∑p≤n1/p=ln⁡(ln⁡n)+O(1)\sum_{p\le n} 1/p = \ln(\ln n) + O(1): An Exposition

preprint2015arXivOpen access

Abstract

It is well known that ∑p≤n1/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 ∑p≤n1/p≤ln⁡(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.

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