The reciprocal sum of divisors of Mersenne numbers

Abstract

p≤z 1/p = log log z + C 1 + o(1) as z → ∞. Hence, f (n) ≥ log log z + C 1 − 1/2 + o(1). Now taking z = 1 2 log x, we deduce from the prime number theorem that (1) max n≤x f (n) ≥ log log log x + C 1 − 1/2 + o(1). Perhaps surprisingly, given the ease with which (1) was established, this lower bound on max n≤x f (n) is close to best possible.