/- Copyright 2025 The Formal Conjectures Authors. Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the License. You may obtain a copy of the License at https://www.apache.org/licenses/LICENSE-2.0 Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License. -/ import FormalConjecturesUtil

Erdős Problem 257

Reference: erdosproblems.com/257

namespace Erdos257

Let $A\subseteq\mathbb{N}$ be an infinite set. Is $$ \sum_{n\in A} \frac{1}{2^n - 1} $$ irrational?

@[category research open, AMS 11] theorem erdos_257 : answer(sorry) (A : Set ), A.Infinite Irrational (∑' n : A, (1 : ) / (2 ^ n.1 - 1)) := True (A : Set ), A.Infinite Irrational (∑' (n : A), 1 / (2 ^ n - 1)) All goals completed! 🐙

Show that $$ \sum_{n} \frac{1}{2^n - 1} = \sum_{n} \frac{d(n)}{2^n}, $$ where $d(n)$ is the number of divisors of $n$.

hr:1 / 2 < 1hpos: (n : ), 0 < n 2 2 ^ nkey:∑' (n : ℕ+), 1 / (2 ^ n - 1) = ∑' (n : ℕ+), (↑n).divisors.card / 2 ^ nhsummL:Summable fun n 1 / (2 ^ n - 1)hsummR:Summable fun n n.divisors.card / 2 ^ n1 / (2 ^ 0 - 1) + ∑' (n : ℕ+), 1 / (2 ^ n - 1) = (Nat.divisors 0).card / 2 ^ 0 + ∑' (n : ℕ+), (↑n).divisors.card / 2 ^ n All goals completed! 🐙

Show that $$ \sum_{n} \frac{d(n)}{2^n} $$ is irrational.

[Er48] Erdős, P., On arithmetical properties of Lambert series. J. Indian Math. Soc. (N.S.) (1948), 63-66.

@[category research solved, AMS 11] theorem erdos_257.variants.tsum_top : Irrational <| ∑' n, n.divisors.card / (2 ^ n : ) := Irrational (∑' (n : ), n.divisors.card / 2 ^ n) All goals completed! 🐙end Erdos257