Product Form of Sum on Completely Multiplicative Function

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

Let $$\forall p \in \mathbb{P}: \left |{f \left({n}\right)}\right| < 1$$, where $$\mathbb{P}$$ is the set of all prime numbers.

Then:
 * $$ \sum_{n=1}^\infty f \left({n}\right) = \prod_{p \in \mathbb{P}} \frac{1}{1 - f \left({p}\right)} \ $$

Proof
So, let $$f \ $$ be a completely multiplicative function such that $$\forall p \in \mathbb{P}: \left |{f \left({p}\right)}\right| < 1$$.

From the sum of an infinite geometric progression, we have that:
 * $$\sum_{i=0}^\infty f \left({p}\right)^n = \frac{1}{1 - f \left({p}\right)}$$

and, as $$\left |{f \left({p}\right)}\right| < 1$$, is absolutely convergent.

Since $$f \ $$ is completely multiplicative, we have that $$\forall k \in \N: f \left({p}\right)^k = f \left({p^k}\right)$$.

So $$\frac{1}{1 - f \left({p}\right)} = \sum_{i=0}^\infty f \left({p}\right)^n = \sum_{i=0}^\infty f \left({p^n}\right)$$.

Thus $$\sum_{i=0}^\infty f \left({p^n}\right)$$ is also absolutely convergent.

So we are able to use the corollary to Product of Sums and form:
 * $$\prod_{p \text{ prime}} \frac{1}{1 - f \left({p}\right)} = \prod_{p \text{ prime}} \left({\sum_{i=0}^\infty f \left({p^n}\right)}\right) \ $$

Expanding the summation:
 * $$\prod_{p \text{ prime}} \frac{1}{1 - f \left({p}\right)} = \prod_{p \text{ prime}} \left({f \left({1}\right) + f \left({p}\right) + f \left({p^2}\right) + f \left({p^3}\right) + \cdots}\right) \ $$

Writing the product explicitly, we obtain:
 * $$\prod_{p \text{ prime}} \frac{1}{1 - f \left({p}\right)} = \left({ f(1) + f(2) + f(4) + f(8)+ \cdots}\right) \left({ f(1) + f(3) + f(9) + f(27) + \cdots}\right) \left({f(1) + f(5) + f(25) + \cdots}\right) \cdots \ $$

Each term of the sum is the sum of $$f \ $$ of the prime powers of a certain prime.

Expanding this product, then, will create a sum, each term of which is precisely $$f \ $$ of some string of distinct primes, each to a certain power.

By the Fundamental Theorem of Arithmetic, such a set is precisely $$\N \ $$. Hence:


 * $$\prod_{p \text{ prime}} \frac{1}{1 - f \left({p}\right)}= \sum_{n=1}^\infty f \left({n}\right) \ $$