Zero Error Small Piece Of Pi

 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.