Sum of Reciprocals of Divisors equals Abundancy Index

Theorem
Let $n$ be a positive integer.

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

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

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