why we do tailor around 1 and not 0
?
Your expression is equal to $\displaystyle \ln(z)$. So, $\displaystyle \lim_{z\to1}\frac{\ln(z)}{z-1}=\lim_{x\to0}\frac{\ln(1-x)}{x}=1$. Another easier way is to write out the terms of your series and notice that $\displaystyle \left|(z-1)-\sum_{n=1}\frac{(-1)^n(z-1)^n}{n}\right|<\varepsilon$ by taking $\displaystyle |z-1|<\delta$ for a suitable $\displaystyle \delta$