Product Form of Sum on Completely Multiplicative Function

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Theorem

Let $f$ be a completely multiplicative arithmetic function.

Let the series $\displaystyle \sum_{n \mathop = 1}^\infty \map f n$ be absolutely convergent.

Then:

$\displaystyle \sum_{n \mathop = 1}^\infty \map f n = \prod_p \frac 1 {1 - \map f p}$

where the infinite product ranges over the primes.


Proof

Define $P$ by:

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


Change the summing variable using:

\(\displaystyle \sum_{v \mathop \in V} \map g {\map h v}\) \(=\) \(\displaystyle \sum_{w \mathop \in \set {\map h v: v \mathop \in V} } \map g w\) 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:

$\displaystyle \map h v = \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{v_p}$


Then:

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

where $\map Q {A, K}$ is defined as:

$\displaystyle \map Q {A, K} := \set {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{-v_p} : v \in \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} }$


Consider:

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

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 $\map P {A, K}$:

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

$\dfrac 1 {1 - \map f p} = \paren {1 + \map f p + \map f {p^2} + \cdots}$

Therefore, in this case we may write:

$\displaystyle \sum_{n \mathop = 1}^\infty \map f n = \prod_p \paren {1 + \map f p + \map f {p^2} + \cdots}$