In October 2026, OpenAI released something like 700 AI-generated math papers. One of those papers was about the Quasi-Riemann Hypothesis, which is about the Riemann zeta function.

I had heard of this function before but had essentially no understanding of it. I asked AI to explain it to me, and there’s quite a simple and beautiful insight that I’d like to reproduce here.

Here’s the surprising and beautiful (IMO) equality1 that we’re going to build up to:

\[\boxed{ \underbrace{\sum_{n=1}^{\infty}\frac{1}{n}}_{\text{All positive integers}} = \underbrace{\prod_{p\text{ prime}}\frac{1}{1-\frac{1}{p}}}_{\text{All prime numbers}} }\]

Do you remember the formula for the sum of an infinite series?

\[1 + s + s^2 + s^3 + \ldots = \frac{1}{1-s}\]

So, for example:

\[1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots = \frac{1}{1-\frac{1}{3}} = \frac{3}{2}\]

Now consider the following product:

\[\frac{1}{1-\frac{1}{2}} \cdot \frac{1}{1-\frac{1}{3}} \cdot \frac{1}{1-\frac{1}{5}}\]

We can rewrite this as:

\[(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots) \\ \cdot (1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots) \\ \cdot (1 + \frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \ldots)\]

Here’s the critical question: What terms will exist when we expand this product into an (infinite) sum? Here are the first few terms of the expanded product:

\[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \ldots\]

In order to see why these terms exist and not others, let’s give a name to the three infinite sums that we’re multiplying together:

  • $p2 = (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots)$
  • $p3 = (1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots)$
  • $p5 = (1 + \frac{1}{5} + \frac{1}{25} + \frac{1}{125} + \ldots)$

Remember that when you expand a product like $p2 \cdot p3 \cdot p5$, you get to pick one term from each of $p2$, $p3$, and $p5$ and then multiply them together. It’s like FOIL but generalized.

Now I’ll show what we picked from $p2$, $p3$, and $p5$ for a few of the terms in the final infinite sum (this one: $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \ldots$):

Term in final sum What we picked from $p2$ From $p3$ From $p5$
$1$ $1$ $1$ $1$
$\frac{1}{2}$ $\frac{1}{2}$ $1$ $1$
$\frac{1}{5}$ $1$ $1$ $\frac{1}{5}$
$\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$ $1$
$\frac{1}{9}$ $1$ $\frac{1}{9}$ $1$
$\frac{1}{10}$ $\frac{1}{2}$ $1$ $\frac{1}{5}$
$\frac{1}{24}$ $\frac{1}{8}$ $\frac{1}{3}$ $1$

Do you see the pattern? For any number $n$ whose prime factorization is of the form $2^a \cdot 3^b \cdot 5^c$, we’ll end up with a term of $\frac{1}{n}$ in our final sum by picking:

  • $\frac{1}{2^a}$ from $p2$
  • $\frac{1}{3^b}$ from $p3$
  • $\frac{1}{5^c}$ from $p5$

Put another way, our final sum will include exactly one $\frac{1}{n}$ term for every natural number $n$ whose prime factorization includes only the primes $2$, $3$, and $5$.

The missing terms (e.g. $\frac{1}{7}$, $\frac{1}{11}$, $\frac{1}{13}$, $\frac{1}{14}$, $\ldots$) are the terms corresponding to numbers with prime factors other than $2$, $3$, and $5$.

So the natural next step is… what if we just included all the primes? Let’s consider the following product:

\[\prod_{p \text{ prime}} \frac{1}{1-\frac{1}{p}}\]

which we can rewrite as:

\[\begin{align} &(1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots) \\ \cdot &(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots) \\ \cdot &(1 + \frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + \ldots) \\ \cdot &(1 + \frac{1}{7} + \frac{1}{7^2} + \frac{1}{7^3} + \ldots) \\ \cdot &(1 + \frac{1}{11} + \frac{1}{11^2} + \frac{1}{11^3} + \ldots) \\ \cdot &\ldots \end{align}\]

If we expand this product, we will get

\[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}+ \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \ldots\]

This product will include every natural number $n$ exactly once (since every natural number has a unique prime factorization), so we can just rewrite this infinite sum as

\[\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}}_{\text{All positive integers}}\]

So summing the reciprocals of the natural numbers is the same as multiplying one over one minus the reciprocal of the primes. How cool / weird / surprising is that!?

  1. Strictly speaking, both sides diverge to infinity, so this isn’t a finite equality. It’s shorthand for Euler’s product formula, which holds when each reciprocal is raised to a power $s > 1$; as $s$ approaches $1$ from above, both sides grow without bound. ↩