CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The complex geometric series is not uniformly convergent on its open unit disc

Statement refuted

Every complex power series converges uniformly on its entire open disc of convergence.

Facts & Assumptions

Given: The geometric power series $\sum_{n \geq 0} z^{n}$ on $D = {∣ z ∣ < 1}$ .

[L1]

Uniform convergence requires one index to work for every point of the domain (Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary).

[L2]

Cauchy–Hadamard gives the geometric series radius $1$ and pointwise convergence for $∣ z ∣ < 1$ (Cauchy-Hadamard for complex power series, including zero and infinite radius).

Counterexample

technique · direct

1.1algebra

For each $n \geq 1$ , $\sup_{z \in D} ∣ z^{n} ∣ = 1$ , although the supremum is not attained: real $z ↑ 1$ makes $z^{n} ↑ 1$ .

2.1step 1.1L1

If the series converged uniformly, its terms would tend uniformly to $0$ , contradicting step 1.1 and the quantifiers in [L1].

3.1step 2.1L2∎

Nevertheless [L2] gives pointwise convergence throughout $D$ , so this is a counterexample to uniform convergence on the whole open disc.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. Ahlfors, Complex Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)