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: :$\displaystyle \map h v = \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{v_p}$ 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: :$\displaystyle \map Q {A, K} := \set {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{-v_p} : v \in \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} }$ 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 $\displaystyle \sum_{n \mathop = 1}^{\infty} \map f n$ 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}}