Solution: Given: \(f(x)=\ln(1-x)\)

About b=0: the maclaurins series is:

\(f(x)=\ln(1-x)=(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...\infty)=(-1)\sum_{n=1}^\infty\frac{x^n}{n}\)

Over [\(-\frac{1}{2},\frac{1}{2}\)] by the Taylor's inequality:

\(|R_n|\leq|\frac{f^{(n+1)}e}{(n+1)!}\cdot|x-b|^{n+1}\)

Now: \(|x-b|=\frac{1}{2}\) and \(f'(x)=\frac{1}{1-x}\)

So: \(f^{n+1}(x)\) over \([\frac{-1}{2},\frac{1}{2}]\) is maximised

\(\Rightarrow(n+1)>+\infty\Rightarrow n>99\)