Using Taylor series to show a function is analytic

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.

-
You know the quotient is going to be analytic everywhere in the disk you specified except possibly at at $z=1$. Divide the power series by $(z-1)$ and show the singularity there is removable. Then plug in 1 to find the value. –  Potato Jul 5 '12 at 3:18

Well, using what you're given: $$\frac{\operatorname{Log} z}{z-1}=\frac{1}{z-1}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}(z-1)^n=\sum_{n=0}^\infty\frac{(-1)^{n}}{n+1}(z-1)^n$$ so our function has a power series around $\,z=1\,$ and is thus analytic in some neighbourhood of this point. For the rest of the points in the complex plane we have no problem, of course, as the function is a quotient of two analytic functions whose denominator never vanishes.
Please do note that we don't even have a problem about the multivaluation of $\,\operatorname{Log}z\,$ since you've defined it as a power series...
It should be $(-1)^n/(n+1)$ in front of $(z-1)^n$ if you start from $n=0$. –  Mercy Jul 6 '12 at 18:48