Hadamard Gap Theorem

SN(z)=∑k=0Nz2kS_N(z) = \sum_{k=0}^{N} z^{2^k} (gap ratio exactly 2): plotting ∣SN(reiθ)∣|S_N(r e^{i\theta})| for r∈{0.90,0.95,0.99}r \in \{0.90, 0.95, 0.99\}, spikes erupt at more and more angles as NN grows — a gappy series with radius exactly 1 is singular at every point of the unit circle. The gap-free contrast ∑zk/(k+1)2\sum z^k/(k+1)^2 stays calm.

nk+1nk=2k+12k=2>1,1R=lim sup⁡n→∞∣an∣1/n=1\frac{n_{k+1}}{n_k} = \frac{2^{k+1}}{2^k} = 2 > 1, \qquad \frac{1}{R} = \limsup_{n\to\infty} |a_n|^{1/n} = 1
max⁡θ∣S12(0.99 eiθ)∣≈6.3139\max_\theta |S_{12}(0.99\,e^{i\theta})| \approx 6.3139

Examples & Exercises

Prove by the Cauchy-Hadamard root test that f(z)=∑k=0∞z2kf(z) = \sum_{k=0}^{\infty} z^{2^k} has radius of convergence exactly 1.

f(z)=∑n=0∞anzn,an=1 (n=2k),  an=0 otherwisef(z) = \sum_{n=0}^{\infty} a_n z^n, \qquad a_n = 1\ (n = 2^k),\; a_n = 0\ \text{otherwise}
1R=lim sup⁡n→∞∣an∣1/n=1  ⇒  R=1\frac{1}{R} = \limsup_{n\to\infty} \lvert a_n \rvert^{1/n} = 1 \;\Rightarrow\; R = 1
N=5N = 5: inside the disc, S5(0.5)≈0.81642150902S_5(0.5) \approx 0.81642150902 converges fast — consistent with the radius-1 proof below.
Show proof
  1. Write f(z)f(z) as an ordinary power series ∑n=0∞anzn\sum_{n=0}^{\infty} a_n z^n: an=1a_n = 1 when n=2kn = 2^k and an=0a_n = 0 otherwise. The Cauchy-Hadamard formula gives 1R=lim sup⁡n→∞∣an∣1/n\frac{1}{R} = \limsup_{n\to\infty} \lvert a_n \rvert^{1/n} (with the conventions 1/0=∞1/0 = \infty, 1/∞=01/\infty = 0).
  2. Since ∣an∣1/n≤1\lvert a_n \rvert^{1/n} \le 1 for every nn, the limsup is at most 1. But along the subsequence n=2kn = 2^k we have ∣a2k∣1/2k=11/2k=1\lvert a_{2^k} \rvert^{1/2^k} = 1^{1/2^k} = 1, so the limsup is at least 1. Hence the limsup equals 1 and R=1R = 1 — exactly 1.
  3. So the series converges absolutely for ∣z∣<1\lvert z \rvert < 1, and diverges for ∣z∣>1\lvert z \rvert > 1 since the general term z2kz^{2^k} does not tend to 0 (∣z∣2k→∞\lvert z \rvert^{2^k} \to \infty). The boundary ∣z∣=1\lvert z \rvert = 1 is where the test goes silent — exactly where the gap theorem speaks.

Prove by the Weierstrass M-test that T(z)=∑k=0∞zk(k+1)2T(z) = \sum_{k=0}^{\infty} \frac{z^k}{(k+1)^2} converges uniformly on the closed unit disc ∣z∣≤1\lvert z \rvert \le 1, so its sum extends continuously to the boundary. Without gaps, no natural boundary appears.

