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:
 * $H \left({n}\right) = \dfrac {n \, \tau \left({n}\right)} {\sigma \left({n}\right)}$

where:
 * $\tau \left({n}\right)$ denotes the $\tau$ (tau) function: the number of divisors of $n$
 * $\sigma \left({n}\right)$ denotes the $\sigma$ (sigma) function: the sum of the divisors of $n$.

Proof
By definition of harmonic mean:


 * $\dfrac 1 {H \left({n}\right)} = \dfrac 1 {\tau \left({n}\right)} \left({\displaystyle \sum_{d \mathop \backslash n} \dfrac 1 d}\right)$

From Sum of Reciprocals of Divisors equals Abundancy Index:


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

and so:


 * $\dfrac 1 {H \left({n}\right)} = \dfrac {\sigma \left({n}\right)} {n \tau \left({n}\right)}$

Hence the result.