English

$\sum_{p\le n} 1/p = \ln(\ln n) + O(1)$: An Exposition

History and Overview 2015-11-17 v2

Abstract

It is well known that pn1/p=ln(ln(n))+O(1)\sum_{p\le n} 1/p =\ln(\ln(n)) + O(1) where pp goes over the primes. We give several known proofs of this. We first present a a proof that ln(ln(n))+O(1)\ge \ln(\ln(n)) + O(1). This is based on Euler's proof that p1/p\sum_p 1/p diverges. We then present three proofs that pn1/pln(ln(n))+O(1)\sum_{p\le n} 1/p \le \ln(\ln(n)) + O(1) The first one, due to Mertens, does not use the prime number theorem. The second and third one do use the prime number theorem and hence are shorter.

Keywords

Cite

@article{arxiv.1511.01823,
  title  = {$\sum_{p\le n} 1/p = \ln(\ln n) + O(1)$: An Exposition},
  author = {William Gasarch and Larry Washington},
  journal= {arXiv preprint arXiv:1511.01823},
  year   = {2015}
}