# The zeta function

Here’s a function that is super important in analytic number theory:

$\zeta(s)=\displaystyle\sum_{n=1}^\infty n^{-s}=\frac1{1^s}+\frac1{2^s}+\frac1{3^s}+\cdots$.

There is way more to be said about this function than I can possible fit in one, or even seventeen posts, so I won’t even attempt to talk about the Riemann hypothesis. Here is a fun fact though:

$\zeta(s)=\displaystyle\sum_{n=1}^\infty n^{-s}=\prod_p\frac1{1-p^{-s}}$,

where the product is over all prime numbers $p$. How does this work? Well, each term $\frac1{1-p^{-s}}$ is the sum of a geometric series. Expanding the sum, we get

$\displaystyle\prod_p\frac1{1-p^{-s}}=(1+2^{-s}+2^{-2s}+\cdots)(1+3^{-s}+3^{-2s}+\cdots)\cdot\dots$

If we were to expand the product of infinite sums, we need to take some entry from each sum in the product. Say we take the $3^{-s}$ term, the $5^{-3s}$ term, and the rest ones. When we multiply all of these together, we get $375^{-s}$. We get each positive integer to show up exactly once (by unique factorization). Cool!

So what is $\zeta(1)$? According to the sum, it’s $1+\frac12+\frac13+\cdots$. This sum does not converge. If you didn’t already know this, there are lots of proofs on the internet.

So if $\zeta(1)$ is infinite, it better be infinite when we use the product definition. It’s obvious that a finite product of finite numbers is finite, so if $\zeta(1)$ is going to be infinite, there had better be infinitely many primes.