A positive integer n is called perfect if the sum of its positive divisors σ(n) is twice n, that is, σ(n)=2n. For example, 6 is a perfect number since the sum of its positive divisors is 1+2+3+6=12, which is twice 6. Prove that if n is a positive perfect integer, then:
p∣n∑p+11<ln2<p∣n∑p−11
where the sums are taken over all prime divisors p of n.
algebranatural logprime numbersnumber theorysum of divisorsTaylor expansion