Taylor series of $\ln(1+(1/x))$

Question

With $\ln(1+x)$ having a Taylor series of $x - (x^2)/2 + (x^3)/3 - (x^4)/4 +\dots$ I was wondering how I would modify it for $\ln(1+(1/x))$.

Thank you for the help.

Answer 1

From properties of logarithmic function: $$ \ln\left(1+\frac{1}{x}\right) = \ln\left(\frac{1 + x}{x}\right) = \ln\left(1 + x\right) - \ln\left(x\right) = \ln\left(1 + x\right) - \ln\left(1 + (x - 1)\right) $$

Answer 2

In order to use the expansion at $x=0$ we need to take $x>0$ and therefore

$$\ln\left(1+\frac{1}{x}\right) = \ln\left(\frac{1}{x}\right)+\ln\left(1+x\right) = \ln\left(\frac{1}{x}\right)+x-\frac12 x^2+\frac13x^3+\ldots = $$