Prove that the function can be continued into a larger domain Prove that the function  $f(z)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{z^n}{n}$ can be continued into a larger domain by means of the series $$\ln2-\frac{1-z}{2}-\frac{(1-z)^2}{2\cdot 2^2}-\frac{(1-z)^3}{3\cdot2^3}-\dots$$

Attempt

First notice that $f(z)$ is similar to $\ln z=\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(z-1)^{n+1}}{n+1}$

Thus I think we can say that $f(z)=\ln(z-1)$

I do not know what can I do from here

I don't understand 

Where does $\ln 2$ come from?

How is f(z) and the series $\ln2-\frac{1-z}{2}-\frac{(1-z)^2}{2\cdot 2^2}-\frac{(1-z)^3}{3\cdot2^3}-\dots$ related?

Any help would be greatly appreciated.
The right formula of $f(z)$ is $\ln(1+z)$, with the domain $(-1,1]$.

The series converge only in $(-1,1]$, but the function $\ln(1+z)$ exist in $(-1,+\infty)$.

The series in your question is just the Taylor series of $f(z)$ at the point $z=1$

$$\ln(1+z)=\ln(2+(z-1))=\cdots$$
The right formula of $f(z)$ is $\ln(1+z)$, with the domain $(-1,1]$.

The series converge only in $(-1,1]$, but the function $\ln(1+z)$ exist in $(-1,+\infty)$.

The series in your question is just the Taylor series of $f(z)$ at the point $z=1$

$$\ln(1+z)=\ln(2+(z-1))=\cdots$$