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Source. Théorème, printed p. 1287 (physical PDF p. 1); the expressions (1), (3), (5) on p. 1287; proof on p. 1289 (PDF p. 3), formulas (14)--(15), from the Lemme of p. 1288. Read on the page images; the scan has no text layer.

Statement

For $q\in\mathbb Z\setminus\{-1,0,1\}$ put, as in the note's (1), (3) and (5) (the first form is defined for every $q\in\mathbb C$ with $|q|>1$),

$$\zeta(q;2)=\sum_{n=1}^{\infty}q^n\Big(\frac{q-1}{q^n-1}\Big)^2 =(q-1)^2\sum_{n=1}^{\infty}\frac{n}{q^n-1} =(q-1)^2\sum_{n=1}^{\infty}\frac{\sigma(n)}{q^n},$$

where $\sigma(n)=\sum_{d\mid n}d$ (the note's $d_1(n)$); Step 1 proves the two equalities. Then $\zeta(q;2)$ is irrational.

Specialization. For $q=2$, $\zeta(2;2)=\sum_{n\ge1}\sigma(n)/2^n$, the number of Problem 250, so that number is irrational. For every integer $q$ with $|q|\ge2$ the factor $(q-1)^2$ is a nonzero integer, so $\sum_{n\ge1}\sigma(n)/q^n$ is irrational for every such $q$; this is the all-base form in which Erdős posed the question in 1948 and 1957.

Premises

• The Lemme of the same note (essential), with its complete author-recorded proof on its page; it consumes Euler's pentagonal number theorem and Théorème 2 of Duverney 1993 (theoreme_2), identified there with their reading depths. The Théorème uses from the Lemme only that $f(1/q)$ and $(1/q)f'(1/q)$ admit no nontrivial $\mathbb Q$-linear relation with $1$.
• Elementary analysis, used without citation: an absolutely convergent double series may be summed in any order; a series of differentiable functions that converges on an interval and whose derivative series converges uniformly there may be differentiated termwise; and $-\log(1-t)\le t/(1-t)$ for $0\le t<1$.

Complete rewritten proof

Throughout, $q\in\mathbb Z$ with $|q|\ge2$.

Step 1 (the three expressions (1), (3), (5)). For $n\ge1$ put $t=q^{-n}$, so $0<|t|\le1/2$. Then $q^n\big((q-1)/(q^n-1)\big)^2=(q-1)^2\,t(1-t)^{-2}$, and $t(1-t)^{-2}=\sum_{j\ge1}jt^j$ for $|t|<1$. Hence

$$\zeta(q;2)=(q-1)^2\sum_{n=1}^{\infty}\sum_{j=1}^{\infty}j\,q^{-nj}.$$

The double series converges absolutely, because $\sum_{n}\sum_{j}j|q|^{-nj}=\sum_n|q|^{-n}(1-|q|^{-n})^{-2}\le4\sum_n|q|^{-n}<\infty$, so it may be summed in any order. Summing over $n$ first, with $\sum_{n\ge1}q^{-nj}=1/(q^j-1)$, gives the note's (3),

$$\zeta(q;2)=(q-1)^2\sum_{j=1}^{\infty}\frac{j}{q^j-1},$$

and grouping the terms by $m=nj$, with $\sum_{j\mid m}j=\sigma(m)$, gives the note's (5),

$$\zeta(q;2)=(q-1)^2\sum_{m=1}^{\infty}\frac{\sigma(m)}{q^m}.$$

(The note reaches (3) by exchanging the sums over $n$ and $k=j-1$, and (5) by citing the Lambert-series expansion, Hardy and Wright, p. 257.) Write $D_q=\sum_{j\ge1}j/(q^j-1)$, so that $\zeta(q;2)=(q-1)^2D_q$ with $(q-1)^2$ a nonzero integer.

