2

Let n denote a positive integer and let $\sigma(n)$ denote the sum of all divisors of $n$, so that $\sigma(n)$ is larger than $n$ (for $n > 1$) but not by much since it's bounded above by $c\ n\log \log n$ .

This invites comparison between situations in which the two are interchanged. As an example, consider the following:

Let $A$ be a subset of the positive integers. Suppose that the sum of $1/n$ , for all $n$ in $A$ , converges. Then the sum of 1/$\sigma(n)$ , over all $n$ in $A$, clearly converges as well.

Is the converse true?

user2052
  • 2,427
  • When you say $s(n)$, do you not consider $n$ in the sum? If you do, consider using $\sigma(n)$. See http://mathworld.wolfram.com/RestrictedDivisorFunction.html . – Vincenzo Oliva Nov 22 '14 at 20:14

0 Answers0