∣zk(k+1)2∣≤1(k+1)2=Mk,∑k=0∞Mk=π26<∞\left\lvert \frac{z^k}{(k+1)^2} \right\rvert \le \frac{1}{(k+1)^2} = M_k, \quad \sum_{k=0}^{\infty} M_k = \frac{\pi^2}{6} < \infty
max⁡θ∣T200(0.99 eiθ)∣≈1.6045<1.65<π26\max_\theta \lvert T_{200}(0.99\,e^{i\theta}) \rvert \approx 1.6045 < 1.65 < \frac{\pi^2}{6}
N=12N = 12: the dashed gray T12T_{12} stays calm at max⁡θ∣T12∣≈1.5551\max_\theta \lvert T_{12} \rvert \approx 1.5551 — without gaps there is no blow-up at the boundary.
Show proof
  1. On ∣z∣≤1\lvert z \rvert \le 1, ∣zk/(k+1)2∣≤1/(k+1)2=:Mk\left\lvert z^k/(k+1)^2 \right\rvert \le 1/(k+1)^2 =: M_k, and ∑Mk=∑k=1∞1/k2=π2/6<∞\sum M_k = \sum_{k=1}^{\infty} 1/k^2 = \pi^2/6 < \infty (the Basel problem). By the Weierstrass M-test, ∑zk/(k+1)2\sum z^k/(k+1)^2 converges uniformly on the closed disc ∣z∣≤1\lvert z \rvert \le 1.
  2. Each partial sum is a polynomial, hence continuous, and a uniform limit of continuous functions is continuous — so the sum T(z)T(z) is continuous on the whole closed disc. That is, TT extends continuously to the boundary ∣z∣=1\lvert z \rvert = 1: every boundary point is a point of continuity, so there is not a single singularity there.
  3. T(z)T(z) and ∑z2k\sum z^{2^k} both have radius exactly 1 yet opposite fates — the gaps make the difference. The measurements agree: max⁡θ∣T200(0.99 eiθ)∣≈1.6045<π2/6≈1.6449\max_\theta \lvert T_{200}(0.99\,e^{i\theta}) \rvert \approx 1.6045 < \pi^2/6 \approx 1.6449, with no hint of blow-up near the boundary.

Sketch the key idea of the Hadamard gap theorem in 2–3 steps: assuming a holomorphic extension past the boundary, explain the mechanism by which the gaps force a contradiction. State explicitly in your answer that this is a sketch of the idea, not the complete proof.

sk(z)=∑j=0kz2j,sk+1(z)−sk(z)=z2k+1s_k(z) = \sum_{j=0}^{k} z^{2^j}, \qquad s_{k+1}(z) - s_k(z) = z^{2^{k+1}}
∣z0∣>1  ⇒  ∣z0∣2k+1→∞\lvert z_0 \rvert > 1 \;\Rightarrow\; \lvert z_0 \rvert^{2^{k+1}} \to \infty
N=12N = 12: 106 spikes crowd the boundary — below is the proof idea behind what this picture suggests (a sketch, not the complete proof).
Show proof
  1. Start with contradiction: suppose f(z)=∑k≥0z2kf(z) = \sum_{k\ge 0} z^{2^k} extended holomorphically past some boundary point ζ\zeta (∣ζ∣=1\lvert \zeta \rvert = 1). Then there is a small disc DD around ζ\zeta on which a holomorphic function agrees with ff inside ∣z∣<1\lvert z \rvert < 1 — so DD pokes outside the unit circle.
  2. [IDEA — the step not proved here] Gap-forced overconvergence: the partial sums sk(z)=∑j=0kz2js_k(z) = \sum_{j=0}^{k} z^{2^j} converge to ff on ∣z∣<1\lvert z \rvert < 1, hence to the extension on D∩{∣z∣<1}D \cap \{\lvert z \rvert < 1\}. Each new partial sum adds just one term yet leaps twice as far in degree (2k+1=2⋅2k2^{k+1} = 2\cdot 2^k) — with no intermediate terms to interfere, a classical overconvergence lemma forces the sequence sks_k to keep converging uniformly on all of DD, including points with ∣z∣>1\lvert z \rvert > 1. That lemma is the technical heart; it is stated here, not proved.
  3. But at any point with ∣z0∣>1\lvert z_0 \rvert > 1, sk+1(z0)−sk(z0)=z02k+1s_{k+1}(z_0) - s_k(z_0) = z_0^{2^{k+1}} and ∣z02k+1∣=∣z0∣2k+1→∞\lvert z_0^{2^{k+1}} \rvert = \lvert z_0 \rvert^{2^{k+1}} \to \infty, so sk(z0)s_k(z_0) cannot converge — a contradiction. Hence no holomorphic extension past any boundary point exists, and every point of the unit circle is singular. To stress again: this is the idea (the skeleton) of the proof, not the complete proof.

7. Power Series