- Suppose that f is holomorphic in an open set containing the closed unit disc, except for a pole at \(z _ { \mathrm { 0 } }\) on the unit circle. Show that if
\[ \sum_ {n = 0} ^ {\infty} a _ {n} z ^ {n} \]
denotes the power series expansion of \(f\) in the open unit disc, then
\[ \lim _ {n \to \infty} \frac {a _ {n}}{a _ {n + 1}} = z _ {0}. \]
Solution
Let the pole of \(f\) at \(z_0\) have order \(m\ge1\). Since \(|z_0|=1\), write the principal part at \(z_0\) as \[ \sum_{k=1}^m \frac{c_{-k}}{(z-z_0)^k}, \qquad c_{-m}\ne0. \] Subtracting this principal part from \(f\) removes the only singularity on the closed unit disc. Thus \[ h(z)=f(z)-\sum_{k=1}^m \frac{c_{-k}}{(z-z_0)^k} \] is holomorphic on a neighborhood of the closed unit disc. Hence there is some \(\rho>1\) such that \(h\) is holomorphic on \(|z|<\rho\), and if \[ h(z)=\sum_{n\ge0}b_nz^n, \] then Cauchy’s estimates give \(b_n=O(\rho^{-n})\).
For \(|z|<1\), \[ \frac1{(z-z_0)^k} =(-1)^kz_0^{-k}\left(1-\frac z{z_0}\right)^{-k} =(-1)^kz_0^{-k}\sum_{n\ge0}\binom{n+k-1}{k-1}z^nz_0^{-n}. \] Therefore \[ a_n=z_0^{-n}P(n)+b_n, \tag{1} \] where \[ P(n)=\sum_{k=1}^m(-1)^kc_{-k}z_0^{-k}\binom{n+k-1}{k-1} \] is a polynomial in \(n\) of degree \(m-1\), with nonzero leading coefficient because \(c_{-m}\ne0\).
Thus \(P(n)\sim Cn^{m-1}\) for some \(C\ne0\), while \(b_n=O(\rho^{-n})\). From (1), \[ \frac{a_n}{a_{n+1}} =z_0\, \frac{P(n)+z_0^n b_n}{P(n+1)+z_0^{n+1}b_{n+1}}. \] Since \(|z_0|=1\), the error terms are exponentially small compared with \(P(n)\), and \[ \frac{P(n)}{P(n+1)}\longrightarrow1. \] Hence \[ \boxed{\displaystyle \lim_{n\to\infty}\frac{a_n}{a_{n+1}}=z_0}. \]