Sum of Reciprocals of Divisors equals Abundancy Index

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Theorem

Let $n$ be a positive integer.

Let $\sigma \left({n}\right)$ be the sigma function of $n$.


Then:

$\displaystyle \sum_{d \mathop \backslash n} \frac 1 d = \frac {\sigma \left({n}\right)} n$

where $\dfrac {\sigma \left({n}\right)} n$ is the abundancy index of $n$.


Proof

\(\displaystyle \sum_{d \mathop \backslash n} \frac 1 d\) \(=\) \(\displaystyle \sum_{d \mathop \backslash n} \frac 1 {\left({\frac n d}\right)}\) Sum Over Divisors Equals Sum Over Quotients
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 n \sum_{d \mathop \backslash n} d\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sigma \left({n}\right)} n\) Definition of Sigma Function

$\blacksquare$