Suppose \(f(z) = \sum_{n=0}^\infty a_n z^n\) is a power series with radius of convergence exactly 1.
Give an example of such a series that converges at every point of the unit circle \(S^1 = \{|z| = 1\}\).
Give an example of such a function \(f\) that is analytic at \(z = 1\), but whose series \(\sum_{n=0}^\infty a_n\) diverges.
Prove that \(f\) cannot be analytically continued to be analytic at every point of \(S^1\).
Solution
Goal: Provide examples illustrating boundary convergence/divergence vs analyticity, and prove that every power series with radius of convergence 1 has at least one singularity on its circle of convergence.
1. Part (a): Example converging everywhere on \(S^1\).
1.1. Define the power series\[f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}.\]
1.2. Radius of convergence:\[R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| = \lim_{n \to \infty} \frac{(n+1)^2}{n^2} = 1.\]
1.3. Convergence on \(S^1\):- For any \(z \in S^1\), \(|z| = 1\), so \(\left| \frac{z^n}{n^2} \right| = \frac{1}{n^2}\).- The series \(\sum_{n=1}^\infty \frac{1}{n^2} < \infty\) converges (\(p\)-series with \(p = 2 > 1\)).- By the Weierstrass \(M\)-test, the series converges absolutely (and uniformly) on the entire closed disk \(\overline{\mathbb{D}} = \{|z| \le 1\}\).
1.4. Thus \(f(z)\) converges at every point of \(S^1\).
2. Part (b): Example analytic at \(z = 1\) with divergent series \(\sum a_n\).
2.1. Define the function\[f(z) = \frac{1}{1 + z} = \sum_{n=0}^\infty (-1)^n z^n \quad \text{for } |z| < 1.\]
2.2. Radius of convergence: \(R = 1\) because the sequence of coefficients is \(a_n = (-1)^n\), so \(\limsup_{n \to \infty} |a_n|^{1/n} = 1\).
2.3. Analyticity at \(z = 1\):- \(f(z) = \frac{1}{1 + z}\) is holomorphic on the open set \(\mathbb{C} \setminus \{-1\}\).- In particular, \(f(z)\) is holomorphic in the open disk \(D(1, 1)\), so \(f\) is analytic at \(z = 1\).
2.4. Divergence of the series at \(z = 1\):- At \(z = 1\), \(\sum_{n=0}^\infty a_n 1^n = \sum_{n=0}^\infty (-1)^n = 1 - 1 + 1 - 1 + \cdots\).- The sequence of terms does not converge to 0 (\(\lim_{n \to \infty} (-1)^n \ne 0\)), so the series diverges by the Divergence Test.
3. Part (c): \(f\) cannot be analytic at every point of \(S^1\).
3.1. Suppose for contradiction that \(f\) can be analytically continued across every point of \(S^1\).
3.2. Local extensions: For each point \(w \in S^1\), there exists an open disk \(D(w, r_w)\) and a holomorphic function \(f_w: D(w, r_w) \to \mathbb{C}\) such that \(f_w(z) = f(z)\) on \(D(w, r_w) \cap \mathbb{D}\).
3.3. Open cover of compact \(S^1\):- The collection \(\{D(w, r_w) \mid w \in S^1\}\) is an open cover of the compact set \(S^1\).- By compactness, there exists a finite subcover \(\{D(w_1, r_1), \dots, D(w_m, r_m)\}\) of \(S^1\).
3.4. Global extension:- Define the open domain \(U = \mathbb{D} \cup \bigcup_{j=1}^m D(w_j, r_j)\).- The functions \(f\) and \(f_{w_j}\) agree on all pairwise intersections by the Identity Theorem for connected domains.- Thus they define a single holomorphic function \(F: U \to \mathbb{C}\) that extends \(f\).
3.5. Enlarged disk:- The domain \(U\) is an open set containing the compact set \(\overline{\mathbb{D}} = \{|z| \le 1\}\).- By the Lebesgue Covering Lemma (or distance from a compact set to the complement), there exists \(\varepsilon > 0\) such that the disk \(D(0, 1 + \varepsilon) \subseteq U\).
3.6. Radius of convergence contradiction:- Since \(F\) is holomorphic on the open disk \(D(0, 1 + \varepsilon)\), its Taylor series centered at 0 converges on all of \(D(0, 1 + \varepsilon)\).- The Taylor coefficients of \(F\) at 0 are given by \(\frac{F^{(n)}(0)}{n!} = \frac{f^{(n)}(0)}{n!} = a_n\).- This implies that the radius of convergence of \(\sum_{n=0}^\infty a_n z^n\) is at least \(1 + \varepsilon > 1\).- This contradicts the hypothesis that the radius of convergence is exactly 1.
3.7. Therefore, \(f\) must have at least one singular point on \(S^1\).
4. Conclusion:
$\sum z^n/n^2$ converges on $S^1$, $\frac{1}{1+z}$ is analytic at 1 with divergent series, and every power series has a singularity on its circle of convergence by compactness and Taylor's theorem.