## The Riemann zeta function

Recall from my previous post that given a function $f(n)$, we define $\zeta_f$, the Dirichlet generating function of $f$, by $\displaystyle \displaystyle \zeta_f(s) = \sum_{n \geq 1} \frac{f(n)}{n^s}.$

We also proved that $\zeta_f \zeta_g = \zeta_{f \ast g}$: the product of Dirichlet generating functions is the Dirichlet generating function of the Dirichlet convolution. Now, consider taking $f(n) = 1$ in the above definition. We get $\displaystyle \displaystyle \zeta_{\mathbf{1}}(s) = \sum_{n \geq 1} \frac{1}{n^s}$

which is also often written as just plain $\zeta(s)$. This function is quite famous: it is the Riemann zeta function. The reason it is so famous is because it is the subject of a famous unproved conjecture, the Riemann hypothesis, which in turn is famous because it has been both difficult to prove—many mathematicians have been attacking it for a long time—and deeply related to many other areas of mathmatics. In particular it is deeply related to prime numbers. If you want to understand the Riemann hypothesis better, I highly recommend reading the (truly wonderful) Secrets of Creation Trilogy, especially the first two volumes, which explain it from first principles. (I have previously written about the Secrets of Creation trilogy on this blog: a review of Volume 1 is here, and here is my review of Volume 2). In this post I just want to help you understand a few cool things about the zeta function.

Remember that $\mu \ast \mathbf{1} = \varepsilon$, so we must have $\zeta_\mu \zeta_{\mathbf{1}} = \zeta_\varepsilon$. Also recall that $\varepsilon(n) = 1$ when $n = 1$ but it equals $0$ otherwise, so in fact $\displaystyle \displaystyle \zeta_\varepsilon(s) = \sum_{n \geq 1} \frac{\varepsilon(n)}{n^s} = 1$

since the only nonzero term is $\varepsilon(1)/1^s = 1/1 = 1$. All together, then, we have $\displaystyle \zeta_\mu(s) \zeta(s) = 1$

and hence $\displaystyle \displaystyle \frac{1}{\zeta(s)} = \zeta_\mu(s) = \sum_{n \geq 1} \frac{\mu(n)}{n^s}$

For example, consider $\zeta(2)$: $\displaystyle \zeta(2) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots$

This converges to something, although a priori it is not obvious what. By writing a simple computer program, or by asking Wolfram Alpha, we can add up, say, 1000 terms and find that it is approximately $1.6439\dots$. Apparently, the reciprocal of this number is given by $\displaystyle \frac{1}{\zeta(2)} = \frac{1}{1^2} + \frac{-1}{2^2} + \frac{-1}{3^2} + \frac{0}{4^2} + \dots$

where each numerator is $\mu(n)$. Again, we can use a computer to check that this sum is approximately $0.6079\dots$: and sure enough, this is approximately $1/1.6439\dots$!

It turns out that $\zeta(2) = \sum_{n \geq 1} 1/n^2$ converges to exactly $\pi^2/6$ (!!!)—hopefully I can write another blog post explaining that (to be honest at the moment I don’t know how to prove it). We also know that $\displaystyle \displaystyle \zeta(4) = \sum_{n \geq 1} \frac{1}{n^4} = \frac{\pi^4}{90}$

and there is actually a formula giving $\zeta(2n)$ for any even positive integer. $\zeta(3) \approx 1.2020\dots$ is called Apéry’s constant, since Roger Apéry proved in 1978 that it is irrational; but we don’t know of any nice formula for it.

I will leave you with a few things to prove that you may find amusing:

• $\zeta(s)^2 = \zeta_\tau(s)$
• $\zeta(s)\zeta(s-1) = \zeta_\sigma(s)$

Recall that $\tau(n)$ is the number of divisors of $n$, and $\sigma(n)$ is the sum of divisors of $n$; to prove these you will want to reference some facts we proved about $\tau$ and $\sigma$ in terms of Dirichlet convolution.

Next time, we’ll see another way to relate the zeta function to prime numbers. 