Product Form of Sum on Completely Multiplicative Function

Theorem
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:

Change the summing variable using:

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:

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:

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 $\map P {A, K}$:

Note
When the function $f$ is multiplicative but not completely multiplicative, the above derivation is still valid, except that we do not have the equality:


 * $\dfrac 1 {1 - \map f p} = \paren {1 + \map f p + \map f {p^2} + \cdots}$

Therefore, in this case we may write:


 * $\ds \sum_{n \mathop = 1}^\infty \map f n = \prod_p \paren {1 + \map f p + \map f {p^2} + \cdots}$