If $x>1$ show that $\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}-\frac{1}{4x^4}+\cdots$
I know from binomial expansion that $(1+x)$ will produce a divergent series in the form of $1-x+x^2-x^3+\cdots$ but I don't know how to apply that in this situation.
Do I just need to integrate $\ln(1-x+x^2+\cdots$)? If so, why integrate? What role does integration have here?
On
Let
$$\begin{align} f(x)&=\ln(1+x)-\ln x\\ &=\ln\left(1+\frac{1}{x}\right) \end{align} $$
By differenting w.r.t. $x$ and writing as sum of infinite GP $$\begin{align}f'(x)&=\frac{1}{1+\frac1x}\cdot\left(\frac{-1}{x^2}\right)\\ &=-\frac{1}{x^2}\left(1-\frac1x+\frac{1}{x^2}-\cdots\right)\\ f'(x)&=-\frac{1}{x^2}+\frac{1}{x^3}-\frac{1}{x^4}+\cdots\\ \end{align}$$ By integrating $$\begin{align} f(x)&=\frac{1}{x}-\frac{1}{2x^2}+\cdots\\ \ln(1+x)-\ln x&=\frac{1}{x}-\frac{1}{2x^2}+\cdots\\ \end{align}$$
$$\ln(1+x)=\ln x+\frac{1}{x}-\frac{1}{2x^2}+\cdots$$
Hint: Expand $\ln(1+x)-\ln(x)=\ln(1+u)$ into a power series in $u=1/x$ with $|u|\lt1$.