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

Question

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? $$\sum_{k=0}^\infty x^k$$

Answer 1

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.

Answer 2

The problem with your approach would be the radius of convergence. So converting it to $-\ln(1-x)$ is the way to go.

Answer 3

The geometric series

$$\sum_{k=0}^{\infty} z^k = \frac{1}{1-z}$$

converges uniformly for $0\leq|z| \leq x < 1$ and can be integrated termwise.

Hence,

$$\ln \frac1{1-x}=-\ln(1-x) = \int_{0}^{x}\frac{dz}{1-z}=\sum_{k=0}^{\infty} \int_{0}^{x}z^kdz =\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1} = \sum_{k=1}^{\infty} \frac{x^{k}}{k}.$$

Make a similar argument for $-1 < x \leq 0.$