[Math] Taylor expansion at $x=0$ of $\ln(1/(1-x))$

calculustaylor expansion

Hello I am having some trouble with the taylor expansion of
$$f(x)= \ln \frac1{1-x}$$

Would it be correct to treat the inner part as the following geometric series? $$\sum_{k=0}^\infty x^k$$

Best Answer

First I would write $$f(x) = \ln\frac{1}{1-x} = - \ln(1-x)$$ now differentiate to find: $$f'(x) = \frac{1}{1-x}$$ Now use the geometric series to expand the derivative.

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