We are almost at the very frontiers of serious number theory research now at the end of the semester. In order to start to understand this, though, we will need to introduce two final concepts:
To do so, let's step back a bit to two things you actually already knew. If we let $p\mid n$ denote the set of primes which divide $n$, then $$\sigma(n)=\prod_{p\mid n}\left(\frac{p^{e+1}1}{p1}\right)=\prod_{p\mid n}\left(1+p+p^2+\cdots+p^e\right)$$ and $$\phi(n)=n\prod_{p\mid n}\left(1\frac{1}{p}\right)\, .$$ Indeed, we explicitly wrote out the product $$\frac{\sigma(n)}{n}=\frac{\prod_{p\mid n} \left(\frac{p^{e+1}1}{p1}\right)}{\prod_{p\mid n} p^{e}} = \prod_{p\mid n}\frac{p1/p^{e}}{p1}$$ in our work on perfect numbers and their friends, though perhaps at this point the product $$\frac{\sigma(n)}{n}=\frac{\prod_{p\mid n}\left(1+p+p^2+\cdots+p^e\right)}{\prod_{p\mid n}p^e}=\prod_{p\mid n}\left(1+p^{1}+p^{2}+\cdots p^{e}\right)$$ might be more useful since it's not a nasty fraction.
But let's not forget that these products over primes are also sums over divisors  either by definition or by theorem. It's clear with $\sigma$, since $$\sigma(n)=\sum_{d\mid n}d\, .$$ In fact, I can 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, but not much: recall that 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 $\phi\star u=N\Rightarrow \phi=N\star \mu$.) But 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. Another way to write it is as $$\frac{\phi(n)}{n}=\sum_{d\mid n}\frac{\mu(d)}{d}$$ (which actually was an arithmetic function we discussed a little bit in February).
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.
But 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? We will define new, related functions, such as $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots$$


This is known as the Riemann zeta function, and is probably one of the most studied, yet most mysterious functions in all of mathematics. Riemann, by the way, died quite young (around 40), having made groundbreaking contributions to analysis (Riemann sums), geometry (Riemannian metrics  later used by Einstein), and one paper that changed the course of number theory. He was a PK, and a quietly devout Lutheran throughout his life.
We'll motivate this function with the case $s=1$.
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Notice how every integer $d$ formable by a product of the prime powers dividing $n$ shows up precisely once (as a reciprocal) in the sum. In some sense, this is just the equivalence we already noted; in another sense, though, this gives us a way into introducing limits.
What would happen if we did, for instance, $$\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots\right)\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots\right)\, ?$$ We should get a sum with exactly one copy of the reciprocal of each number divisible by only 2 and 3, e.g. $$\sum_{2\mid n\text{ or }3 \mid n}\frac{1}{n}\, .$$
Click to the left again to hide and once more to show the dynamic interactive window 
There is no reason this wouldn't continue to work for many prime factors.
Because every integer is uniquely represented as a product of prime powers (FTA!), this implies that we might multiply out the lefthand side of the infinite product of infinite sums to get $$\prod_{p}\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\cdots\right)=\sum_{n=1}^{\infty}\frac{1}{n}\, .$$ Even better, since each of the multiplied terms on the left is a geometric series, we can write $$\prod_{p}\left(\frac{1}{11/p}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\, .$$
The only problem with all this is that both of these things clearly diverge! So using $=$ is probably not the best idea. Nonetheless, Euler's intuition is spot on, and we will be able to fix this quite satisfactorily. For now, we can say that the harmonic series $$\zeta(1)=\sum_{n=1}^{\infty}\frac{1}{n}\text{ "equals" }\prod_{p}\left(\frac{1}{11/p}\right)=\prod_{p}\left(\frac{1}{1p^{1}}\right)\; .$$
To make this rigorous, we start talking convergence.
But why is the product the same thing?
In order to see this, let's instead show that our other main example of a sum over divisors equalling a product over primes works.
Click to the left again to hide and once more to show the dynamic interactive window 
This is the same setup, but for the sum/product combination $$\sum_{d\mid n}\frac{\mu(d)}{d}=\prod_{p\mid n}\left(1\frac{1}{p}\right)\, .$$ This would lead to the series $$\sum_{n=1}^\infty \frac{\mu(n)}{n^s}\; .$$
We call a series like this a Dirichlet series.
Definition:
In general, for an arithmetic function $f(n)$, its Dirichlet series is $$F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\, .$$
Three questions to see if we understand this:
Note that this already indicates some level of connection between arithmetic functions  connections which may not have been evident otherwise.
The very important thing to note about such series is that they often can indeed be expanded as infinite products as well. Something like $$\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_p (\text{ something involving }f(p)\text{ and }p^s)$$ If this is possible, then we say that the series has an Euler product form.
A potential Euler product, then, is one for the Dirichlet series of the Moebius function  $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\prod_{p}\left(1\frac{1}{p^s}\right)=\prod_p (1p^{s})$$  wherever these make sense.
We'll prove this works.
What is particularly cool about this is that now we see that the Euler products for the Riemann $\zeta$ function and this function are simply multiplicative inverses:
That is, we have that
This is not a coincidence.

0.0741344280534404 0.607927101854027 0.0741344280534404 0.607927101854027 
These are deep and amazing results, which we will exploit with many applications next time.
