Maclaurin Series of $\ln(2-e^{-x}) $

Question

I tried to solve this by using the series for $e^{-x}$ and $\ln(1+u)$ $$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+...\\ \ln(2-e^{-x})=\ln[2-(1-x+\frac{x^2}{2}-\frac{x^3}{6})]=\ln(1+x-\frac{x^2}{2}+\frac{x^3}{6})\\ \ln(1+u)=u-\frac{u^2}{2}+\frac{u^3}{3}-\frac{u^4}{4}+...;\mbox{using }u=(x-\frac{x^2}{2}+\frac{x^3}{6})\\ \ln(2-e^{-x})=(x-\frac{x^2}{2}+\frac{x^3}{6})-\frac{(x-\frac{x^2}{2}+\frac{x^3}{6})^2}{2}+\frac{(x-\frac{x^2}{2}+\frac{x^3}{6})^3}{3}-\frac{(x-\frac{x^2}{2}+\frac{x^3}{6})^4}{4}$$ Unfortunately after adding up the terms this is not equal to the series for $\ln{(2-e^{-x})}$

Answer 1

Just as Yves Daoust commented, what you obtained is correct up to the third order.

If you use an extra term $$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^5\right)$$ and make (just as you did) $$u=x-\frac{x^2}{2}+\frac{x^3}{6}-\frac{x^4}{24}$$ you should get, using the expansion you wrote up to $u^4$, $$\log(2-e^{-x})=x-x^2+x^3-\frac{13 x^4}{12}+\frac{5 x^5}{4}+\cdots$$ which is correct up to the fifth order.

If you use the expansions up to the sixth order for both $e^{-x}$ and $\log(1+u)$, you should arrive to $$\log(2-e^{-x})=x-x^2+x^3-\frac{13 x^4}{12}+\frac{5 x^5}{4}-\frac{541 x^6}{360}+\cdots$$ which is correct up to the sixth order.

This is always one of the problems when composing series from series.

Answer 2

an alternative approach would be to use differentiation: $$y=\ln(2-e^{-x})$$ $$\rightarrow e^y=2-e^{-x}$$ $$\rightarrow e^yy'=e^{-x}$$ $$\rightarrow y'=e^{-x-y}$$ $$\rightarrow y''=(-1-y')e^{-x-y}$$ $$\rightarrow y''=-(y'+(y')^2)$$ $$\rightarrow y'''=-(y''+2y'y'')$$

then you can keep differentiating and evaluating the derivatives to compile the series