CounterexampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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The series $\sum_{n\ge0}\iota(n!)x^n$ converges only at $x=0$ and has radius zero

Statement

The real power series $\sum_{n\ge0}\iota(n!)x^n$ has radius $0$ : it converges at its centre $x=0$ and diverges at every $x\ne0$ .

Facts & Assumptions

Given: Coefficients $a_n:=\iota(n!)$ .

[L1]

Factorials eventually exceed every fixed geometric progression (For every real $x$ , $x^k/k! \to 0$ , The factorial $n!$ and the falling factorial $n^{\underline{k}}$ , defined by recursion in $\mathbb{N}$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[L2]

Cauchy–Hadamard assigns radius $0$ when the coefficient-root limit superior is $+\infty$ (Cauchy–Hadamard: the reciprocal radius is $\limsup_{k\to\infty}|a_{k+1}|^{1/(k+1)}$ , with the zero and infinite cases included).

Verification

technique · direct

1.1

For every real $M>0$ , [L1] applied to a geometric base larger than $M$ shows that $\iota((k+1)!)^{1/(k+1)}>M$ eventually. Hence the coefficient-root limit superior is $+\infty$ .

2.1

By [L2], the radius is $0$ . At $x=0$ only the constant term survives, while for $x\ne0$ the terms fail to tend to zero, so the stated convergence set follows.

step 1.1L2∎

Depends on

Used by

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Sources

Cauchy-Hadamard theorem, Encyclopedia of Mathematics (standard reference, not scraped)
Power series, Encyclopedia of Mathematics (standard reference, not scraped)

The series ∑n≥0ι(n!)xn\sum_{n\ge0}\iota(n!)x^n∑n≥0​ι(n!)xn converges only at x=0x=0x=0 and has radius zero