Manipulating Taylor series with $\operatorname{Log} z$

Question

I'm trying to prove the following:

$$ \operatorname{Log}z = \sum_{n=1}^{\infty} \frac{\left(1-\frac1z\right)^n}{n} $$

for $\left|1-\frac1z\right|<1$.

Does anyone have advice on where to begin? I'm trying to manipulate the series for $1/z$ but can't seem to get around using $1-1/z$. Any ideas?

Answer 1

Substitue $w=1-1/z$. You need to show

$$\log\frac{1}{1-w}=\sum_{n=1}^\infty\frac{w^n}{n}$$

for $|w|<1$.