Proof of Euler Product Formula

We prove that Dirichlet series of multiplicative functions admit Euler product representations, establishing the bridge between additive and multiplicative structures.

TheoremEuler Product (Restated)

Let $f$ be a multiplicative function and suppose $F(s) = \sum_{n=1}^{\infty} f(n)/n^s$ converges absolutely for $\Re(s) > \sigma$ . Then:

F(s) = \prod_{p \text{ prime}} \left(\sum_{k=0}^{\infty} \frac{f(p^k)}{p^{ks}}\right)

for $\Re(s) > \sigma$ .

ProofProof via Fundamental Theorem of Arithmetic

Step 1: Finite Products

Fix a finite set of primes $\mathcal{P} = \{p_1, \ldots, p_N\}$ . Consider the partial product:

P_N(s) = \prod_{p \in \mathcal{P}} \left(\sum_{k=0}^{\infty} \frac{f(p^k)}{p^{ks}}\right)

Expanding the product:

P_N(s) = \sum_{a_1, \ldots, a_N \geq 0} \frac{f(p_1^{a_1}) \cdots f(p_N^{a_N})}{p_1^{a_1 s} \cdots p_N^{a_N s}}

Since $f$ is multiplicative and $p_1^{a_1}, \ldots, p_N^{a_N}$ are pairwise coprime:

f(p_1^{a_1} \cdots p_N^{a_N}) = f(p_1^{a_1}) \cdots f(p_N^{a_N})

Therefore:

P_N(s) = \sum_{n \in S_N} \frac{f(n)}{n^s}

where $S_N$ is the set of positive integers whose prime factors all lie in $\mathcal{P}$ .

Step 2: Convergence to Full Sum

For $\Re(s) > \sigma$ , absolute convergence gives:

\left| F(s) - P_N(s) \right| = \left| \sum_{n \notin S_N} \frac{f(n)}{n^s} \right| \leq \sum_{n \notin S_N} \frac{|f(n)|}{n^{\Re(s)}}

As $N \to \infty$ , the set $S_N$ includes all integers up to arbitrarily large bounds, so the tail sum vanishes:

\sum_{n \notin S_N} \frac{|f(n)|}{n^{\Re(s)}} \to 0

Step 3: Absolute Convergence of Product

We need to verify the infinite product converges. For $\Re(s) > \sigma$ :

\sum_{p} \sum_{k=1}^{\infty} \frac{|f(p^k)|}{p^{k\Re(s)}} < \infty

This follows from absolute convergence of $F(s)$ , since:

\sum_{n=1}^{\infty} \frac{|f(n)|}{n^{\Re(s)}} \geq \sum_{p} \sum_{k=1}^{\infty} \frac{|f(p^k)|}{p^{k\Re(s)}}

Therefore the product $\prod_p (1 + \sum_{k=1}^{\infty} f(p^k)/p^{ks})$ converges absolutely.

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ExampleCompletely Multiplicative Case

If $f$ is completely multiplicative, then $f(p^k) = f(p)^k$ , so:

\sum_{k=0}^{\infty} \frac{f(p^k)}{p^{ks}} = \sum_{k=0}^{\infty} \left(\frac{f(p)}{p^s}\right)^k = \frac{1}{1 - f(p)p^{-s}}

This gives the simplified Euler product:

F(s) = \prod_{p} \frac{1}{1 - f(p)p^{-s}}

Remark

The proof exploits unique prime factorization: every integer $n$ can be uniquely written as $\prod_p p^{a_p}$ , and multiplicativity ensures $f(n) = \prod_p f(p^{a_p})$ . The Euler product is a continuous analogue of this discrete decomposition.