Erdős Problem #290 — Decreasing Denominators of Harmonic Block Sums

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For every natural number a≥1a\geq 1, there exists a natural number b>ab>a such that the denominator, in lowest terms, of

∑a≤n≤b+11n\sum_{a\leq n\leq b+1}\frac1n

is strictly smaller than the denominator, in lowest terms, of

∑a≤n≤b1n.\sum_{a\leq n\leq b}\frac1n.
References

Progress summary

Refreshed
Claimed solved

The question is settled: every starting point eventually has infinitely many extensions whose reduced denominator is smaller.

The problem asks whether, for every starting point aa, some longer consecutive reciprocal sum has a smaller denominator in lowest terms. Paul Erdős and Ronald Graham posed the question; Peter Shiu and Wouter van Doorn gave independent affirmative answers.

Known results

  • For a=1a=1, denominator drops occur infinitely often; in particular, dn<dn−1d_n<d_{n-1} when n=p(p−1)n=p(p-1) for the relevant primes pp.
  • For general aa, the least drop point b(a)b(a) satisfies b(a)≤4.374(a−1)b(a)\le 4.374(a-1) for all a≥6a\ge 6.

Van Doorn’s 2024 resolution

Van Doorn proved that every a∈Na\in\mathbb{N} has infinitely many b>ab>a for which the reduced denominator of ∑i=ab1/i\sum_{i=a}^{b}1/i is smaller than that of ∑i=ab−11/i\sum_{i=a}^{b-1}1/i. He also obtained 0.54<lim inf⁡a→∞(b(a)−a)/log⁡a<0.610.54<\liminf_{a\to\infty}(b(a)-a)/\log a<0.61.

Current status (as of June 2026): The ordinary harmonic-interval problem is resolved affirmatively, with infinitely many drops for every aa; quantitative bounds are known for the first drop.

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