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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The lacunary power series with factorial exponents has radius one and diverges at 1

Example

Define $c_{m} = 1$ when $m = n!$ for some $n \geq 2$ , and $c_{m} = 0$ otherwise. Then $\sum_{m \geq 0} c_{m} z^{m}$ has radius $1$ and diverges at $z = 1$ .

Facts & Assumptions

Given: The coefficient sequence $(c_{m})$ in the Example.

[L1]

If $L = {lim sup}_{k \to \infty} ∣ c_{k + 1} ∣^{1 / (k + 1)}$ , Cauchy–Hadamard gives radius $1 / L$ when $0 < L < + \infty$ (Cauchy-Hadamard for complex power series, including zero and infinite radius).

[L2]

A convergent real series has terms tending to $0$ (If a series converges then its terms tend to $0$ ).

[L3]

The factorial satisfies $F (0) = 1$ and $F (σ (n)) = F (n) \cdot σ (n)$ , with $n! : = F (n)$ , and $n! \neq 0$ for every $n$ (The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $N$ ).

Verification

technique · direct

1.1L3algebra

Every coefficient root is $0$ or $1$ , since each $c_{m}$ is $0$ or $1$ . By [L3] every $n!$ is a nonzero natural, hence $n! \geq 1$ , and the recursion clause then gives $σ (n)! = n! \cdot σ (n) \geq σ (n)$ ; so $m! \geq m$ for every $m \geq 1$ and the set ${n! : n \geq 2}$ of indices carrying $c_{m} = 1$ is unbounded. The value $1$ therefore occurs at arbitrarily large indices, so the root limsup is $1$ .

2.1step 1.1L1L2∎

By [L1] the radius is $1$ . At $z = 1$ , the terms $c_{m}$ do not tend to $0$ because $c_{n!} = 1$ for every $n \geq 2$ , so [L2] gives divergence.

Depends on

Used by

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Dependency tree · two levels

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