Product Form of Sum on Completely Multiplicative Function

Theorem
Let $f$ be a completely multiplicative function.

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

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

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

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

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

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

where $p$ ranges over the primes.

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

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

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:


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