Abundance Average of Naturals

This is a classic problem, anyhow we want to use the identity that $\dfrac{\sigma(n)}{n} =\sum_{d|n} d^{-1}$ The proof is left as an exercise to the reader.

Now the sum is just $\sum_{N\geq n\geq 1} \sum_{d|n} d^{-1}=\sum_{N\geq d\geq 1} d^{-1} \left\lfloor\frac{N}{d}\right\rfloor$ We used the fact that each $d$ appears $\left\lfloor\frac{N}{d}\right\rfloor$ , i.e it divides that many $n$ between $1$ and $N$ . Notice that $\dfrac{N}{d}\geq \left\lfloor\dfrac{N}{d}\right\rfloor\geq \dfrac{N}{d}-1$ Hence the limit is $\lim_{N\to \infty} \dfrac{1}{N}\sum_{N\geq d\geq 1} d^{-1} \frac{N}{d}\geq \lim_{N\to \infty} \dfrac{1}{N}\sum_{N\geq d\geq 1} d^{-1}\left\lfloor\frac{N}{d}\right\rfloor\geq \lim_{N\to \infty} \dfrac{1}{N}\sum_{N\geq d\geq 1} d^{-1} \left(\frac{N}{d}-1\right)$ Clearly both sides converge to $\zeta(2)$ , so the limit is just $\boxed{\dfrac{\pi^2}{6}}$