Theorem: Product Form of Sum on Completely Multiplicative Function

Let $f$ be a completely multiplicative arithmetic function.
Let the series $\ds \sum_{n \mathop = 1}^\infty \map f n$ be absolutely convergent.
Then:
:$\ds \sum_{n \mathop = 1}^\infty \map f n = \prod_p \frac 1 {1 - \map f p}$
where the infinite product ranges over the primes.

Proof:
Define $P$ by:
{{begin-eqn}}
{{eqn | l = \map P {A, K}
      | o = :=
      | r = \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \frac {1 - \map f p^{K + 1} } {1 - \map f p}
      | c = where $\mathbb P$ denotes the set of prime numbers
}}
{{eqn | r = \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \paren {\sum_{k \mathop = 0}^K \map f p^k}
      | c = Sum of Geometric Sequence
}}
{{eqn | r = \sum_{v \mathop \in \prod \limits_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} } \paren {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \map f p^{v_p} }
      | c = Product of Summations is Summation Over Cartesian Product of Products
}}
{{eqn | r = \sum_{v \mathop \in \prod \limits_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} } \map f {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{v_p} }
      | c = as $f$ is completely multiplicative
}}
{{end-eqn}}
Change the summing variable using:
{{begin-eqn}}
{{eqn | l = \sum_{v \mathop \in V} \map g {\map h v} 
      | r = \sum_{w \mathop \in \set {\map h v: v \mathop \in V} } \map g w
      | c = where $h$ is a one to one mapping
}}
{{end-eqn}}
The Fundamental Theorem of Arithmetic guarantees a unique factorization for each positive natural number.
Therefore this function is one to one:
:$\ds \map h v = \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{v_p}$
Then:
{{begin-eqn}}
{{eqn | l = \map P {A, K}
      | r = \sum_{n \mathop \in \map Q {A, K} } \map f n
      | c = change of summing variable
}}
{{end-eqn}}
where $\map Q {A, K}$ is defined as:
:$\ds \map Q {A, K} := \set {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{-v_p} : v \in \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} }$
Consider:
{{begin-eqn}}
{{eqn | l = W
      | r = \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \map Q {A, K}
}}
{{eqn | r = \set {\prod_{p \mathop \in \mathbb P} p^{-v_p}: v \in \prod_{p \mathop \in \mathbb P} \set {0 \,.\,.\, \infty} }
      | c = 
}}
{{end-eqn}}
The construction defines it as the set of all possible products of positive powers of primes.
From the definition of a prime number, every positive natural number may be expressed as a prime or a product of powers of primes:
:$k \in \N^+ \implies k \in W$
and also every element of W is a positive natural number:
:$k \in W \implies k \in \N^+$
So $W = \N^+$.
Then taking limits on $\map P {A, K}$:
{{begin-eqn}}
{{eqn | l = \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \map P {A, K}
      | r = \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \frac {1 - \map f p^{K + 1} } {1 - \map f p}
      | c = taking limits of both sides of the definition of $\map P {A, K}$
}}
{{eqn | r = \prod_{p \mathop \in \mathbb P} \frac 1 {1 - \map f p}
      | c = $\map f p^{K + 1} \to 0$, because $\ds \sum_{n \mathop = 1}^\infty \map f n$ is convergent
}}
{{eqn | r = \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \sum_{n \mathop \in \map Q {A, K} } \map f n
      | c = from the expression for $\map P {A, K}$
}}
{{eqn | r = \sum_{n \mathop \in \N^+} \map f n
      | c = substituting for $\N^+$: order of summation is not defined
}}
{{eqn | r = \sum_{n \mathop = 1}^\infty \map f n
      | c = absolutely convergent, so the order does not alter the limit
}}
{{end-eqn}}
{{qed}}