I am studying complex variable by myself and I want to prove that:
$F_2(z)=\cfrac{1}{4}\pi i-\cfrac{1}{2}ln2+\left(\cfrac{z-i}{1-i}\right)+\cfrac{1}{2}\left(\cfrac{z-i}{1-i}\right)^2+\cfrac{1}{3}\left(\cfrac{z-i}{1-i}\right)^3+\ldots$ converges if $|z-1|<\sqrt{2}$.
I have proved that $F_1(z)=z+\cfrac{1}{2}z^2+\cfrac{1}{3}z^3+\ldots$ is a convergent series and since $F_2(z)$ is an analytic prolongation of $F_1(z)$ my intuition says that $F_2(z)$ is convergent too. But I don't know how to prove it.
We can ignore the first two terms and use the ratio test here $$\lim_{n\rightarrow \infty} \left|\frac{(z-i)^{n+1}}{(1-i)^{n+1}}\cdot \frac{(1-i)^{n}}{(z-i)^n}\right| = \left|\frac{(z-i)}{(1-i)}\right|.$$ In order for this series to converge absolutely we want $$\left|\frac{(z-i)}{(1-i)}\right| < 1.$$ That pretty much give what I think your looking for.