Sum of Reciprocals of Divisors equals Abundancy Index

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

$\blacksquare$