Series Expansion for $\ln(x)$ I have a mathematics assignment, which requires me to proof that $$\ln\frac{2}{3} = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{2^{n}n}$$.
I know, I can solve this by proving $\ln x$ = $\sum_{n=1}^{\infty }\frac{1}{n}\left ( \frac{x-1}{x} \right )^{n}$, but I don't know how to prove this, so can anybody offer some help?

Using the definition of geometric series 
$$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^{k} $$
Integrating both sides 
$$\int\frac{1}{1-x}=\sum_{k=0}^{\infty}\frac{(-1)^k x^{k+1}}{k+1} $$
So $$\ln (1-x)=-\sum_{k=1}^{\infty} \frac { x^k}{k} $$
$$\ln (y)= \sum_{k=1}^{\infty} \frac{1}{k} \left( \frac{y-1}{y} \right) ^{k}  $$
