# Euler Products

## Series with Multiplicative Structure

Euler products arise when an infinite series has coefficients controlled by multiplication. The simplest and most important example is the zeta series

$$
\sum_{n=1}^{\infty}\frac{1}{n^s},
$$

where $s$ is a real number greater than $1$. This condition ensures absolute convergence.

The key observation is that every positive integer factors uniquely into primes:

$$
n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}.
$$

Thus a sum over all positive integers can sometimes be decomposed into independent choices of prime powers.

## The Basic Euler Product

For $s>1$, the geometric series gives

$$
1+\frac1{p^s}+\frac1{p^{2s}}+\frac1{p^{3s}}+\cdots =
\left(1-\frac1{p^s}\right)^{-1}.
$$

Multiplying this identity over all primes gives

$$
\prod_p \left(1-\frac1{p^s}\right)^{-1}.
$$

Expanding the product means choosing one term from each prime factor. A typical choice gives

$$
\frac1{p_1^{a_1s}}\frac1{p_2^{a_2s}}\cdots\frac1{p_k^{a_ks}} =
\frac1{(p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k})^s}.
$$

By unique factorization, every positive integer occurs exactly once. Therefore

$$
\sum_{n=1}^{\infty}\frac1{n^s} =
\prod_p \left(1-\frac1{p^s}\right)^{-1}.
$$

This identity is Euler’s product formula for the zeta function.

## Finite Approximation

To avoid formal manipulation of infinite products, first restrict to primes at most $x$. Define

$$
P_x(s)=\prod_{p\leq x}\left(1-\frac1{p^s}\right)^{-1}.
$$

Expanding the finite product gives a finite-prime sum:

$$
P_x(s) =
\sum_{\substack{n\geq 1\\ p\mid n\Rightarrow p\leq x}}\frac1{n^s}.
$$

The summation is over all positive integers whose prime factors are at most $x$. These are called $x$-smooth integers. As $x\to\infty$, more primes are allowed, and the sum approaches the full zeta series.

Since the series converges absolutely for $s>1$, this limiting process is justified.

## Multiplicative Functions

Euler products also apply to Dirichlet series of multiplicative functions. An arithmetic function $f:\mathbb{N}\to\mathbb{C}$ is multiplicative if

$$
f(mn)=f(m)f(n)
$$

whenever $\gcd(m,n)=1$.

For such a function, the Dirichlet series

$$
\sum_{n=1}^{\infty}\frac{f(n)}{n^s}
$$

often factors as

$$
\prod_p \left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\frac{f(p^3)}{p^{3s}}+\cdots\right),
$$

provided the series and product converge absolutely.

This formula is the analytic form of multiplicativity. The values of $f$ on prime powers determine the whole series.

## Examples

For the constant function $f(n)=1$, the Euler product becomes the zeta product:

$$
\sum_{n=1}^{\infty}\frac1{n^s} =
\prod_p \left(1-\frac1{p^s}\right)^{-1}.
$$

For the Möbius function $\mu(n)$, whose values are

$$
\mu(p^0)=1,\qquad \mu(p)=-1,\qquad \mu(p^k)=0\quad(k\geq2),
$$

we get

$$
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s} =
\prod_p \left(1-\frac1{p^s}\right).
$$

Thus

$$
\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s} =
\frac1{\zeta(s)}
$$

for $s>1$.

For Euler’s totient function $\varphi(n)$, one obtains

$$
\sum_{n=1}^{\infty}\frac{\varphi(n)}{n^s} =
\frac{\zeta(s-1)}{\zeta(s)},
$$

valid in the region where the series converges absolutely, namely $s>2$.

## Arithmetic Meaning

Euler products express a central principle of number theory: global arithmetic is built from local prime data. Each factor in the product describes what happens at a single prime $p$. The full product combines all primes into one analytic object.

This viewpoint becomes fundamental in later topics. Dirichlet $L$-functions, Dedekind zeta functions, modular $L$-functions, and automorphic $L$-functions all have Euler products. In each case, the product records arithmetic information prime by prime.

Euler products therefore form one of the main bridges between prime factorization and analytic number theory.