Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31
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The series sum z to the n over n squared is continuous on the closed disc but singular at 1

Statement refuted

Refuted claim. If a power-series sum extends continuously to the closed disc of convergence, then every boundary point is regular.

The witness is

f(z)=n1znn2.

This sum is continuous on D but the boundary point 1 is singular.

Facts & Assumptions

Given: The series f(z)=n1zn/n2.

[L1]

The p-series n11/n2 converges (For rational p>0, 1/kp converges iff p>1).

[L2]

A convergent numerical majorant makes a complex function series converge uniformly on the domain of the bound (Weierstrass M-test for complex-valued function series).

[L3]

The coefficients 1/n2 give radius 1, and Pringsheim makes the positive boundary point singular because those coefficients are nonnegative (Cauchy-Hadamard for complex power series, including zero and infinite radius, Pringsheim's theorem for power series with nonnegative coefficients).

Counterexample

technique · direct
1.1

On the closed unit disc one has zn/n21/n2. By [L1] and [L2], the series for f converges uniformly on D, so its sum is continuous there.

L1L2
2.1

Fact [L3] gives radius 1 and shows that the boundary point 1 is singular. Therefore continuity on the closed disc does not force regularity at all boundary points, and the displayed claim is false.

L3step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources