# Understood Euler Product Formula

## Before I understood this formula

When I was a high school student, it’s hard for me to imagine a product whose index looped through all prime numbers because primes don’t appear in a regular way: between 1 and 100, there’re 25 primes, but between 900 and 1000, there’re 14.

Even though it’s easier to imagine the infinite sum whose $n$-th term is $n^{-s}$, without learning the $p$-Test and the Comparison Test for the convergence of infinite sums, I couldn’t understand why the infinite sum in the following equality exists.

$\sum_{k = 1}^{\infty} \frac{1}{k^s} = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}$

## A heuristic way to understand it

1. Note that the geometric series converge (absolutely).
2. Borrow the proofs about the Comparison Test and the $p$-Test to convince yourself that the infinite sum on the LHS of the formula is well-defined.
3. Know something about convergent infinite products.1
4. Convince yourself that the infinite product on the RHS exists.2
5. Recall the Fundamental Theorem of Arithmetic.

I think that the last item is the trickiest step. Writing the following lines, I understood this equation.

\begin{align} & 1^{-s} + 2^{-s} + 3^{-s} + \cdots \\ =& (1^{-s} + 3^{-s} + 5^{-s} + \cdots) (1^{-s} + 2^{-s} + 2^{-2s} + \cdots) \label{step1} \\ =& (1^{-s} + 3^{-s} + 5^{-s} + \cdots) \cdot \frac{1}{1 - 2^{-s}} \label{step2} \\ =& (1^{-s} + 5^{-s} + 7^{-s} + \cdots) (1^{-s} + 3^{-s} + 3^{-2s} + \cdots) \cdot \frac{1}{1 - 2^{-s}} \label{step3} \\ =& (1^{-s} + 5^{-s} + 7^{-s} + \cdots) \cdot \frac{1}{1 - 3^{-s}} \cdot \frac{1}{1 - 2^{-s}} \label{step4} \end{align}

Steps \eqref{step1} (resp. \eqref{step3}) holds because for each $k^{-s}$ in the leftmost bracket, powers of 2 (resp. 3) can be taken out from $k$. In steps \eqref{step2} and \eqref{step4}, the formula for the sum of geometric series is applied.

## Refining the above thoughts

I’ll end this post by wrapping up the above ideas by summation and product signs.

\begin{aligned} & \sum_{k = 1}^{\infty} \frac{1}{k^s} \\ =& \sum_{k \in \N} \frac{1}{k^s} \\ =& \prod_{p \text{ prime}} \sum_{k \in \N} \frac{1}{p^{s k}} \\ =& \prod_{p \text{ prime}} \sum_{k = 1}^{\infty}\frac{1}{p^{s k}}\\ =& \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}} \end{aligned}

1. One may refer to the convergence criteria of infinite products on Wikipedia

2. There’s more than one way to do it. When I tried to do this for the first time, I used the Mean Value Theorem on logarithms to establish a standard result.

$\frac{x}{1 + x} < \log(1 + x) < x \quad \forall\, x > 0$