Geometric power-series remainder ($\frac{1}{1-z}-{\sum }_{n=0}^N{z}^n=\frac{z^{N+1}}{1-z}$)

For the degree-$N$ partial sum $P_N(z)=1+z+\cdots +z^N$ of the geometric series, the finite identity $(1-z)P_N(z)=1-z^{N+1}$ givesFor any fixed $0<r<1$, the triangle inequality then gives the uniform error boundEquality is attained at the positive real point $z=r$, so this is the exact worst-case error on that closed disk. It also proves uniform convergence on each such disk, not just pointwise convergence.

Given a positive tolerance $\epsilon$, choosing $N$ with $r^{N+1}/(1-r)\le \epsilon$ turns a convergence statement into a finite stopping rule. For $\epsilon =10^{-3}$, degree ten suffices throughout $|z|\le 1/2$. Nearer the boundary, $r=0.9$ requires degree $87$ and $r=0.99$ requires degree $1145$. The required degree depends on both the tolerance and the region, not just on the name of the function.