Product Form of Sum on Completely Multiplicative Function

Theorem
Let $f$ be a completely multiplicative arithmetic function.

Let the series $\displaystyle \sum_{n \mathop = 1}^\infty f \left({n}\right)$ be absolutely convergent.

Then:


 * $\displaystyle \sum_{n \mathop = 1}^\infty f \left({n}\right) = \prod_p \frac 1 {1 - f \left({p}\right)}$

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.
 * $\displaystyle h(v) = \prod_{p \in P}^{p \le A} p^{v_p} $

The change of variable gives,

Define Q by,

Then,

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. Then,
 * $\displaystyle k \in \mathbb{N} \implies k \in W $

also every element of W is a positive natural number
 * $\displaystyle k \in W \implies k \in \mathbb{N} $

So $ W = \mathbb{N} $.

Then taking limits on P:

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


 * $\dfrac 1 {1 - f \left({p}\right)} = \left({ 1 + f \left({p}\right) + f \left({p^2}\right) + \cdots }\right)$

Therefore, in this case we may write:


 * $\displaystyle \sum_{n \mathop = 1}^\infty f \left({n}\right) = \prod_p \left({1 + f \left({p}\right) + f \left({p^2}\right) + \cdots}\right)$