Section24.2The Riemann Zeta Function

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.

Definition24.2.1Riemann zeta function
We define the zeta (\(\zeta\)) function as the sum of the infinite series \[\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=1+\frac{1}{2^s}+\frac{1}{3^s}+\cdots\] as a function of \(s\).

Several notes are in order about this.

  • Riemann, the quietly devout son of a Lutheran pastor, made ground-breaking contributions in nearly every area of mathematics. In analysis (Riemann sums), geometry (Riemannian metrics, later used by Einstein), in function theory (Riemann surfaces) — and one paper that changed the course of number theory. He died quite young (around 40).
  • For now we'll keep the domain of \(\zeta\) to be only the \(s\) where this series converges.
  • Later, we'll see that it will be useful to think about for other values of \(s\).

Subsection24.2.1Motivating the Zeta Function

We'll motivate this function with the case \(s=1\).

Start with our last equation (for \(\sigma\)), \[\prod_{p\mid n}\left(1+\frac{1}{p}+\frac{1}{p^{2}}+\cdots +\frac{1}{p^{e}}\right)=\sum_{d\mid n}\frac{1}{d}\] Let's 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. 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. \[\sum_{2\mid n\text{ or }3 \mid n}\frac{1}{n}\, .\]

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 left-hand 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 \[\prod_{p}\left(\frac{1}{1-1/p}\right)=\sum_{n=1}^{\infty}\frac{1}{n}\, .\]

Subsection24.2.2Being careful

So much for Euler's contribution, a very impressive one. The only problem with all this is that both of these things clearly diverge! Oops.

So using the simple equality (\(=\)) for this 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 \[\zeta(1)=\sum_{n=1}^{\infty}\frac{1}{n}\text{ equals }\prod_{p}\left(\frac{1}{1-1/p}\right)=\prod_{p}\left(\frac{1}{1-p^{-1}}\right)\; .\]

To make this rigorous, we start talking convergence. Recall this informal version of the integral test for series.

How does this apply? The improper integral in this case is \[\int_1^{\infty} x^{-s}\; dx\] which evaluates to \[\frac{-x^{-s+1}}{1-s}\biggr|_1^{\infty}=\frac{1}{1-s}\left(1-\lim_{x\to\infty}\frac{1}{x^{s-1}}\right)\, .\] For \(s\) a real number, this converges precisely when \(s>1\) (since that keeps \(x\) in the denominator).

So \(\zeta(s)\) converges for all \(s>1\).

But why is the (infinite) product equal to this infinite sum too? Is is even meaningful?

One has to carefully set up the convergence here, too. It is NOT true in general that if a partial product equals a partial sum, then the “full” sum is the “full” product.

But if we can show that the product converges to the sum, then both will converge. Then it will make sense to say that \[\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}=\prod_p \left(\frac{1}{1-p^{-s}}\right)\]