Erdős Problem #1051 — Is it true that if is a sequence of integers with then is irrational?
Is it true that if is a sequence of integers with then is irrational?
References
Primary source
Additional references
UnsolvedMath, Erdős Problems set, ULAM AI, licensed CC BY 4.0.
Progress summary
The conjecture has been proved: the stated rapid growth condition forces the series to be irrational, and later work identifies a near-sharp threshold.
Erdős posed the problem with Graham in 1980, asking whether sufficiently rapid growth of an increasing integer sequence forces the reciprocal-product series to be irrational.
Known results
- Erdős, 1988: noted affirmative results when tends to infinity sufficiently rapidly and asked for the strongest such theorem.
- For the original condition , the series is irrational by a proof using tail estimates and Mahler’s criterion.
January 2026 solution and follow-up
Aletheia, identified as a DeepMind system powered by Gemini Deep Think, produced an autonomous affirmative solution, later formalized in Lean by Barreto. Barreto, Kang, Kim, Kovač, and Zhang subsequently proved irrationality under the weaker condition , with , and constructed rational examples when for every .
Current status (as of June 2026): The original problem is settled affirmatively; the follow-up work gives a substantially sharper growth threshold and matching rational constructions.
Solutions 0
No solutions have been posted yet.