Step 2 (the product: convergence, positivity and the logarithmic derivative (14)). Fix $0<\rho<1$ and let $x\in[-\rho,\rho]$. For every $n\ge1$, $1-x^n\ge1-\rho^n>0$, so $\ell_n(x)=\log(1-x^n)$ is defined and $|\ell_n(x)|\le-\log(1-\rho^n)\le\rho^n/(1-\rho)$. By the M-test, $\sum_n\ell_n$ converges uniformly on $[-\rho,\rho]$ to a function $g$, and $\prod_{n\le N}(1-x^n)=\exp\big(\sum_{n\le N}\ell_n(x)\big)\to e^{g(x)}$. So the product (6) converges at every $x\in(-1,1)$ to $f(x)=e^{g(x)}>0$; in particular

$$f(1/q)>0 .$$

The derivatives $\ell_n'(x)=-nx^{n-1}/(1-x^n)$ satisfy $|\ell_n'(x)|\le n\rho^{n-1}/(1-\rho)$ on $[-\rho,\rho]$, a summable bound, so $\sum_n\ell_n'$ converges uniformly there and $g$ is differentiable on $(-\rho,\rho)$ with $g'=\sum_n\ell_n'$. Since $\rho<1$ was arbitrary, $f=e^g$ is differentiable on $(-1,1)$ with $f'=g'f$, which is the note's (14):

$$x\frac{f'(x)}{f(x)}=xg'(x)=-\sum_{n=1}^{\infty}\frac{nx^n}{1-x^n} \qquad(|x|<1).$$

This $f'$ is the derivative used in the Lemme: by Euler's theorem (premise (E) on the Lemme page) $f$ coincides on $(-1,1)$ with the power series (7), so both descriptions of $f$ have the same derivative.

Step 3 (the value (15)). Put $x=1/q\in(-1,1)$ in (14). Since $f(1/q)>0$ by Step 2, the division is legitimate and

$$\frac{(1/q)f'(1/q)}{f(1/q)}=-\sum_{n=1}^{\infty}\frac{nq^{-n}}{1-q^{-n}} =-\sum_{n=1}^{\infty}\frac{n}{q^n-1}=-D_q .$$

The note writes (15) without remarking on $f(1/q)\ne0$.

Step 4 (conclusion). Suppose $\zeta(q;2)$ were rational. Then $D_q=\zeta(q;2)/(q-1)^2$ is rational, and (15) gives

$$0\cdot1+D_q\cdot f(1/q)+1\cdot(1/q)f'(1/q)=0,$$

a linear relation over $\mathbb Q$ among $1$, $f(1/q)$ and $(1/q)f'(1/q)$ with a nonzero coefficient. This contradicts the Lemme. Hence $\zeta(q;2)$ is irrational, and by (5) so is $\sum_{n\ge1}\sigma(n)/q^n$. $\blacksquare$ (The note: "Le théorème résulte donc immédiatement du lemme et de (3).")

Remarks

• What the reconstruction supplies beyond the printed text: the absolute convergence behind the exchanges of summation in (3) and (5); the convergence, positivity and differentiability of the product in Step 2, which the note takes for granted when it states (14) on p. 1289; the nonvanishing $f(1/q)\ne0$ needed for (15). The route is the note's.
• (5) is not needed for the irrationality of $\zeta(q;2)$; it is what identifies $\zeta(q;2)$ with Erdős's form and with the site's series.
• The note says the deduction of the Théorème from the Lemme follows the route of Bundschuh and Väänänen (Compositio Math. 91 (1994), no. 2, 175--199; the note's reference [1] prints pp. 175--201) for $\sum_{n\ge1}1/(q^n-1)$. The argument gives irrationality only; for irrationality measures see Zudilin 2002 and Smet and Van Assche 2009.

Verification

This full reconstruction is independently reviewed; verdict refutation-failed; grade pass. It contains every deduction of the note's proof of the Théorème (p. 1289, formulas (14)--(15), with (3) and (5) of p. 1287) together with the expansions listed under Remarks; the Lemme it consumes has its own complete, independently reviewed proof and its own external premises. A fresh-context whole-claim review of this page and of the Lemme page is filed under this card's evidence/verify/, with the distinct grade beside it. The proof is therefore independently accepted compilation proof coverage relative to Euler's pentagonal number theorem, not proved here, and to Théorème 2 of Duverney 1993, whose statement and proof were checked. The reviewed text is the copy evidence/assets/reviewed_pages/theoreme.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 2026-10-07, the page range of the Bundschuh and Väänänen paper under Remarks, where the reviewed copy repeats the note's misprinted 175--201; that change touches neither the statement nor the proof. The result is relied on for the status of Problem 250 as a refereed publication (Zbl 0843.11034), the first published proof that $\sum_{n\ge1}\sigma(n)/2^n$ is irrational, independently of this reconstruction.

Bears on. #250: this theorem at $q=2$ settles the problem's exact question in the affirmative.