Product Form of Sum on Completely Multiplicative Function

Theorem

Let $f$ be a completely multiplicative arithmetic function.

Suppose that the series $\displaystyle \sum_{n=1}^\infty f(n)$ is absolutely convergent.

Then:

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

where $p$ ranges over the primes.

Proof

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

$\displaystyle \sum_{n=1}^\infty f(n)$

converges absolutely.

First, note that because $f(p^k) = f(p)^k$ this implies that $f(p) < 1$ for all primes $p$.

Therefore, 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.

We also have, for all primes $p$:

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

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

Now consider the finite product of absolutely convergent series:

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

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 we can write

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

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

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

where $\mathcal B$ is the set of 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 = 1}^\infty f(n) - P(x) \right| \leq \sum_{n > x} \left|f(n)\right| $

By assumption the right hand side falls to zero as $x \to \infty$, so passing to this limit we have the result.

$\blacksquare$

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:

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

Therefore, in this case we may write

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