function with isolated singularity on the unit circle and coefficients of its taylor expansion 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?
 A: For convenience, take $z_0 = 1$.
Let $\Gamma$ be the positively oriented circle $|z|=1+r$, where $r > 0$ and $f$ is analytic in
 $\{z: |z| \le 1+s\} \backslash \{1\}$ for some $s > r$.  Let
$$ c_n = \frac{1}{2\pi i} \oint_\Gamma \frac{f(z)}{z^{n+1}}\ dz,\ b_n = \text{Res}(f(z)/z^{n+1}; z=1)$$
By the residue theorem, $a_n = c_n - b_n$.  Now $|c_n| < M (1+r)^{-n-1}$ for some constant $M$.
Suppose the Laurent series of $f(z)$ in $\{0 < |z - 1| < r\}$ is
$\sum_{k=-\infty}^\infty d_k (z - 1)^k$.  Since $z^{-n-1} = \sum_{j=0}^\infty {{n+j} \choose n} (1-z)^j$, we get
$$ b_n = \sum_{j=0}^\infty {{n+j} \choose n} (-1)^j d_{-j-1}$$
If $1$ is a pole of order $m$, this is a finite sum and the dominant term  is for $j=m-1$, namely
$\displaystyle {{n+m-1} \choose n} (-1)^{m-1} d_{-m} = \frac{n^{m-1}}{(m-1)!} d_{-m} + O(n^{m-2})$
and we get
$$\frac{a_n}{a_{n+1}} = \frac{n^{m-1} d_{-m} + O(n^{m-2})}{(n+1)^{m-1} d_{-m} + O(n^{m-2})} = 1 + O(1/n)$$
But it's not at all clear to me what will happen in the case of an essential singularity.
It seems entirely possible, for example, that some cancellations will make infinitely many $a_n = 0$.
EDIT: Consider $f(z) = \exp(1/(z-1))$.  This has its only singularity at $z=1$. 
I claim that there are  infinitely many positive $a_n$ and infinitely many negative $a_n$, which implies that $a_n/(a_{n+1}) \le 0$ infinitely often.
Since 
each derivative  $f^{(n)}(z)$ is a rational function of $z$ (whose only pole is at $1$) times $f(z)$, we have $f^{(n)}(z) \to 0$ as $z \to 1-$.  But if only finitely many coefficients $a_j$ were positive or only finitely many were negative, all the coefficients of $f^{(n)}(z)$ would have the same sign for sufficiently large $n$, and then
$\lim_{z \to 1-} f^{(n)}(z)$ could not be $0$.
