The most important such infinite process is the following fundamental function. It is one of the most studied, yet most mysterious functions in all of mathematics.
We define the zeta function (denoted \(\zeta\)) as the sum of the infinite series \begin{equation*}\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\end{equation*} as a function of \(s\).
For now we'll keep the domain of \(\zeta\) to be only the \(s\) where this series converges. Later, in Subsection 25.3.1, we'll see that it will be useful to think about what \(\zeta\) might mean for other values of \(s\).
Here we plot the function for a few positive values of \(s\).
Riemann, the quietly devout son of a Lutheran pastor, made ground-breaking contributions in nearly every area of mathematics. He did it in analysis (Riemann sums), in geometry (Riemannian metrics, later used by Einstein), in function theory (Riemann surfaces) – and in one paper that changed the course of number theory. He died quite young (around 40).
The motivation for this definition comes from this function with the case \(s=1\).
We begin with the second formula in Fact 24.1.1:\begin{equation*}\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}\end{equation*} Try computing both sides of this and seeing how they come together for a few fairly composite \(n\), like 12, 16, 18, 20, or 30.
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. This gives us a way into introducing limits.
What would happen if we introduced infinity in each term of the product, for instance? \begin{equation*}\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)\end{equation*} By analogy, we should get a sum with exactly one copy of the reciprocal of each number divisible by only 2 and 3, e.g. \begin{equation*}\sum_{2\mid n\text{ or }3 \mid n}\frac{1}{n}\, .\end{equation*}
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 (Fundamental Theorem of Arithmetic), this implies that we might multiply out the left-hand side of an infinite product of infinite sums to get \begin{equation*}\prod_{p}\left(1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+\cdots\right)=\sum_{n=1}^{\infty}\frac{1}{n}\, .\end{equation*} Since each of the multiplied terms on the left is a geometric series, we can simplify the product slighlty to write \begin{equation*}\prod_{p}\left(\frac{1}{1-1/p}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\, .\end{equation*}
So much for Euler's contribution, a very impressive one. The only problem with all this is that both of these things clearly diverge!
Thus we cannot use a simple equality (\(=\)) for this discussion. Nonetheless, Euler's intuition is spot on, and we will be able to fix this issue quite satisfactorily. For now, we can say is that, in some sense, the harmonic series is also an infinite product: \begin{equation*}\zeta(1)=\sum_{n=1}^{\infty}\frac{1}{n}\mathrel{\text{ “}\mathord{=}{” }}\prod_{p}\left(\frac{1}{1-1/p}\right)=\prod_{p}\left(\frac{1}{1-p^{-1}}\right)\; .\end{equation*}
To make this rigorous, we should start talking about convergence. Recall this informal version of the integral test for series.
Assume \(f\) is a positive decreasing function going to zero as \(x\to \infty\). Then the series \(\sum_{i=1}^n f(i)\) converges if and only if the integral \(\int_1^{\infty}f(x)dx\) converges.
How does this apply to our situation? The improper integral in this case is \begin{equation*}\int_1^{\infty} x^{-s}\; dx\end{equation*} which evaluates to \begin{equation*}\frac{-x^{-s+1}}{1-s}\biggr|_1^{\infty}=\frac{1}{1-s}\left(1-\lim_{x\to\infty}\frac{1}{x^{s-1}}\right)\, .\end{equation*} For \(s\) a real number, this converges precisely when \(s>1\) (since that keeps \(x\) in the denominator).
The infinite sum \(\zeta(s)\) converges for all \(s>1\).
But why is the (infinite) product equal to this infinite sum too? Is this product even meaningful? After all, it is not true in general that if a partial product equals a partial sum, then the ‘full’ sum is the ‘full’ product.
One has to carefully set up the convergence. If we can show that the product converges to the sum, then both will converge. Then it will make sense to say that \begin{equation*}\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_p \left(\frac{1}{1-p^{-s}}\right)\end{equation*}
