Let $f$ be a function holomorphic in an open set containing the closed unit disc $D(0,1)$, except at the point $z_0$ with $|z_0|=1$, where $f$ has an isolated singularity. If $a_n$ are the coefficients of the Taylor expansion of $f$ centered at the origin, show that
$\displaystyle\lim_{n \to\infty}\frac{a_n}{a_{n+1}}=z_0$
I considered first the case: $z_0$=simple pole. Then i can write the Laurent series for $f$ at $z_0$
$f(z)=\frac{c}{z-z_0}+c_0+c_1(z-z_0)+\ldots$ I put
$g(z)=f(z)-\frac{c}{z-z_0}$
Then $g$ is holomorphic in an open set containing the closed unit disc $\overline{D(0,1)}$ and we have $g(z)=\sum a_n z^n+\frac{c}{z_0}\sum(\frac{z}{z_0})^n=\sum(a_n+\frac{c}{z_0^{n+1}})z^n$ The power series of $g$ has radius of convergence $>1$, hence i can substitute 1 in RHS of last equation, and i get a numeric convergent series. The necessary condition for convergence says that its general term has to be infinitesimal $\displaystyle\lim_{n\to\infty}(a_n+\frac{c}{z_0^{n+1}})=0$ from which follows $\frac{a_n}{a_{n+1}}\rightarrow z_0$
the question is:how can i adapt this argument to deal with cases where z_0 is not a simple pole?