Sum of Reciprocals of Divisors equals Abundancy Index

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n$ be a positive integer.

Let $\map \sigma n$ denote the sigma function of $n$.


Then:

$\displaystyle \sum_{d \mathop \divides n} \frac 1 d = \frac {\map \sigma n} n$

where $\dfrac {\map \sigma n} n$ is the abundancy index of $n$.


Proof

\(\displaystyle \sum_{d \mathop \divides n} \frac 1 d\) \(=\) \(\displaystyle \sum_{d \mathop \divides n} \frac 1 {\paren {\frac n d} }\) Sum Over Divisors Equals Sum Over Quotients
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 n \sum_{d \mathop \divides n} d\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\map \sigma n} n\) Definition of Sigma Function

$\blacksquare$