Product Form of Sum on Completely Multiplicative Function

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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

Define $P$ by:

\(\displaystyle P \left(A, K \right)\) \(:=\) \(\displaystyle \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \frac {1 - f \left({p}\right)^{K + 1} } {1 - f \left({p}\right)}\) where $\mathbb P$ denotes the set of prime numbers
\(\displaystyle \) \(=\) \(\displaystyle \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \left({\sum_{k \mathop = 0}^K f \left({p}\right)^k}\right)\) Sum of Geometric Progression
\(\displaystyle \) \(=\) \(\displaystyle \sum_{v \mathop \in \prod \limits_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \left\{ {0 \,.\,.\, K}\right\} } \left({\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } f \left({p}\right)^{v_p} }\right)\) Product of Summations is Summation Over Cartesian Product of Products
\(\displaystyle \) \(=\) \(\displaystyle \sum_{v \mathop \in \prod \limits_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \left\{ {0 \,.\,.\, K}\right\} } f \left({\prod_{p \mathop \in P}^{p \mathop \le A} p^{v_p} }\right)\) as $f$ is completely multiplicative


Change the summing variable using:

\(\displaystyle \sum_{v \mathop \in V} g \left({h \left({v}\right)}\right)\) \(=\) \(\displaystyle \sum_{w \mathop \in \left\{ {h \left({v}\right): v \mathop \in V}\right\} } g \left({w}\right)\) where $h$ is a one to one mapping


The Fundamental Theorem of Arithmetic guarantees a unique factorization for each positive natural number.

Therefore this function is one to one:

$\displaystyle h \left({v}\right) = \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{v_p}$


Then:

\(\displaystyle P \left({A, K}\right)\) \(=\) \(\displaystyle \sum_{n \mathop \in Q \left({A, K}\right)} f \left({n}\right)\) change of summing variable

where $Q \left({A, K}\right)$ is defined as:

$Q \left({A, K}\right) := \left\{ {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{-v_p} : v \in \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \left\{ {0 \,.\,.\, K}\right\} }\right\}$


Consider:

\(\displaystyle W\) \(=\) \(\displaystyle \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } Q \left({A, K}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left\{ {\prod_{p \mathop \in \mathbb P} p^{-v_p}: v \in \prod_{p \mathop \in \mathbb P} \left\{ {0 \,.\,.\, \infty}\right\} }\right\}\)

The construction defines it as the set of all possible products of positive powers of primes.

From the definition of a prime number, every positive natural number may be expressed as a prime or a product of powers of primes:

$k \in \N^+ \implies k \in W$

and also every element of W is a positive natural number:

$k \in W \implies k \in \N^+$

So $W = \N^+$.


Then taking limits on $P \left({A, K}\right)$:

\(\displaystyle \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } P \left({A, K}\right)\) \(=\) \(\displaystyle \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \frac{1 - f \left({p} \right)^{K + 1} } {1 - f \left({p}\right)}\) taking limits of both sides of the definition of $P(A, K)$
\(\displaystyle \) \(=\) \(\displaystyle \prod_{p \mathop \in \mathbb P} \frac 1 {1 - f \left({p}\right)}\) $f \left({p}\right)^{K + 1} \to 0$, because $\displaystyle \sum_{n \mathop = 1}^{\infty} f \left({n}\right)$ is convergent
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \sum_{n \mathop \in Q \left({A, K}\right)} f \left({n}\right)\) from the expression for $P \left({A, K}\right)$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop \in \N^+} f \left({n}\right)\) substituting for $\N^+$: order of summation is not defined
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty f \left({n}\right)\) absolutely convergent, so the order does not alter the limit

$\blacksquare$


Note


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When the function $f$ is multiplicative but not completely multiplicative, the above derivation is still valid, except that 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:

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