Sum of Reciprocals of Divisors of Perfect Number is 2

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$


Sources