logarithms - Show that $ln(1+x)=ln x+frac{1}{x}-frac{1}{2x^2}+frac{1}{3x^3}-frac{1}{4x^4}+cdots$ when $x>1$

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?

Answer

Hint: Expand $\ln(1+x)-\ln(x)=\ln(1+u)$ into a power series in $u=1/x$ with $|u|\lt1$.