Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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A complex power series converges absolutely and uniformly on every closed subdisc strictly inside its disc of convergence

Statement

Let ∑n≥0cn(z−a)n have radius R. For every real r with 0≤r<R, the series converges absolutely and uniformly on the closed disc ∣z−a∣≤r.

Facts & Assumptions

Given: A complex power series of radius R and 0≤r<R.

[L1]

The Cauchy–Hadamard theorem gives absolute convergence for ∣z−a∣<R, divergence for ∣z−a∣>R, and no boundary assertion (Cauchy-Hadamard for complex power series, including zero and infinite radius).

[L2]

The complex M-test gives uniform and pointwise absolute convergence under a convergent real majorant series (Weierstrass M-test for complex-valued function series).

[L3]

For complex numbers, ∣zw∣=∣z∣∣w∣ (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct
1.1

By [L1], the real series ∑∣cn∣rn converges, since it is the modulus series at any point whose distance from a is r; for r=0 it has only the constant contribution.

L1
1.2

If ∣z−a∣≤r, then [L3] gives ∣cn(z−a)n∣≤∣cn∣rn.

L3algebra
2.1

Apply [L2] to the majorants of step 1.1 and the bound of step 1.2. This also covers R=+∞ and makes no assertion when r=R.

step 1.1step 1.2L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 81 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources