gpt pro 5.2 attempt on Math-735

Erdős Problem #257

1) FORMAL RESTATEMENT

Let $A\subseteq\mathbb{N}:=\{1,2,3,\dots\}$ be an infinite set. Define \[ S(A):=\sum_{n\in A}\frac{1}{2^n-1}. \] Since $0<\frac{1}{2^n-1}\le 2^{-(n-1)}$ for $n\ge 2$, the sum converges absolutely for every $A$.

The problem asks whether the following statement is true: \[ (\dagger)\qquad \forall\text{ infinite }A\subseteq\mathbb{N}:\ S(A)\notin\mathbb{Q}. \]

2) QUICK LITERATURE/CONTEXT CHECK

(From the standard summary and cited papers.)

• For $A=\mathbb{N}$, one has \[S(\mathbb{N})=\sum_{n\ge1}\frac{1}{2^n-1}=\sum_{m\ge1}\frac{d(m)}{2^m},\] where $d(m)$ is the divisor-counting function; Erdős $1948$ proved this is irrational [Er48].
• Tao--Teräväinen $2025$ proved irrationality when $A$ is the set of primes, and more generally when $A$ is the set of prime powers [TaTe25].
• Erdős $1968$ proved $S(A)$ is irrational whenever $(a,b)=1$ for all distinct $a,b\in A$ and $\sum_{n\in A} \frac{1}{n}<\infty$ [Er68d].
• Erdős also speculated a far-reaching "bounded perturbation" generalisation $\sum_{n\in A}1/(2^n-t_n)$; this was disproved by Kovač--Tao $2024$, who give bounded $t_n$ for which such a sum can be rational [KoTa24].
• As of late 2025/early 2026 the general statement $(\dagger)$ remains listed as open on the Erdős Problems website [ErP257].

3) ATTACK PLAN

Two natural routes:

1. Lambert-series / base-2 expansion route. Rewrite $S(A)$ as a Lambert-type series $\sum f_A(m)/2^m$ where $f_A(m)$ counts divisors of $m$ lying in $A$; then try to show the base-$2$ expansion cannot be eventually periodic.
2. Construction route (for a counterexample). Try to choose an infinite $A$ whose induced coefficients $f_A(m)$ lead to an eventually periodic base-$2$ expansion after carrying.

I did not manage to complete either route for arbitrary infinite $A$.

4) WORK

Rewriting as a Lambert-type series.

For every $n\ge1$, \[ \frac{1}{2^n-1}=\frac{2^{-n}}{1-2^{-n}}=\sum_{k=1}^{\infty}2^{-kn} \] (geometric series). Hence, by Tonelli's theorem (all terms are nonnegative), \[ S(A)=\sum_{n\in A}\sum_{k\ge1}2^{-kn} =\sum_{m\ge1}\left(\#\{(n,k):n\in A,\ k\ge1,\ kn=m\}\right)\,2^{-m}. \] If we define \[ f_A(m):=\#\{d\in A: d\mid m\}, \] then the map $(n,k)\mapsto m=kn$ identifies the above count with $f_A(m)$, giving the identity \begin{equation}\label{wsref-0-65-71-3a-6c-61-6d-62-65-72-74} S(A)=\sum_{m\ge1}\frac{f_A(m)}{2^m}. \end{equation} In particular, when $A=\mathbb{N}$ we have $f_A(m)=d(m)$.

Tiny numerical experiments (illustrative only).

Below are high-precision numerical values (computed from long partial sums):

• $S(\mathbb{N})\approx 1.60669515241529176378\dots$ (the Erdős--Borwein constant).
• For $A=2\mathbb{N}$ (even indices), $S(A)=\sum_{k\ge1}\frac{1}{2^{2k}-1}\approx 0.42109768603342377729\dots$.
• For $A=\{2^k:k\ge0\}$, $S(A)=\sum_{k\ge0}\frac{1}{2^{2^k}-1}\approx 1.40393682788217832057\dots$.

These computations do not distinguish rational/irrational and are included only as sanity checks.

A simple fully proved special case (conditional on $S(\mathbb{N})$).

If $A$ is cofinite (i.e. $\mathbb{N}\setminus A$ is finite), then \[ S(A)=S(\mathbb{N})-\sum_{n\in\mathbb{N}\setminus A}\frac{1}{2^n-1}. \] The second term is a finite sum of rationals and hence rational. Therefore:

If $S(\mathbb{N})$ is irrational (known by Erdős [Er48]), then $S(A)$ is irrational for every cofinite $A$.

This does not address sparse $A$.

5) VERIFICATION

• Identity (eq:lambert) is correct: each $2^{-m}$ coefficient counts divisors $d\in A$ of $m$.
• The problem remains that $f_A(m)$ can be irregular and may be large, and carrying in base $2$ complicates direct periodicity arguments.

6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here. The Lambert-series identity (eq:lambert) holds for every $A\subseteq\mathbb{N}$ (proved above). Also, if $A$ is cofinite then $S(A)$ is irrational assuming Erdős' theorem that $S(\mathbb{N})$ is irrational [Er48].

(ii) First gap. A proof of $(\dagger)$ for arbitrary infinite $A$ would need a method to rule out eventual periodicity (in base $2$) of the sum (eq:lambert) after carries, using only that $A$ is infinite. I did not find such an argument.

(iii) Top 3 next moves.

1. Try to extend Erdős' 1968 method [Er68d] to drop the pairwise-coprime hypothesis, perhaps by controlling overlaps in divisor structure of $A$.
2. Seek a structural dichotomy for infinite $A$ (e.g. "contains a dense multiplicative substructure" vs "very sparse") and prove irrationality in each regime.
3. Attempt an explicit counterexample search in the style of Kovač--Tao [KoTa24], but with the rigid choice $t_n\equiv1$; i.e. try to design $A$ so that the base-$2$ expansion of (eq:lambert) becomes eventually periodic.

(iv) Minimal counterexample structure. A counterexample would be an infinite $A$ such that $S(A)\in\mathbb{Q}$. Any such $A$ must be infinite (by hypothesis) and cannot be cofinite (since $S(\mathbb{N})$ is irrational). It would likely need to be extremely sparse and carefully engineered so that the induced coefficients $f_A(m)$ in (eq:lambert) produce eventual periodicity after carrying.

7) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 35%.