Harmonic Mean of Divisors in terms of Divisor Count and Divisor Sum

Theorem
Let $n \in \Z_{>0}$ be a positive integer.

The harmonic mean of the divisors of $n$ is given by:
 * $\map H n = \dfrac {n \, \map \tau n} {\map \sigma n}$

where:
 * $\map \tau n$ denotes the $\tau$ (tau) function: the number of divisors of $n$
 * $\map \sigma n$ denotes the $\sigma$ (sigma) function: the sum of the divisors of $n$.

Proof
By definition of harmonic mean:


 * $\dfrac 1 {\map H n} = \dfrac 1 {\map \tau n} \paren {\ds \sum_{d \mathop \divides n} \dfrac 1 d}$

From Sum of Reciprocals of Divisors equals Abundancy Index:


 * $\ds \sum_{d \mathop \divides n} \frac 1 d = \frac {\map \sigma n} n$

and so:


 * $\dfrac 1 {\map H n} = \dfrac {\map \sigma n} {n \map \tau n}$

Hence the result.