For each natural number n≥1, let m(n) be the least natural number m such that ∑_n≤ k≤ mfrac1k≥1, and define epsilon(n)=∑_n≤ k≤ m(n)frac1k-1. Then liminf_n→∞ n^2 epsilon(n)=0.

Erdős Problem #314 — Overshoots of Harmonic Sums

For each natural number $n\geq1$ , let $m(n)$ be the least natural number $m$ such that

\sum_{n\leq k\leq m}\frac1k\geq1,

and define

\epsilon(n)=\sum_{n\leq k\leq m(n)}\frac1k-1.

Then

\liminf_{n\to\infty} n^2\epsilon(n)=0.

References

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