TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-29

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.

A canonical product converges when the $(p + 1)$ -power reciprocal sum converges

Statement

Let $(a_{n})_{n \geq 1}$ be a sequence of nonzero complex numbers with no finite accumulation point, and fix an integer $p \geq 0$ . If

$\sum_{n \geq 1} ∣ a_{n} ∣^{- (p + 1)} < \infty,$

then the canonical product

$\prod_{n \geq 1} E_{p} (z / a_{n})$

converges normally on $C$ and therefore defines an entire function whose zeros are exactly the points $a_{n}$ , counted with multiplicity.

Facts & Assumptions

Given: The sequence $(a_{n})$ and the integer $p \geq 0$ .

[F1]

The factor estimate gives $∣ 1 - E_{p} (w) ∣ \leq e ∣ w ∣^{p + 1}$ for $∣ w ∣ \leq 1$ (The unit-disc estimate for Weierstrass elementary factors).

[F2]

Canonical products are the fixed-genus Weierstrass products (Weierstrass products, canonical products, and genus).

[F3]

A normally convergent holomorphic product defines a holomorphic function with exactly the zeros contributed by the finitely many nonzero exceptional factors on each compact set (Normally convergent products define holomorphic functions with the expected zeros).

Proof

technique · direct

1.1givenchoose

Fix a compact disc $K_{R} = {z : ∣ z ∣ \leq R}$ . Since $(a_{n})$ has no finite accumulation point, $∣ a_{n} ∣ \to \infty$ , so there is $N$ with $∣ a_{n} ∣ > R$ for $n \geq N$ ; then $∣ z / a_{n} ∣ \leq 1$ on $K_{R}$ for every $n \geq N$ .

2.1F1step 1.1algebra

For $z \in K_{R}$ and $n \geq N$ , [F1] gives $∣ 1 - E_{p} (z / a_{n}) ∣ \leq ∣ z / a_{n} ∣^{p + 1} \leq R^{p + 1} ∣ a_{n} ∣^{- (p + 1)}$ . Because the reciprocal power series converges, the Weierstrass $M$ -test makes $\sum_{n \geq N} \sup_{z \in K_{R}} ∣ 1 - E_{p} (z / a_{n}) ∣$ convergent.

3.1F2F3step 2.1algebra∎

By [F2], the product $\prod_{n \geq 1} E_{p} (z / a_{n})$ is a canonical product, and step 2.1 is exactly the normal-convergence condition on the arbitrary compact disc $K_{R}$ . Hence [F3] makes the product entire, with zeros exactly at the points $a_{n}$ and with their multiplicities.

Depends on

Used by

Dependency tree · two levels

13 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

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 5 Infinite products (standard reference, not scraped)

A canonical product converges when the (p+1)(p+1)-power reciprocal sum converges