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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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n0n!xn is a formal power series that is not rational over Q

Statement refuted

Every formal power series over Q is rational.

The series F(x)=n0n!xn is a counterexample.

Facts & Assumptions

Given: The formal series F(x)=n0n!xnQx.

[L1]

A formal series over a field is rational if and only if its coefficient sequence satisfies an eventual constant-coefficient recurrence (A coefficient sequence is eventually linearly recurrent if and only if its formal generating function is rational).

[L2]

A nonzero polynomial of degree d over an integral domain has at most d distinct roots (A nonzero polynomial of degree n over an integral domain has at most n distinct roots).

Counterexample

technique · contradiction by polynomial degree
1.1

Suppose F were rational. By [L1] its coefficient sequence would satisfy an eventual constant-coefficient recurrence. That recurrence cannot have order zero, since an eventual order-zero recurrence makes the sequence eventually zero while n!0 in Q for every n. So there would be d1, coefficients c1,,cdQ with cd0, and an index N such that (n+d)!+c1(n+d1)!++cdn!=0 for every nN.

assume-contraL1
2.1

Divide the relation by the nonzero integer n!. It says that the polynomial P(z)=(z+1)(z+d)+c1(z+1)(z+d1)++cd1(z+1)+cd vanishes at every integer z=nN.

step 1.1algebra
3.1

The polynomial P has degree d and leading coefficient 1, so it is nonzero. But step 2.1 gives it more than d distinct rational roots, contradicting [L2].

step 2.1L2algebra
4.1

Therefore the coefficient sequence is not eventually recurrent and [L1] shows that F(x) is not rational. This argument is entirely formal and uses no convergence claim.

step 3.1L1discharge-contradiction

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