I need need to calculate $log_{10} 11$ with an accuracy of at least $0.0001$ with Maclaurin Series.
I thought it was not so difficult, but I can not move forward in reasoning.
I forgot to clarify what should be presented in the form of a sum of rational fractions
Hint
Write $$\log_{10}(11)=\log_{10}(10)+\log_{10}\left(1+\frac 1 {10}\right)=1+\log_{10}\left(1+\frac 1 {10}\right)=1+\frac{\log_e\left(1+\frac 1 {10}\right)}{\log_e(10)}$$ Now, use Taylor series in particular the very fast converging $$ \log \left(\frac{1+x}{1-x}\right)=2\left(\frac{x}1+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\cdots\right)$$ Setting $$\left(\frac{1+x}{1-x}\right)=\frac{11}{10}\implies x=\frac 1{21}$$