Suppose that \(f\) is holomorphic on an open set containing the closed unit disk \(\overline{\mathbb{D}} = \{z \in \mathbb{C} \mid |z| \le 1\}\) except for a single pole at \(z_0 \in \partial\mathbb{D}\) (\(|z_0| = 1\)). Let \(\sum_{n=0}^\infty a_n z^n\) be the Maclaurin series expansion of \(f\) in \(\mathbb{D}\). Prove that: \[\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = z_0.\]
Solution
Let the pole at \(z_0\) have order \(m\). Its principal part can be written \[ Q(z)=\sum_{j=1}^m\frac{c_j}{(z_0-z)^j}, \qquad c_m\ne0. \] After subtracting \(Q\), the function \(g=f-Q\) is holomorphic on a neighborhood of the closed unit disk. Hence there is some \(R>1\) such that \(g\) is holomorphic on \(D_R(0)\).
1. For \(|z|<1\),\[ \frac1{(z_0-z)^j} =\sum_{n=0}^\infty \binom{n+j-1}{j-1}z_0^{-(n+j)}z^n. \]Thus, if \(q_n\) denotes the \(n\)th Taylor coefficient of \(Q\),\[ q_n=\sum_{j=1}^m c_j \binom{n+j-1}{j-1}z_0^{-(n+j)}. \]The term \(j=m\) has highest polynomial degree in \(n\), so\[ q_n= \frac{c_m}{(m-1)!}n^{m-1}z_0^{-(n+m)} \left(1+O(n^{-1})\right). \]
2. Write \(g(z)=\sum b_nz^n\). Cauchy’s estimates on any circle of radius \(r\) with \(1<r<R\) give \(b_n=O(r^{-n})\). Hence the Taylor coefficients \(a_n=q_n+b_n\) of \(f\) satisfy\[ a_n= \frac{c_m}{(m-1)!}n^{m-1}z_0^{-(n+m)} \left(1+O(n^{-1})\right), \]because the exponentially decaying \(b_n\) is negligible compared with the nonzero polynomial-size leading term.
Therefore, for all sufficiently large \(n\), \(a_{n+1}\ne0\), and \[ \frac{a_n}{a_{n+1}} =z_0\left(\frac{n}{n+1}\right)^{m-1} \frac{1+O(n^{-1})}{1+O((n+1)^{-1})} \longrightarrow z_0. \]