# Puzzling property of Taylor series and simple poles

1. Dec 5, 2011

### Grothard

Let $f(z)$ be a function that is analytic for all $|z|≤1$, with the exception of $z_0$, which lies on the circle $|z|=1$. $f(z)$ has a first order pole at $z_0$. If $Ʃ a_n z^n$ is the Maclaurin expansion of the function, then $z_0 = lim_{n \to \infty} \frac{a_n}{a_{n+1}}$.

I have failed to show that this is true. I think the key might lie in the fact that there exist points in the radius of convergence of the Maclaurin series AND in the radius of convergence of the Laurent series about $z_0$, allowing us to set the two series equal to each other and fix a $z$ value. I can't manage to separate the $z_0$ term in that equation, though.
Any thoughts? Is this a well-known property?

Last edited: Dec 5, 2011