CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)audited 2026-08-28

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Ignoring summability in the Euler product leads to illegal coefficient manipulations

Statement refuted

One may compute the coefficient of $x^{3}$ in the Euler product by the naive rule

$[x^{3}] \prod_{m \geq 1} (1 - x^{m})^{- 1} = \sum_{m \geq 1} [x^{3}] (1 + x^{m} + x^{2 m} + \dots) .$

Facts & Assumptions

Given: the formal factors $(1 - x^{m})^{- 1} = 1 + x^{m} + x^{2 m} + \dots$ .

[F1]

The coefficient of $x^{3}$ in the true Euler product counts partitions of $3$ , so it is $3$ .

[L1]

A locally finite product may be rearranged only by genuine regrouping of the same summable family; the infinite product is defined by stabilization of its finite partial products modulo each $x^{N}$ (Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products).

Counterexample

technique · direct

1.1

On the right-hand side, only the $m = 1$ and $m = 3$ summands contribute a nonzero $x^{3}$ coefficient, so the naive calculation gives $1 + 1 = 2$ .

givenalgebra

2.1

But [F1] gives the true coefficient as $3$ , corresponding to the partitions $3$ , $2 + 1$ , and $1 + 1 + 1$ . The naive rule loses the mixed contribution $x \cdot x^{2}$ coming from two different factors, so it is not a genuine regrouping of the locally finite product expansion licensed by [L1]. Therefore the displayed identity is false, and the missing summability control is exactly what permits the legal coefficientwise product.

step 1.1F1L1∎

Depends on

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