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
Let $f$ be a completely multiplicative function such that the sum:


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

converges absolutely.

As $f$ is completely multiplicative:
 * $f \left({p^k}\right) = f \left({p}\right)^k$

This implies that $f \left({p}\right) < 1$ for all primes $p$.

Thus for all primes $p$:

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

Now consider the finite product of absolutely convergent series:


 * $\displaystyle P \left({x}\right) = \prod_{p \mathop \le x} \frac 1 {1 - f \left({p}\right)} = \prod_{p \le x} \left({ 1 + f \left({p}\right) + f \left({p^2}\right) + \cdots}\right)$

Here we may freely rearrange the terms, so expanding the product we see that each term has the form:


 * $\displaystyle f \left(p_1^{e_1}\right)\cdots f \left(p_k^{e_k}\right) = f\left( p_1^{e_1}\cdots p_k^{e_k} \right)$

with the $e_i$ integers and each $p_i$ not larger than $x$.

Therefore, by the Fundamental Theorem of Arithmetic:


 * $\displaystyle P \left({x}\right) = \sum_{n \mathop \in \mathcal A_x} f \left({n}\right)$

where $\mathcal A_x$ is the set of all natural numbers with no prime factor larger than $x$.

So:


 * $\displaystyle \left| \sum_{n \mathop = 1}^\infty f \left({n}\right) - P \left({x}\right) \right| \le \sum_{n \mathop \in \mathcal B} \left|f \left({n}\right)\right|$

where $\mathcal B$ is the set of natural numbers with at least one prime factor larger than $x$.

Clearly $\mathcal B$ is contained in the integers larger than $x$, so:


 * $\displaystyle \left| \sum_{n \mathop = 1}^\infty f \left({n}\right) - P \left({x}\right) \right| \le \sum_{n \mathop > x} \left|f \left({n}\right)\right|$

By assumption the falls to zero as $x \to \infty$.

Passing to this limit we have the result.

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:


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