Let $\sum^\infty_{n=0}a_nz^n$ have radius of convergence $r > 0$. Assume that the function $f(z)$ to which it converges has exactly one singular point $z_0$ on $\{z \in \mathbb{C} \vert |z| = r\}$, and that $z_0$ is a simple pole. Show that $\lim_{n\to\infty}(a_n/a_{n+1})$ exists and equals $z_0$.
It looks like ratio test, but I don't know how to go one. Please help.
A few details missing here but this is the general idea:
Note that we can write $f(z) = {g(z) \over z-z_0}$, where $g$ is entire and $g(z_0) \neq 0$. Let $g(z) = \sum_{k=0} g_k z^k$.
For $|z| < |z_0|$ we have ${ 1\over z-z_0} = - {1 \over z_0} {1 \over 1-{z \over z_0}}= - {1 \over z_0} (1+({z \over z_0}) + ({z \over z_0})^2+\cdots)$.
From the multiplication (cf. convolution) we can equate coefficients to get (with $f(z) = \sum_{k=0} f_k z^k$) $f_k = - {1 \over z_0} {1 \over z_0^k} (g_kz_0^k+ \cdots + g_0)$.
Note that $g_kz_0^k+ \cdots + g_0 \to g(z_0) \neq 0$.
Hence ${f_k \over f_{k+1}} = z_0 { g_kz_0^k+ \cdots + g_0 \over g_{k+1}z_0^{k+1}+ + g_kz_0^k+ \cdots + g_0 }$, and ${f_k \over f_{k+1}} \to z_0 $.
