Suppose that \(f\) is holomorphic in an open set containing the closed unit disc, except for a pole at \(z_0\) on the unit circle. Let \(\displaystyle f(z) = \sum_{n = 1}^\infty c_n z^n\) denote the the power series in the open disc. Show that
\(c_n \neq 0\) for all large enough \(n\)’s, and
\(\displaystyle \lim_{n \rightarrow \infty} \frac{c_n}{c_{n+1}}= z_0\).
Solution
Goal: Let \(f\) be holomorphic in an open neighborhood \(U \supset \overline{\mathbb{D}}\) except for a single pole at \(z_0 \in \partial \mathbb{D}\) (\(|z_0| = 1\)). Let \(f(z) = \sum_{n=0}^\infty c_n z^n\) be its Taylor expansion in \(\mathbb{D}\). Prove that:
\(c_n \neq 0\) for all sufficiently large \(n\).
\(\lim_{n \to \infty} \frac{c_n}{c_{n+1}} = z_0\).
Step 1: Laurent Decomposition and Principal Part Expansion
1. Isolate the pole at \(z_0\) of order \(k \geq 1\).
1.1. Near \(z_0\), the Laurent principal part is: \[P(z) = \sum_{j=1}^k \frac{A_j}{(z - z_0)^j} = \frac{A_k}{(z - z_0)^k} + \frac{A_{k-1}}{(z - z_0)^{k-1}} + \dots + \frac{A_1}{z - z_0},\] where \(A_k \neq 0\) and \(k \geq 1\) is the order of the pole.
A pole of order \(k\) has a Laurent expansion whose principal part is a finite sum of negative powers from \(-k\) to \(-1\), with the coefficient \(A_k\) of the most negative power non-zero.
1.2. Define \(g(z) = f(z) - P(z)\).
Subtracting the principal part from \(f\) removes the singular terms, leaving a function \(g\) to be analyzed.
1.3. The singularity of \(g\) at \(z_0\) is removable, so \(g\) is holomorphic on all of \(U \supset \overline{\mathbb{D}}\).
By construction \(g\) has no negative powers in its Laurent expansion at \(z_0\), so its singularity is removable by Riemann’s removable singularity theorem; hence \(g\) extends holomorphically across \(z_0\) to all of \(U\).
1.4. Thus \(g(z) = \sum_{n=0}^\infty d_n z^n\) is a power series with radius of convergence \(R_g > 1\).In particular, there exists \(\rho > 1\) such that \(|d_n| = O(\rho^{-n})\) as \(n \to \infty\).
Since \(g\) is holomorphic on \(U \supset \overline{\mathbb{D}}\), its Taylor series at \(0\) converges on a disk of radius \(R_g > 1\); the Cauchy–Hadamard formula then gives \(|d_n| \le \rho^{-n}\) for some \(\rho\) with \(1 < \rho < R_g\).
1.5. Q.E.D.
2. Expand the principal part \(P(z)\) inside \(\mathbb{D}\).
2.1. For \(|z| < |z_0| = 1\), using the binomial series for \((1 - z/z_0)^{-j}\): \[\frac{1}{(z - z_0)^j} = \frac{(-1)^j}{z_0^j \left(1 - \frac{z}{z_0}\right)^j} = \frac{(-1)^j}{z_0^j} \sum_{n=0}^\infty \binom{n+j-1}{j-1} \left(\frac{z}{z_0}\right)^n = \frac{(-1)^j}{z_0^j} \sum_{n=0}^\infty \binom{n+j-1}{j-1} z_0^{-n} z^n.\]
Factor \(z - z_0 = -z_0(1 - z/z_0)\) and expand \((1 - z/z_0)^{-j}\) by the negative binomial series, valid for \(|z/z_0| < 1\).
2.2. The dominant contribution to the \(n\)-th Taylor coefficient comes from the highest-order term \(j = k\): \[\frac{A_k}{(z - z_0)^k} = \frac{(-1)^k A_k}{z_0^k} \sum_{n=0}^\infty \binom{n+k-1}{k-1} z_0^{-n} z^n.\]
This is 1 specialized to \(j = k\) and multiplied by \(A_k\).
2.3. Note that \(\binom{n+k-1}{k-1} = \frac{(n+k-1)\cdots(n+1)}{(k-1)!} = \frac{n^{k-1}}{(k-1)!}\left(1 + O(1/n)\right)\) is a polynomial in \(n\) of degree \(k-1\).
The binomial coefficient \(\binom{n+k-1}{k-1}\) is a polynomial in \(n\) of degree \(k-1\) with leading coefficient \(\frac{1}{(k-1)!}\), so it equals \(\frac{n^{k-1}}{(k-1)!}(1 + O(1/n))\).
