CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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Every formal coefficient exists while convergence radius vanishes

Statement refuted

Define $C ⟦ g ⟧$ to be the set of coefficient sequences $(a_{n})_{n \geq 0}$ , written $\sum_{n \geq 0} a_{n} g^{n}$ , with coefficientwise equality and addition and Cauchy multiplication: the coefficient of $g^{n}$ in a product is $\sum_{k = 0}^{n} a_{k} b_{n - k}$ , a finite sum. Formal membership makes no numerical convergence assertion. For numerical evaluation use $g^{0} : = 1$ for every $g$ , including $g = 0$ ; numerical convergence means convergence of the partial sums in $C$ . In the extended nonnegative reals $[0, + \infty]$ , the numerical radius of a coefficient sequence is

$R : = \sup {r \in [0, + \infty) : \sum_{n \geq 0} a_{n} g^{n} converges for every g \in C with ∣ g ∣ < r},$

where the supremum is taken in the extended order, so $R = + \infty$ when the set is unbounded. The claim to refute is that every element of $C ⟦ g ⟧$ has $R > 0$ .

For $n \in N_{0}$ , define $n! : = \prod_{k = 1}^{n} k$ , with the empty product $0! : = 1$ ; thus $(n + 1)! = (n + 1) n!$ . View these integer products as coefficients in $C$ through the canonical embedding $Z ↪ C$ .

Facts & Assumptions

Given: A formal variable $g$ and the coefficient sequence $a_{n} = n!$ for all $n \geq 0$ .

[A1]

The formal operations and coefficientwise equality are as defined in the Statement; the coefficient sequence itself does not assert convergence after numerical evaluation.

[F1]

Each $n!$ is a positive integer, hence a nonzero coefficient in $C$ ; positivity follows by induction from $0! = 1$ and $(n + 1)! = (n + 1) n!$ with $n + 1 > 0$ .

Counterexample

1.1givenA1F1algebra

Each $n!$ is a finite complex number by [F1], so the sequence $(n!)_{n \geq 0}$ defines an element $S (g) = \sum_{n \geq 0} n! g^{n}$ of $C ⟦ g ⟧$ by [A1], irrespective of numerical convergence.

1.2givenF1algebra

Fix any $g \neq 0$ and put $t_{n} = n! g^{n}$ . By [F1], $n! \neq 0$ , so $t_{n} \neq 0$ for every $n$ ; the factorial recurrence in the Statement therefore gives $∣ t_{n + 1} ∣ / ∣ t_{n} ∣ = (n + 1) ∣ g ∣$ , which is greater than $2$ for all sufficiently large $n$ . Hence $∣ t_{n} ∣$ eventually grows by a factor greater than $2$ at every step, so $t_{n}$ does not tend to zero.

2.1A1F1step 1.2algebra∎

If the numerical series $\sum_{n} t_{n}$ converged, its partial sums $s_{N} : = \sum_{n = 0}^{N} t_{n}$ would converge, so $t_{n} = s_{n} - s_{n - 1} \to 0$ as $n \to \infty$ ; step 1.2 rules this out for every nonzero $g$ . At $g = 0$ , [F1] and the convention $g^{0} = 1$ leave only the constant term, so the series equals $0! = 1$ . Thus the series converges at no nonzero $g$ . For every real $r > 0$ , the point $g = r / 2$ lies in $∣ g ∣ < r$ and gives divergence, so no positive $r$ satisfies the defining property of the radius; $r = 0$ satisfies it vacuously. Therefore the defining set is exactly ${0}$ , its supremum in $[0, + \infty]$ is $R = 0$ , and the claimed positive radius fails.

Remarks

This example separates coefficientwise formal algebra from numerical convergence. It proves no convergence, summability, or divergence claim for any particular QFT perturbation series. A truncation error requires a separate estimate for a specified analytic model; formal coefficient identities alone do not provide one.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.