Consider the series $f_0(z)=\sum _{k=1}^{\mathrm{\infty }}\frac{(-1{)}^{k-1}(z-1{)}^k}{k}$. This series converges to the natural log on the disk $B_1(1)$. Now consider it's derivative

This series is defined everywhere on the punctured plane. Therefore we can analytically continue $\frac{1}{z}$ in 1;; the punctured plane, as long as the disks don't touch the origin. But this means that we can also analytically continue the logarithm in 1;; the punctured plane, along this disk chain.