# Sum of Reciprocals of Divisors of Perfect Number is 2

## Theorem

Let $n$ be a perfect number.

Then:

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

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

## Proof

 $\ds \sum_{d \mathop \divides n} d$ $=$ $\ds \map \sigma n$ Definition of Sigma Function $\ds \leadsto \ \$ $\ds \dfrac 1 n \sum_{d \mathop \divides n} d$ $=$ $\ds \dfrac {\map \sigma n} n$ $\ds \leadsto \ \$ $\ds \sum_{d \mathop \divides n} \frac d n$ $=$ $\ds \dfrac {\map \sigma n} n$ $\ds \leadsto \ \$ $\ds \sum_{d \mathop \divides n} \frac 1 d$ $=$ $\ds \dfrac {\map \sigma n} n$

The result follows by definition of perfect number:

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

$\dfrac {\sigma \left({n}\right)} n = 2$

where $\sigma: \Z_{>0} \to \Z_{>0}$ is the sigma function.

$\blacksquare$