2.4. For any \(j < k\), \(\binom{n+j-1}{j-1} = O(n^{k-2})\), which is of lower polynomial order as \(n \to \infty\).
The binomial coefficient \(\binom{n+j-1}{j-1}\) is a polynomial in \(n\) of degree \(j - 1 \le k - 2\), which grows strictly slower than \(n^{k-1}\).
2.5. Q.E.D.
Step 2: Asymptotics of the Coefficients \(c_n\)
3. Asymptotic formula for \(c_n\).
3.1. The total coefficient \(c_n\) is the sum of the coefficients from \(P(z)\) and \(g(z)\): \[c_n = \frac{(-1)^k A_k}{z_0^{n+k}} \binom{n+k-1}{k-1} + \sum_{j=1}^{k-1} \frac{(-1)^j A_j}{z_0^{n+j}} \binom{n+j-1}{j-1} + d_n.\]
Since \(f(z) = P(z) + g(z)\), the \(n\)-th Taylor coefficient of \(f\) is the sum of the \(n\)-th coefficients of \(P\) (from 2) and of \(g\) (which is \(d_n\)).
3.2. Factoring out the dominant term \(B_n = \frac{(-1)^k A_k}{z_0^{n+k}} \frac{n^{k-1}}{(k-1)!}\): \[c_n = \frac{(-1)^k A_k}{z_0^{n+k}} \frac{n^{k-1}}{(k-1)!} \left( 1 + O\left(\frac{1}{n}\right) + O\left(n^{-(k-1)} \rho^{-n}\right) \right) = \frac{(-1)^k A_k}{(k-1)! z_0^k} \cdot \frac{n^{k-1}}{z_0^n} \left( 1 + O\left(\frac{1}{n}\right) \right).\]
The \(j = k\) term dominates: the lower-order terms \(j < k\) contribute \(O(n^{k-2})\), which is \(O(1/n)\) relative to \(n^{k-1}\), and the remainder \(d_n = O(\rho^{-n})\) with \(\rho > 1\) decays exponentially, which is subsumed by \(O(1/n)\).
3.3. Q.E.D.
Step 3: Proof of (1) and (2)
4. Proof of (1): \(c_n \neq 0\) for all large enough \(n\).
4.1. Let \(C = \frac{(-1)^k A_k}{(k-1)! z_0^k} \neq 0\) (since \(A_k \neq 0\)).
\(A_k \neq 0\) because the pole is of order \(k\), and \(z_0 \neq 0\) since \(|z_0| = 1\), so \(C \neq 0\).
4.2. By <1>3.<2>2, \(\lim_{n \to \infty} \frac{c_n}{C n^{k-1} z_0^{-n}} = 1\).
The asymptotic formula gives \(c_n = C n^{k-1} z_0^{-n}(1 + O(1/n))\), and the factor \(1 + O(1/n)\) tends to \(1\).
4.3. Since the limit is \(1 \neq 0\), there exists \(N \in \mathbb{N}\) such that for all \(n \geq N\), \(\left| \frac{c_n}{C n^{k-1} z_0^{-n}} \right| \geq \frac{1}{2} > 0\).
By the definition of a limit equal to \(1\), taking \(\varepsilon = 1/2\) gives an \(N\) beyond which the ratio is within \(1/2\) of \(1\), hence at least \(1/2\) in modulus.
4.4. Therefore, \(c_n \neq 0\) for all \(n \geq N\).
If \(c_n = 0\) then the ratio would be \(0\), contradicting the lower bound \(1/2 > 0\); hence \(c_n \neq 0\).
4.5. Q.E.D.
5. Proof of (2): \(\lim_{n \to \infty} \frac{c_n}{c_{n+1}} = z_0\).
5.1. Using the asymptotic formula from <1>3.<2>2 for \(c_n\) and \(c_{n+1}\): \[\frac{c_n}{c_{n+1}} = \frac{C \frac{n^{k-1}}{z_0^n} \left(1 + O(1/n)\right)}{C \frac{(n+1)^{k-1}}{z_0^{n+1}} \left(1 + O(1/(n+1))\right)} = z_0 \cdot \left(\frac{n}{n+1}\right)^{k-1} \cdot \frac{1 + O(1/n)}{1 + O(1/n)}.\]
Substituting the asymptotic formulas for \(c_n\) and \(c_{n+1}\) and cancelling the common constant \(C\) and the powers of \(z_0\) gives the stated ratio.
5.2. Since \(\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{k-1} = 1^{k-1} = 1\), and \(\lim_{n \to \infty} \frac{1 + O(1/n)}{1 + O(1/n)} = 1\): \[\lim_{n \to \infty} \frac{c_n}{c_{n+1}} = z_0 \cdot 1 \cdot 1 = z_0.\]
The limit of a product is the product of the limits; each factor tends to \(1\), so the product tends to \(z_0\).
5.3. Q.E.D.