From the discussion A simple way to obtain $\prod_{p \in \mathbb{P}} \frac{1}{1-p^{-s}} = \sum_{n =1}^{\infty} \frac{1}{n^s}$ on math.stackexchange, I learnt an elementary proof of Euler product formula; in fact, I found it so remarkable that I decided to write a post on it.

We only need the two following basic lemmas; only sketchs of proof are given, we refer to Jean Jacod’s book, Probability essentials, for more information.

Definition: $\{ E_i \mid i \in I \}$ is a family of independent events if for every finite subset $J \subset I$,

$\displaystyle P \left( \bigcap\limits_{j \in J} E_j \right)= \prod\limits_{j \in J} P(E_j)$.

Lemma: Let $P$ be a probability measure and $\{ E_i \mid i \geq 1 \}$ be a family of independent events. Then

$\displaystyle P \left( \bigcap\limits_{i \geq 1} E_i \right) = \prod\limits_{i \geq 1} P(E_i)$.

Sketch of proof. $\left( \bigcap\limits_{i=1}^n E_i \right)$ is a decreasing sequence of events, hence

$\displaystyle P \left( \bigcap\limits_{i \geq 1} E_i \right) = \lim\limits_{n \to + \infty} P \left( \bigcap\limits_{i=1}^n E_i \right)= \lim\limits_{n \to + \infty} \prod\limits_{i=1}^n P(E_i) = \prod\limits_{i \geq 1} P(E_i)$. $\square$

Lemma: If $\{ E_i \mid i \in I \}$ is a familiy of independent events, then $\{ E_i^c \mid i \in I \}$ so is.

Sketch of proof. Using inclusion-exclusion principle, prove by induction on $n$ that every subfamily $\{E_{i_1}, \dots E_{i_n} \}$ of cardinality $n$ satisfies

$P \left( \bigcap\limits_{k=1}^n E_{i_k}^c \right)= \prod\limits_{k=1}^n P \left( E_{i_k}^c \right)$.

For $n=2$, we have

$\begin{array}{lcl} P(E_1^c \cap E_2^c ) & = & 1- P(E_1 \cup E_2)= 1- P(E_1) - P(E_2 ) + P(E_1 \cap E_2) \\ \\ & = & 1- P(E_1) - P(E_2) + P(E_1) P(E_2) = ( 1 - P(E_1)) (1- P(E_2)) \\ \\ & = & P(E_1^c) P(E_2^c) \hspace{1cm} \square \end{array}$

Theorem: Let $s > 1$. Then $\displaystyle \zeta(s):= \sum\limits_{n \geq 1} \frac{1}{n^s} = \prod\limits_{p \in \mathbb{P}} \frac{1}{1-p^{-s}}$.

Proof. Let $X : \mathbb{N} \to \mathbb{N}$ be a random variable and $P$ be a probability measure such that

$\displaystyle P(X=n)= \frac{\zeta(s)^{-1}}{n^s}$;

for example, take $X= \mathrm{Id}$ and $P ( \{n \})= \frac{\zeta(s)^{-1}}{n^s}$. Let $E_k$ be the event “$X$ is divisible by $k$“. Notice that

$\displaystyle P(E_k)= \sum\limits_{i \geq 1} P(X=ik) = \sum\limits_{i \geq 1} \frac{\zeta(s)^{-1}}{n^s} = k^{-s}$.

Therefore, for any primes $k_1 , \dots, k_r$,

$\displaystyle P \left( \bigcap\limits_{j=1}^r E_{k_j} \right)= P \left( E_{\prod\limits_{j=1}^r k_j} \right) = \prod\limits_{j=1}^r k_j^{-s} = \prod\limits_{j=1}^r P(E_{k_j})$.

We deduce that $\{ E_k \mid k \ \text{prime} \}$ is a family of independent events. Because $X=1$ if and only if $X$ is not divisible by any prime, we have

$\displaystyle \frac{1}{\zeta(s)} = P(X=1)= P \left( \bigcap\limits_{p \in \mathbb{P}} E_p^c \right) =\prod\limits_{p \in \mathbb{P}} \left( 1- P(E_p) \right) = \prod\limits_{p \in \mathbb{P}} \left( 1-p^{-s} \right)$. $\square$

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