Prove that $lim_{n\rightarrow \infty} a_n/a_{n+1} = z_0.$ I am studying for a complex qual and I am having a little difficulty with the following problem.  Any help that you could give me would be awesome.  Thank you.

Let $f(z) = \sum_{n=0}^\infty a_nz^n$ be the Taylor series around $0$ of a function which is analytic in $\mathbb{C}$ \ ${z_0},$ $z_0\neq 0$ and has only a simple pole at $z_0$. Prove that
$lim_{n\rightarrow \infty} a_n/a_{n+1} = z_0.$
I decided to add another answer rather than delete or change the last since the first answer is nice ill leave it, but here is a complete solution, independent of previous considerations.
$(z-z_0)f(z)$ is an entire function with power expansion 
$$\sum (a_n -a_{n+1}z_0)z^{n+1}$$ now as this series converges for all $z$, and the original series has a finite radius of convergence we must have 
$$\lim |\frac{a_n -a_{n+1}z_0}{a_{n+1}}|=0$$

And this is exactly what was to be proved.
This is only a partial answer, but a power series converges on the largest radius where the function is defined. The series cannot converge at this point $z_0$. Therefore the radius of convergence is $|z_0|$ now by the ratio test this gives, 

$$\lim\limits_{n\to \infty} \frac{|a_n|}{|a_{n+1}|}=|z_0|$$

Of course I have not used that $z_0$ is a simple pole so somehow that must be involved in proving the better limit.