Sum of Reciprocals of Divisors equals Abundancy

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Theorem

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 $n$.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \sum_{d \backslash n} \frac 1 d$	$=$	$\displaystyle \sum_{d \backslash n} \frac 1 {\left({\frac n d}\right)}$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Sum Over Divisors Equals Sum Over Quotients
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac 1 n \sum_{d \backslash n} d$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \frac {\sigma \left({n}\right)} n$	$\displaystyle $	$\displaystyle $	$\displaystyle $	from the definition of the sigma function.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{d \backslash n} \frac 1 d\)	\(=\)	\(\displaystyle \sum_{d \backslash n} \frac 1 {\left({\frac n d}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Sum Over Divisors Equals Sum Over Quotients
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1 n \sum_{d \backslash n} d\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\sigma \left({n}\right)} n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the definition of the sigma function.