If $A=\mathbb{N}$ then this series is $\sum_{n}\frac{d(n)}{2^n}$, where $d(n)$ is the number of divisors of $n$, which Erdős [Er48] proved is irrational. In general, if $f_A(n)$ counts the number of divisors of $n$ which are elements of $A$ then\[\sum_{n\in A}\frac{1}{2^n-1}=\sum_n \frac{f_A(n)}{2^n}.\]The case when $A$ is the set of primes is [69].

Erdős [Er68d] proved this sum is irrational whenever $(a,b)=1$ for all $a\neq b\in A$ and $\sum_{n\in A}\frac{1}{n}<\infty$ (and thought that the condition $(a,b)=1$ could be dropped by complicating his proof).

There is nothing special about $2$ here, and this sum is likely irrational with $2$ replaced by any integer $t\geq 2$.

In [Er88c] Erdős goes further and speculates that $\sum_{n\in A}\frac{1}{2^n-t_n}$ is irrational for every infinite set $A$ and bounded sequence $t_n$ (presumably of integers, and presumably excluding the case when $t_n=0$ for all $n$). This was disproved by Kovač and Tao [KoTa24], and in the comments Kovač has sketched a proof that there exists some choice of $t_n$ with $1\leq t_n\leq 6$ such that this sum is rational.

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This page was last edited 31 October 2025.

6 comments on this problem

• Erdős asked about variants of this problem even earlier, see for instance this paper from 1965.
  

  
  A simpler variant of this problem, where one has a bit more freedom and can choose, say, infinitely many terms of the form $\frac{1}{2^n-1}$ or $\frac{1}{11^m-1}$ has an easy negative answer, simply because the corresponding infinite subsums fill out a union of non-degenerate intervals (from which one simply picks a rational number); see Theorem 2.3 in this paper.
  

  
  The set\[ S = \Big\{ \sum_{n\in A} \frac{1}{2^n-1} : A\subseteq\mathbb{N} \Big\} \]appearing in [257] now has empty interior, so the same trick no longer applies. However, the answer is very likely negative again. Namely, the above set $S$ still has positive Lebesgue measure, i.e., it is a fat Cantor set. Thus, $S$ "almost surely" contains, say, at least one of the dyadic rationals $1/2,1/4,1/8,\ldots$. However, this argument alone cannot be turned into a rigorous proof, since there exist fat Cantor sets containing no rational numbers; a nonconstructive proof was given by Boes, Darst, and Erdős.

  Vjeko_Kovac — 10:55 on 09 Aug 2025

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• The question about subseries of $\sum 1/(t^n-1)$ has been asked for every integer $t\geq2$, but I believe that the case $t=2$ is quite special. In the case $t=2$ the corresponding subsums form a fat (i.e., positive-measure) Cantor set, and it is very likely that it contains many rationals. If its gaps were even smaller (as for, say, $\sum 1/(2^n-2^{-2^n})$), then one could easily prove that it does not even miss many rationals within range. In the cases $t\geq3$ the subsums form a zero-measure Cantor set, which is quite possibly irrational.

  Vjeko_Kovac — 19:36 on 25 Aug 2025

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• Let \(q ≥ 2\) be an integer. We can use the following reference to settle some special cases:
  Yohei Tachiya, “Irrationality of Certain Lambert Series,” Tokyo J. Math. 27 (2004), 73–85.
  PDF: https://projecteuclid.org/journals/tokyo-journal-of-mathematics/volume-27/issue-1/Irrationality-of-Certain-Lambert-Series/10.3836/tjm/1244208475.pdf
  
