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Source. Lemme, printed p. 1288 (physical PDF p. 2); proof pp. 1288--1289 (PDF pp. 2--3), formulas (6)--(13). Read on the page images; the scan has no text layer.

Statement

Let $f(x)=\prod_{n=1}^{\infty}(1-x^n)$ for $|x|<1$ (formula (6)). If $q\in\mathbb Z\setminus\{-1,0,1\}$, then the numbers $1$, $f(1/q)$ and $(1/q)f'(1/q)$ are linearly independent over $\mathbb Q$.

Here $f'$ is the derivative of the function $f$ on $(-1,1)$; Step 1 below records why it is the termwise derivative of the series (7).

Premises

(E) Euler's pentagonal number theorem. For every real $x$ with $|x|<1$,

$$\prod_{n=1}^{\infty}(1-x^n) =1+\sum_{n=1}^{\infty}(-1)^n x^{n(3n+1)/2} +\sum_{n=1}^{\infty}(-1)^n x^{n(3n-1)/2}.$$

This is the note's (7), stated there for $|x|<1$ and cited to Chandrasekharan, Elliptic functions (1985), p. 124, and Exton, q-Hypergeometric functions and applications (1983), p. 229, as a consequence of Jacobi's triple product. Reading depth: the statement was checked against the printed formula (7), which is the classical identity $1-x-x^2+x^5+x^7-x^{12}-x^{15}+\cdots$; neither cited proof was read here and no proof is reconstructed. Only real $x$ is used below.

(T2) Théorème 2 of Duverney 1993, theoreme_2 (statement p. 176, proof section 2, p. 178, of that paper), used exactly as stated there, with the present $q$, the coefficients $a(n)$ of Step 2, $r(n)=n^2$ and $n_k=k(3k+1)/2$ for every $k\ge1$; Step 4 checks its hypotheses. Reading depth: claims checked on the page image; the half-page proof was followed step by step and is sketched on the linked page.

Elementary analysis, used without citation: a power series may be differentiated termwise inside its interval of convergence, and an absolutely convergent series has the same sum after any rearrangement.

Complete rewritten proof

Step 0 (reduction to the irrationality of one number). Suppose $c_0+c_1f(1/q)+c_2\cdot(1/q)f'(1/q)=0$ with $c_0,c_1,c_2\in\mathbb Q$ not all zero. Multiplying by a common denominator we may take $c_0,c_1,c_2\in\mathbb Z$. If $c_1=c_2=0$ then $c_0=0$, a contradiction; so $(c_1,c_2)\ne(0,0)$ and $c_1f(1/q)+c_2\cdot(1/q)f'(1/q)=-c_0\in\mathbb Q$. The Lemme therefore follows from the assertion the note proves ("Il suffit de prouver que ..."):

$(\ast)$ for all integers $a,b$ not both zero, the number $\alpha_q=a\,f(1/q)+b\cdot(1/q)f'(1/q)$ is irrational.

Fix such $a,b$ and suppose, for a contradiction, that $\alpha_q=\eta/\delta$ with $\eta,\delta\in\mathbb Z$, $\delta\ne0$.

Step 1 (the expansion (9)). Since $|q|\ge2$, $x=1/q$ lies in $(-1,1)\setminus\{0\}$. By (E), $f$ agrees on $(-1,1)$ with the power series on the right of (7), which converges for $|x|<1$ because its coefficients are $0$ or $\pm1$. So $f$ is differentiable on $(-1,1)$, $f'$ is the termwise derivative of that series, and multiplying by $x$ gives the note's (8):

$$xf'(x)=\sum_{n=1}^{\infty}(-1)^n\frac{n(3n+1)}{2}x^{n(3n+1)/2} +\sum_{n=1}^{\infty}(-1)^n\frac{n(3n-1)}{2}x^{n(3n-1)/2}.$$

Evaluating (7) and (8) at $x=1/q$ and forming $a\,f(1/q)+b\cdot(1/q)f'(1/q)$ gives the note's (9):

$$\alpha_q=a+\sum_{n=1}^{\infty}(-1)^n\Big(a+b\frac{n(3n+1)}{2}\Big)q^{-n(3n+1)/2} +\sum_{n=1}^{\infty}(-1)^n\Big(a+b\frac{n(3n-1)}{2}\Big)q^{-n(3n-1)/2}.$$

Both series converge absolutely, since $|q|\ge2$, the coefficients are $O(n^2)$ and the exponents are at least $n(3n-1)/2\ge n$.

