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Aproximate ln(2) using perfect numbers

Source: Brazilian Mathematical Olympiad 2024, Level U, Problem 1

October 12, 2024
algebranatural logprime numbersnumber theorysum of divisorsTaylor expansion

Problem Statement

A positive integer nn is called perfect if the sum of its positive divisors σ(n)\sigma(n) is twice nn, that is, σ(n)=2n\sigma(n) = 2n. For example, 66 is a perfect number since the sum of its positive divisors is 1+2+3+6=121 + 2 + 3 + 6 = 12, which is twice 66. Prove that if nn is a positive perfect integer, then: ∑p∣n1p+1<ln⁡2<∑p∣n1p−1 \sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1} where the sums are taken over all prime divisors pp of nn.
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