Sum of Reciprocals of Divisors equals Abundancy Index

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Theorem

Let $n$ be a positive integer.

Let $\map {\sigma_1} n$ denote the divisor sum function of $n$.


Then:

$\ds \sum_{d \mathop \divides n} \frac 1 d = \frac {\map {\sigma_1} n} n$

where $\dfrac {\map {\sigma_1} n} n$ is the abundancy index of $n$.


Proof

\(\ds \sum_{d \mathop \divides n} \frac 1 d\) \(=\) \(\ds \sum_{d \mathop \divides n} \frac 1 {\paren {\frac n d} }\) Sum Over Divisors Equals Sum Over Quotients
\(\ds \) \(=\) \(\ds \frac 1 n \sum_{d \mathop \divides n} d\)
\(\ds \) \(=\) \(\ds \frac {\map {\sigma_1} n} n\) Definition of Divisor Sum Function

$\blacksquare$