Step 2 (the coefficient sequence (10)). For $m\ge1$ put $p_m^-=m(3m-1)/2$ and $p_m^+=m(3m+1)/2$; these are the generalized pentagonal numbers, and $n_k=p_k^+$. Then

$$p_m^+-p_m^-=m,\qquad p_{m+1}^--p_m^+=\frac{(m+1)(3m+2)-m(3m+1)}{2}=2m+1,$$

so $1=p_1^-<p_1^+<p_2^-<p_2^+<\cdots$: the exponents occurring in (9) are pairwise distinct. Define $a(0)=a$, $a(p_m^{\pm})=(-1)^m(a+bp_m^{\pm})$ for $m\ge1$, and $a(n)=0$ for every other $n\ge0$. Every $a(n)$ is an integer, and $a(n)=(-1)^m(a+bn)$ whenever $n\in\{p_m^-,p_m^+\}$. The series $\sum_{n\ge0}a(n)q^{-n}$ is a rearrangement, with zero terms inserted, of the absolutely convergent right side of (9), so it converges absolutely to the same value: this is the note's (10),

$$\alpha_q=\sum_{n=0}^{\infty}a(n)q^{-n}.$$

Step 3 (the zero runs around $n_k$). From the gaps in Step 2, for every $k\ge1$:

• (Z1) $a(n_k+j)=0$ for $1\le j\le2k$, since $n_k=p_k^+$ and the next exponent is $p_{k+1}^-=n_k+2k+1$. This contains the note's (11), which uses only $1\le j\le k$.
• (Z2) $a(n_k-j)=0$ for $1\le j\le k-1$, since the exponent before $n_k$ is $p_k^-=n_k-k$. This is the run $a(n_k-1)=\cdots=a(n_k-k+1)=0$ that the note invokes after (12); for $k=1$ it is empty. The coefficient $a(n_k-k)=a(p_k^-)=(-1)^k(a+bp_k^-)$ is not claimed to vanish.

Step 4 (the hypotheses of (T2) with $r(n)=n^2$). The note asserts "$|a(n)|\le n^2$ pour $n$ assez grand" and applies the criterion; the hypotheses are checked one by one.

• (a) $a(n_k)=(-1)^k(a+bn_k)$. If $b\ne0$, then $a+bn_k=0$ for at most one $k$, because $k\mapsto n_k$ is strictly increasing; if $b=0$, then $a\ne0$ and $a(n_k)=(-1)^ka\ne0$ for every $k$. Either way $a(n)\ne0$ for infinitely many $n$.
• (b) Let $c=|a|+|b|\ge1$ and $n\ge c$. If $a(n)\ne0$ then $a(n)=\pm(a+bn)$, so $|a(n)|\le|a|+|b|n\le n|a|+n|b|=cn\le n^2$; if $a(n)=0$ the bound is trivial. So $|a(n)|\le r(n)$ for $n\ge c$. (b$_1$) $r(n)=n^2>0$ for $n\ge1$; the criterion uses $r$ only at indices $n\ge n_k+k+1$ with $k$ large, so $r(0)=0$ is immaterial. (b$_2$) $r(n+1)/r(n)=(1+1/n)^2\to1<2\le|q|$.
• (c) Take every $k\ge1$ with $n_k=k(3k+1)/2$. (c$_1$) is (Z1). For (c$_2$), $n_k+k+1=(3k^2+3k+2)/2\le4k^2$ for $k\ge1$, so $r(n_k+k+1)/|q|^k\le16k^4/2^k\to0$.

