac{\theta_0}{2^{i}}=\cfrac{\pi}{2^2}+\cfrac{\pi}{2^3}+...+\cfrac{\pi}{2^{i+2}}\)
\(S^{i\rightarrow\infty}_{\theta_i}=\cfrac{\pi}{2^2}(1+\cfrac{1}{2}+\cfrac{1}{2^2}+...+\cfrac{1}{2^{i}}...)\)
\(S^{i\rightarrow\infty}_{\theta_i}=\cfrac{\pi}{2^2}(2)\)
\(S^{i\rightarrow\infty}_{\theta_i}=\cfrac{\pi}{2}\)
So, the last correction is,
\(\epsilon_{i\rightarrow\infty}=M_{i\rightarrow\infty}(\cfrac{1}{sin\theta_{i\rightarrow\infty}}-1)^2tan\theta_{i\rightarrow\infty}\)
\(\epsilon_{i\rightarrow\infty}=M_{i\rightarrow\infty}(\cfrac{1}{sin(\frac{\pi}{2})}-1)^2tan(\frac{\pi}{2})=0\)
no matter what \(M_{i\rightarrow\infty}\) and \(tan(\frac{\pi}{2})\) are. The smallest piece of \(\pi\) corrected is zero.
