First, if we let \(p\mid n\) denote the set of primes which divide \(n\), then we have the following product representation of two familiar arithmetic functions.

• \[\sigma(n)=\prod_{p\mid n}\left(\frac{p^{e+1}-1}{p-1}\right)=\prod_{p\mid n}\left(1+p+p^2+\cdots+p^e\right)\]
• \[\phi(n)=n\prod_{p\mid n}\left(1-\frac{1}{p}\right)\, .\]

As a related example, we explicitly wrote out the product for the abundancy index earlier. \[\frac{\sigma(n)}{n}=\frac{\prod_{p\mid n} \left(\frac{p^{e+1}-1}{p-1}\right)}{\prod_{p\mid n} p^{e}} = \prod_{p\mid n}\frac{p-1/p^{e}}{p-1}\] We can write it alternately as follows to avoid fractions: \[\frac{\sigma(n)}{n}=\frac{\prod_{p\mid n}\left(1+p+p^2+\cdots+p^e\right)}{\prod_{p\mid n}p^e}=\prod_{p\mid n}\left(1+p^{-1}+p^{-2}+\cdots p^{-e}\right)\]

On the other hand, these products over primes are also sums over divisors; this is true either by definition or by theorem, depending on the case.

It's clear with \(\sigma\) that this is the case, since \[\sigma(n)=\sum_{d\mid n}d\, .\] We can even cleverly add up the divisors in the opposite order to get the slightly more felicitous \[\sigma(n)=\sum_{d\mid n}\frac{n}{d}=n\sum_{d\mid n}\frac{1}{d}\, .\]

With \(\phi\) we have something to prove to make this connection, but not much.

• Since \[\sum_{d\mid n}\phi(d)=n,\,\] we have (via Möbius inversion) that \[\sum_{d\mid n}d\mu\left(\frac{n}{d}\right)=\phi(n)\, .\] We also wrote this as \(\phi\star u=N\Rightarrow \phi=N\star \mu\).
• What is interesting about this is that \(\star\) is commutative, so it is also true that \[\phi(n)=\mu\star N=\sum_{d\mid n}\mu(d)\left(\frac{n}{d}\right)=n\sum_{d\mid n}\frac{\mu(d)}{d},\,\] which after all does look a little cleaner of a sum over divisors.

Now, in some sense we already knew all this; great, some arithmetic functions can be represented either as a sum over divisors or as a product over primes, depending on what you need from them.

The genius of Euler was to directly connect these ideas: \[\frac{\phi(n)}{n}=\sum_{d\mid n}\frac{\mu(d)}{d}=\prod_{p\mid n}\left(1-\frac{1}{p}\right)\quad\text{ and }\quad \frac{\sigma(n)}{n}=\prod_{p\mid n}\left(1+\frac{1}{p}+\frac{1}{p^{2}}+\cdots+ \frac{1}{p^{e}}\right)=\sum_{d\mid n}\frac{1}{d}\, .\] Well, almost. His real genius was to take them to the limit!

Now, you can't really take these things to limits – you would get massive divergence. So what do you do? To analyze this, we will define new, related functions that preserve the summation, but allow convergence.