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

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Equal radii do not determine convergence on the boundary circle

Statement refuted

Two complex power series with the same radius of convergence have the same convergence behaviour at every point of their common boundary circle.

Facts & Assumptions

Given: The series $\sum z^{n}$ and $\sum z^{n} / n^{2}$ .

[L1]

Cauchy–Hadamard gives absolute convergence inside the radius, divergence outside it, and no assertion on the boundary (Cauchy-Hadamard for complex power series, including zero and infinite radius).

[L2]

For real $p$ , the real series $\sum_{n \geq 1} 1 / n^{p}$ converges exactly when $p > 1$ (The p-series for a real exponent p converges exactly when p is greater than one).

[L3]

If the terms of a real series do not tend to $0$ , that real series diverges (If a series converges then its terms tend to $0$ ).

Counterexample

technique · direct

1.1L1algebra

Both coefficient sequences have root limsup $1$ , so [L1] gives radius $1$ to both series.

1.2L2L3

At $z = 1$ , the first series has constant term sequence $1$ and diverges by [L3], while the second converges by [L2] with $p = 2$ .

2.1step 1.1step 1.2∎

Thus equal radii do not determine even convergence at the boundary point $1$ .

Depends on

Used by

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Dependency tree · two levels

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