Product Form of Sum on Completely Multiplicative Function

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

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

Proof
From the sum of an infinite geometric progression:
 * $$\prod_{p \text{ prime}} \frac{1}{1 - f \left({p}\right)} = \prod_{p \text{ prime}} \left({f \left({p}\right)^0 + f \left({p}\right)^1 + f \left({p}\right)^2 + \cdots}\right) \ $$

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

Hence:
 * $$\prod_{p \text{ prime}} \left({f \left({1}\right) + f \left({p}\right) + f \left({p}\right)^2 + \dots }\right) = \prod_{p \text{ prime}} \left({f \left({1}\right) + f \left({p}\right) + f \left({p^2}\right) + \dots }\right) \ $$

Writing the product more explicitly, we obtain


 * $$\prod_{p \text{ prime}} \frac{1}{1 - f \left({p}\right)} = \left({ f(1) + f(2) + f(4) + f(8)+ \dots }\right) \left({ f(1) + f(3) + f(9) + f(27) + \dots }\right) \left({f(1) + f(5) + f(25) + \dots }\right) \dots \ $$

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:


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