IMO 1983 SL 2

Origin: BEL

Problem

Let n be a positive integer. Let \sigma(n) be the sum of the natural divisors d of n (including 1 and n). We say that an integer m \geq1 is superabundant (P.Erd¨os, 1944) if \forallk \in{1, 2, . . ., m −1}, \sigma(m) m

\sigma(k) k . Prove that there exists an infinity of superabundant numbers.

Solution

By definition, \sigma(n) =  d|n d =  d|n n/d = n  d|n 1/d, hence \sigma(n)/n =  d|n 1/d. In particular, \sigma(n!)/n! =  d|n! 1/d \geqn k=1 1/k. It follows that the sequence \sigma(n)/n is unbounded, and consequently there exist an infinite number of integers n such that \sigma(n)/n is strictly greater than \sigma(k)/k for k < n.