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

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

Proof
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