Sum of Reciprocals of Divisors of Perfect Number is 2

Theorem

Then:

$\ds \sum_{d \mathop \divides n} \dfrac 1 d = 2$

That is, the sum of the reciprocals of the divisors of $n$ is equal to $2$.

$\ds \sum_{d \mathop \divides n} d$

$=$

$\ds \map {\sigma_1} n$

$\ds \leadsto \ \ $

$\ds \dfrac 1 n \sum_{d \mathop \divides n} d$

$=$

$\ds \dfrac {\map {\sigma_1} n} n$

$\ds \leadsto \ \ $

$\ds \sum_{d \mathop \divides n} \frac d n$

$=$

$\ds \dfrac {\map {\sigma_1} n} n$

$\ds \leadsto \ \ $

$\ds \sum_{d \mathop \divides n} \frac 1 d$

$=$

$\ds \dfrac {\map {\sigma_1} n} n$

The result follows by definition of perfect number:

A perfect number $n$ is a (strictly) positive integer such that:

$\dfrac {\map {\sigma_1} n} n = 2$

where $\sigma_1: \Z_{>0} \to \Z_{>0}$ is the divisor sum function.

$\blacksquare$