[proofplan] We first absorb the factor $n^{-s}$ into a new multiplicative function $g(n)=f(n)n^{-s}$. For a finite set of primes $P$, absolute convergence lets us multiply the finitely many local [power series](/page/Power%20Series), and unique prime factorisation identifies the resulting terms with exactly those integers whose prime factors all lie in $P$. Finally, the absolute convergence of the full Dirichlet series controls the complementary tail and shows that these finite prime products converge to the full series. [/proofplan] [step:Encode the Dirichlet coefficients as an absolutely summable multiplicative function] Define the function \begin{align*} g: \mathbb{N} &\to \mathbb{C} \\ n &\mapsto f(n)n^{-s}. \end{align*} Since $f$ is multiplicative and $n^{-s}=\exp(-s\log n)$, if $\gcd(m,n)=1$, then \begin{align*} g(mn) &= f(mn)(mn)^{-s} \\ &= f(m)f(n)m^{-s}n^{-s} \\ &= g(m)g(n). \end{align*} Also $g(1)=f(1)1^{-s}=1$. Thus $g$ is multiplicative. The hypothesis is precisely \begin{align*} \sum_{n=1}^{\infty} |g(n)| < \infty. \end{align*} For each prime $p \in \mathcal{P}$, the local absolute series is a subseries of the absolutely convergent series $\sum_{n=1}^{\infty}|g(n)|$: \begin{align*} \sum_{a=0}^{\infty} |g(p^a)| \leq \sum_{n=1}^{\infty} |g(n)| <\infty. \end{align*} Therefore \begin{align*} \sum_{a=0}^{\infty} f(p^a)p^{-as} = \sum_{a=0}^{\infty} g(p^a) \end{align*} converges absolutely for every prime $p$. [/step] [step:Expand each finite prime product by unique prime factorisation] Let $P \subset \mathcal{P}$ be a finite set of primes. Define the set of $P$-smooth positive integers \begin{align*} \mathbb{N}_P := \{n \in \mathbb{N} : \text{if } p \in \mathcal{P} \text{ and } p \mid n, \text{ then } p \in P\}. \end{align*} We prove that \begin{align*} \prod_{p \in P}\left(\sum_{a=0}^{\infty} g(p^a)\right) = \sum_{n \in \mathbb{N}_P} g(n). \end{align*} Because $P$ is finite and each local series $\sum_{a=0}^{\infty}|g(p^a)|$ converges, the finite product of local absolute sums is finite: \begin{align*} \prod_{p \in P}\left(\sum_{a=0}^{\infty} |g(p^a)|\right)<\infty. \end{align*} Hence the finite product of the local series may be expanded as the absolutely convergent multiple series \begin{align*} \prod_{p \in P}\left(\sum_{a=0}^{\infty} g(p^a)\right) = \sum_{(a_p)_{p \in P} \in \mathbb{N}_0^P} \prod_{p \in P} g(p^{a_p}), \end{align*} where $\mathbb{N}_0:=\{0,1,2,\dots\}$ and $\mathbb{N}_0^P$ denotes the set of functions $P \to \mathbb{N}_0$, written $p \mapsto a_p$. For each exponent vector $(a_p)_{p \in P} \in \mathbb{N}_0^P$, define \begin{align*} n((a_p)_{p \in P}) := \prod_{p \in P} p^{a_p}. \end{align*} The factors $p^{a_p}$ are pairwise coprime as $p$ varies through $P$, so repeated use of the multiplicativity of $g$ gives \begin{align*} \prod_{p \in P} g(p^{a_p}) = g\left(\prod_{p \in P}p^{a_p}\right) = g(n((a_p)_{p \in P})). \end{align*} By unique prime factorisation, the map \begin{align*} \mathbb{N}_0^P &\to \mathbb{N}_P \\ (a_p)_{p \in P} &\mapsto \prod_{p \in P} p^{a_p} \end{align*} is a bijection. Therefore \begin{align*} \prod_{p \in P}\left(\sum_{a=0}^{\infty} g(p^a)\right) = \sum_{n \in \mathbb{N}_P} g(n). \end{align*} [guided] Fix a finite set $P \subset \mathcal{P}$. The goal is to understand exactly which integers are produced when we multiply the local factors indexed by the primes in $P$. Define \begin{align*} \mathbb{N}_P := \{n \in \mathbb{N} : \text{if } p \in \mathcal{P} \text{ and } p \mid n, \text{ then } p \in P\}. \end{align*} Thus $\mathbb{N}_P$ consists of the positive integers whose prime divisors are all contained in $P$; the integer $1$ belongs to $\mathbb{N}_P$ because it has no prime divisors. First, the product of local series is legitimate because there are only finitely many primes in $P$, and each local series is absolutely convergent. More precisely, \begin{align*} \prod_{p \in P}\left(\sum_{a=0}^{\infty} |g(p^a)|\right)<\infty. \end{align*} This absolute convergence permits the finite product of series to be expanded as the absolutely convergent multiple series \begin{align*} \prod_{p \in P}\left(\sum_{a=0}^{\infty} g(p^a)\right) = \sum_{(a_p)_{p \in P} \in \mathbb{N}_0^P} \prod_{p \in P} g(p^{a_p}), \end{align*} where $\mathbb{N}_0:=\{0,1,2,\dots\}$ and $\mathbb{N}_0^P$ is the set of exponent choices $p \mapsto a_p$. Now fix an exponent vector $(a_p)_{p \in P} \in \mathbb{N}_0^P$, and define the corresponding integer \begin{align*} n((a_p)_{p \in P}) := \prod_{p \in P} p^{a_p}. \end{align*} Because distinct prime powers are coprime, multiplicativity of $g$ gives \begin{align*} \prod_{p \in P} g(p^{a_p}) = g\left(\prod_{p \in P}p^{a_p}\right) = g(n((a_p)_{p \in P})). \end{align*} The reason this captures every term exactly once is unique prime factorisation: every integer in $\mathbb{N}_P$ has a unique expression as $\prod_{p \in P}p^{a_p}$ with $a_p \in \mathbb{N}_0$, and every such product lies in $\mathbb{N}_P$. Hence the map \begin{align*} \mathbb{N}_0^P &\to \mathbb{N}_P \\ (a_p)_{p \in P} &\mapsto \prod_{p \in P} p^{a_p} \end{align*} is a bijection. Reindexing the absolutely convergent multiple series through this bijection yields \begin{align*} \prod_{p \in P}\left(\sum_{a=0}^{\infty} g(p^a)\right) = \sum_{n \in \mathbb{N}_P} g(n). \end{align*} [/guided] [/step] [step:Use absolute convergence to pass from finite prime sets to all primes] Let \begin{align*} S := \sum_{n=1}^{\infty} g(n). \end{align*} For a finite set $P \subset \mathcal{P}$, define the finite prime product \begin{align*} E_P := \prod_{p \in P}\left(\sum_{a=0}^{\infty} g(p^a)\right). \end{align*} By the previous step, \begin{align*} E_P = \sum_{n \in \mathbb{N}_P} g(n). \end{align*} Therefore \begin{align*} S-E_P = \sum_{n \in \mathbb{N}\setminus \mathbb{N}_P} g(n), \end{align*} where the series on the right is absolutely convergent as a subseries of $\sum_{n=1}^{\infty}|g(n)|$. Hence \begin{align*} |S-E_P| \leq \sum_{n \in \mathbb{N}\setminus \mathbb{N}_P} |g(n)|. \end{align*} Let $\varepsilon>0$. Since $\sum_{n=1}^{\infty}|g(n)|$ converges, there exists $N \in \mathbb{N}$ such that \begin{align*} \sum_{n>N} |g(n)| < \varepsilon. \end{align*} Let $P_N$ be the finite set of primes dividing at least one integer $n \in \{1,\dots,N\}$. If $P \subset \mathcal{P}$ is finite and $P_N \subset P$, then every integer $n \leq N$ belongs to $\mathbb{N}_P$. Thus \begin{align*} \mathbb{N}\setminus \mathbb{N}_P \subset \{n \in \mathbb{N}: n>N\}, \end{align*} and consequently \begin{align*} |S-E_P| \leq \sum_{n \in \mathbb{N}\setminus \mathbb{N}_P} |g(n)| \leq \sum_{n>N}|g(n)| <\varepsilon. \end{align*} This proves that the net $(E_P)_{P \subset \mathcal{P},\, P \text{ finite}}$ converges to $S$ as $P$ increases by inclusion. Substituting back $g(n)=f(n)n^{-s}$ gives \begin{align*} \lim_P \prod_{p \in P} \left(\sum_{a=0}^{\infty} f(p^a)p^{-as}\right) = \sum_{n=1}^{\infty} f(n)n^{-s}. \end{align*} This is exactly the asserted Euler product formula. [/step]