Step 5 (the exact relation (12)). By (T2) applied to $x=\alpha_q=\eta/\delta$, there is $k_0$ such that for every $k\ge k_0$

$$\eta\,q^{n_k}-\delta\sum_{n=0}^{n_k}a(n)\,q^{n_k-n}=0.$$

(The note prints "si $x_q=\eta/\delta$" here; $x_q$ is a misprint for the $\alpha_q$ of (10).)

Step 6 (divisibility: $q^k$ divides $\delta a(n_k)$). Fix $k\ge k_0$ and split the sum in (12) at $n_k-k$:

$$\delta a(n_k)=\eta\,q^{n_k}-\delta\sum_{n=0}^{n_k-k}a(n)\,q^{n_k-n} -\delta\sum_{n=n_k-k+1}^{n_k-1}a(n)\,q^{n_k-n}.$$

The last sum vanishes by (Z2). In the first sum every exponent $n_k-n$ is at least $k$, and $n_k\ge k$. So every term on the right is an integer multiple of $q^k$, and $q^k$ divides $\delta a(n_k)$ in $\mathbb Z$. The sign of $q$ plays no role.

Step 7 (growth, and the contradiction). By the note's (13), $\delta a(n_k)=\delta(-1)^k\big(a+bk(3k+1)/2\big)$, so

$$|\delta a(n_k)|\le|\delta|\,(|a|+|b|)\,n_k\le2|\delta|\,(|a|+|b|)\,k^2,$$

using $n_k\le2k^2$. A nonzero integer multiple of $q^k$ has absolute value at least $|q|^k\ge2^k$, and $2^k>2|\delta|(|a|+|b|)k^2$ for all large $k$. Hence $\delta a(n_k)=0$ for all large $k$, and since $\delta\ne0$, $a+bn_k=0$ for all large $k$. Two such values $k<k'$ give $b\,(n_{k'}-n_k)=0$, so $b=0$, and then $a=0$, contradicting $(a,b)\ne(0,0)$. This proves $(\ast)$ and the Lemme. $\blacksquare$ (The note: "Puisque $a$ et $b$ ne sont pas tous les deux nuls, ceci est impossible, et le lemme est démontré.")

Remarks

• What the reconstruction supplies beyond the printed text, without changing the note's route: Step 0 spells out the note's "il suffit"; Step 2 orders the exponents so that (10) is well defined; Step 3 counts the gaps behind (11) and behind the run of $k-1$ zeros; Step 4 checks the hypotheses of (T2), which the note asserts in one sentence; Steps 6 and 7 expand "on déduit de (12) que $q^k$ divise $\delta a(n_k)$" and "ceci est impossible".
• Two harmless imprecisions in the note: "$x_q=\eta/\delta$" for $\alpha_q$ before (12), and (11) records $k$ zeros after $n_k$ where $2k$ are available; (T2) needs only $k$.
• The hypothesis $|q|\ge2$ enters three times: $1/q\in(-1,1)$ (Step 1), (b$_2$) and (c$_2$) (Step 4), and $|q|^k\ge2^k$ (Step 7). Nothing uses the sign of $q$.

Verification

This full reconstruction is independently reviewed; verdict refutation-failed; grade pass. It contains every deduction of the note's proof of the Lemme (pp. 1288--1289, formulas (6)--(13)) together with the expansions listed under Remarks. Its external premises are (E), used as stated with no proof inspected, and (T2), whose statement and half-page proof were checked on the 1993 paper's page image. A fresh-context whole-claim review of this page and of the Théorème page is filed under this card's evidence/verify/, with the distinct grade beside it: statement fidelity against the page images, every essential deduction rederived, and both external premises checked at the reading depths above. The proof is therefore independently accepted compilation proof coverage relative to (E), not proved here, and to (T2). The reviewed text is the copy evidence/assets/reviewed_pages/lemme.md, which the repository does not hold; the current page differs from it only in the desc field, this Verification section, the updated field and, since 2026-09-17, the reading-depth label of (T2) under Premises, restated from "statement checked" to "claims checked" in the read-status vocabulary of docs/anatomy.md; that change touches neither the statement nor the proof.

Bears on. #250, through the Théorème of the same note.