  Tachiya’s Theorem 1 states that for any period-2 sequence \({a_n}\) (not identically zero),\[
  \sum_{n\ge1}\frac{a_n}{1-q^n}\notin\mathbb{Q}.
  \] Taking \(a_n=\mathbf{1}_{\{n\text{ is even}\}}\) (or \(a_n=\mathbf{1}_{\{n\text{ is odd}\}}\)), which are period-2 and not identically zero, we get\[
  \sum_{n\in 2\mathbb{N}}\frac{1}{q^n-1}
  = -\sum_{n\ge1}\frac{a_n}{1-q^n}\notin\mathbb{Q},
  \qquad
  \sum_{n\in (2\mathbb{N}-1)}\frac{1}{q^n-1}
  = -\sum_{n\ge1}\frac{a_n}{1-q^n}\notin\mathbb{Q}.
  \] Thus, for \(A = 2\mathbb{N}\) (positive even integers) or \(A = 2\mathbb{N} - 1\) (positive odd integers), the subseries\[
  \sum_{n\in A}\frac{1}{q^n-1}
  \]is irrational (in particular for \(q=2\) posed in the problem).

  Quanyu Tang — 14:10 on 05 Sep 2025

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  • Well, ok, but the case of even positive integers is a consequence of a much older paper by Erdős, by writing\[ \sum_{n=1}^{\infty} \frac{1}{q^{2n}-1} = \sum_{n=1}^{\infty} \frac{1}{(q^{2})^{n}-1}. \]Similarly, the case of odd positive integers is a consequence of an older paper by Borwein, by writing\[ \sum_{n=1}^{\infty} \frac{1}{q^{2n-1}-1} = q \sum_{n=1}^{\infty} \frac{1}{(q^{2})^{n}-q}. \]

    Vjeko_Kovac — 16:36 on 05 Sep 2025

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    • Thanks for pointing this out — you are right that the even- and odd-positive-integer cases follow from the much earlier results of Erdős and Borwein, respectively.
      What Tachiya’s Theorem 1 offers here is a uniform way to handle any nontrivial period-2 coefficient sequence, not just the pure even or pure odd subsets.
      Beyond these two special cases in this problem, I am not sure if it yields further consequences here, though.

      Quanyu Tang — 17:48 on 05 Sep 2025

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• The last claim in the comments, on $\sum_{n\in A}\frac{1}{2^n-t_n}$ being irrational for integers $(t_n)_{n=1}^{\infty}$ such that $0<|t_n|<C$, is easily seen to be false, even when $A=\mathbb{N}$. This has already been disproved under [264], in this paper - see the more general Theorem 2.5, but a very short argument for $\sum_{n=1}^{\infty}\frac{1}{2^n+b_n}$ is also sketched in a paragraph on p.17.
  
  However, the question was written quite ambiguously in [Er88c] (as Erdős forgot to say that $t_n\neq0$), so one might also interpret it by requiring $0<t_n<C$. In that case the negative answer also easily follows from the ideas of the same paper, but this is not explicitly written there. Thus, I sketch the same very simple argument here, without taking any extra credit. :-)
  
  Consider the set of all sums $\sum_{n=100}^{\infty}\frac{1}{2^n-t_n}$, where each $t_n$ ranges over $1,2,3,4,5,6$. They form a non-degenerate interval (and thus can be equal to some rational numbers), which can be seen by choosing $t_n$ inductively, so that the series sums to a desired target value. Namely, in each step, choosing $t_n$ achieves an approximation precision dictated by the largest gap:\[ \frac{1}{2^n-t_n-1} -\frac{1}{2^n-t_n} = \frac{1}{(2^n-t_n-1)(2^n-t_n)} \leq \frac{1}{(2^n-6)^2}, \]while the values of the series' tail can fully cover an interval of length\[ \sum_{k=n+1}^{\infty} \Big( \frac{1}{2^k-6} -\frac{1}{2^k-1} \Big) > \frac{1}{2^{n+1}-6} -\frac{1}{2^{n+1}-1} = \frac{5}{(2^{n+1}-6)(2^{n+1}-1)} > \frac{1}{(2^n-6)^2} \]for $n\geq100$.
  
  If one doesn't feel like working out the details, then rather just apply Lemma 4 (a) from this paper to the above collection of sums.

  (The site has been updated to address this comment.)
  Vjeko_Kovac — 13:29 on 30 Oct 2025

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