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Source. Théorème 2, printed p. 176 (physical PDF p. 2); proof in section 2, p. 178 (PDF p. 4). Read on the page images; the scan has no text layer.

Statement

Let $q\in\mathbb Z$ with $|q|\ge2$, and let $(a(n))_{n\in\mathbb N}$ be a sequence in $\mathbb Z^{\mathbb N}$ with the following properties:

• (a) $a(n)\ne0$ for infinitely many $n$;
• (b) for $n$ large enough, $|a(n)|\le r(n)$, where (b$_1$) $r(n)>0$ and (b$_2$) $\limsup r(n+1)/r(n)<|q|$;
• (c) there are infinitely many integers $k\in\mathbb N$, and integers $n_k\in\mathbb N$, such that (c$_1$) $a(n_k+1)=a(n_k+2)=\cdots=a(n_k+k)=0$ and (c$_2$) $\lim_{k\to\infty}r(n_k+k+1)/|q|^k=0$.

Let $x=\sum_{n=0}^{\infty}a(n)q^{-n}$. Then, if $x=\alpha/\beta\in\mathbb Q$, one has for $k$ large enough

$$\alpha q^{n_k}-\beta\sum_{n=0}^{n_k}a(n)q^{n_k-n}=0. \qquad (4)$$

The conclusion is an exact arithmetic relation, not yet a contradiction; the Remarque on p. 178 notes that when $q\ge2$ and all $a(n)\ge0$ it already yields the irrationality of $x$, while in general one can only hope for a contradiction of arithmetic type from (4).

Proof (p. 178), as a pointer and sketch

If $\beta x-\alpha=0$ then

$$\alpha q^{n_k}-\beta\sum_{n=0}^{n_k}a(n)q^{n_k-n} =\beta q^{n_k}\sum_{n=n_k+1}^{\infty}\frac{a(n)}{q^n} =\beta q^{n_k}\sum_{n=n_k+k+1}^{\infty}\frac{a(n)}{q^n},$$

the second equality by (c$_1$) (the paper's (11)). By (b) and (b$_2$) choose $\eta\in\,]0,|q|[$ with $r(n+1)/r(n)\le\eta$ for large $n$; since $n_k\to\infty$ with $k$ (the paper: "en vertu de (a)"; if $n_k$ stayed bounded along infinitely many $k$, then (c$_1$) would force $a(n)=0$ for every large $n$, against (a)), for large $k$ the absolute value of the left side is at most

$$|\beta|\,|q|^{n_k}\sum_{n=n_k+k+1}^{\infty}\frac{r(n_k+k+1)\,\eta^{\,n-(n_k+k+1)}}{|q|^n} \le\frac{|\beta|}{|q|-\eta}\cdot\frac{r(n_k+k+1)}{|q|^k},$$

which tends to $0$ by (c$_2$). The left side is an integer, so it vanishes for $k$ large enough.

Use in Duverney 1995

The Lemme of Duverney 1995 applies the theorem with $r(n)=n^2$ and $n_k=k(3k+1)/2$ to the coefficients of $a\,f(1/q)+b\cdot(1/q)f'(1/q)$, $f(x)=\prod(1-x^n)$; the note's (12) is this theorem's (4). The hypotheses are checked in Step 4 of that page.

Coverage

Claims checked on the page image; the half-page proof was read and is recorded as a sketch. The statement and the proof of section 2 were also checked, on the page images of pp. 176 and 178, by the independent review of the Duverney 1995 reconstruction, passed by its distinct grade; that review covers only this theorem and its proof, not Théorème 1 or sections 3--5, and does not make this page a complete rewritten proof. The review retained a copy of the reviewed text under that card's evidence/assets/reviewed_pages/, which the library no longer holds. The current page differs from the reviewed text in this Coverage section, the updated field and the desc, whose "polynomially bounded" coefficients were corrected to hypotheses (b) and (c$_2$).

Bears on. #250, only as the external criterion consumed by Duverney 1995; it is not a result about the problem's series.