Suppose $f(z) = \frac{\operatorname{Log} z}{z-1}$ when $z \neq 1$ and $f(1) = 1$. Using the fact that $\operatorname{Log} z = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} (z-1)^n$ show that $f(z)$ is analytic.
What I've got: Since $\operatorname{Log} z$ can be represented as a power series, it's analytic (well, for $|z-1| < 1$). $\frac{1}{z-1}$ is analytic everywhere except at $z = 1$. I guess what's throwing me off is the $f(1) = 1$. I want to look at the limit of $f(z)$ as it approaches $1$, but that doesn't seem to be helping me, as it isn't 1.