Product Form of Sum on Completely Multiplicative Function
Theorem
Let $f$ be a completely multiplicative arithmetic function.
Let the series $\displaystyle \sum_{n \mathop = 1}^\infty \map f n$ 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:
\(\ds \map P {A, K}\) | \(:=\) | \(\ds \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \frac {1 - \map f p^{K + 1} } {1 - \map f p}\) | where $\mathbb P$ denotes the set of prime numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \paren {\sum_{k \mathop = 0}^K \map f p^k}\) | Sum of Geometric Sequence | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{v \mathop \in \prod \limits_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} } \paren {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \map f p^{v_p} }\) | Product of Summations is Summation Over Cartesian Product of Products | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{v \mathop \in \prod \limits_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \set {0 \,.\,.\, K} } \map f {\prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } p^{v_p} }\) | as $f$ is completely multiplicative |
Change the summing variable using:
\(\ds \sum_{v \mathop \in V} \map g {\map h v}\) | \(=\) | \(\ds \sum_{w \mathop \in \set {\map h v: v \mathop \in V} } \map g w\) | 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:
\(\ds \map P {A, K}\) | \(=\) | \(\ds \sum_{n \mathop \in \map Q {A, K} } \map f n\) | 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:
\(\ds W\) | \(=\) | \(\ds \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \map Q {A, K}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\prod_{p \mathop \in \mathbb P} p^{-v_p}: v \in \prod_{p \mathop \in \mathbb P} \set {0 \,.\,.\, \infty} }\) |
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 $\map P {A, K}$:
\(\ds \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \map P {A, K}\) | \(=\) | \(\ds \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \prod_{\substack {p \mathop \in \mathbb P \\ p \mathop \le A} } \frac {1 - \map f p^{K + 1} } {1 - \map f p}\) | taking limits of both sides of the definition of $\map P {A, K}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \prod_{p \mathop \in \mathbb P} \frac 1 {1 - \map f p}\) | $\map f p^{K + 1} \to 0$, because $\displaystyle \sum_{n \mathop = 1}^{\infty} \map f n$ is convergent | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{\substack {A \mathop \to \infty \\ K \mathop \to \infty} } \sum_{n \mathop \in \map Q {A, K} } \map f n\) | from the expression for $\map P {A, K}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop \in \N^+} \map f n\) | substituting for $\N^+$: order of summation is not defined | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \map f n\) | absolutely convergent, so the order does not alter the limit |
$\blacksquare$
Note
A part of this page has to be extracted as a theorem: Extract this to a page in its own right. |
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:
- $\displaystyle \sum_{n \mathop = 1}^\infty \map f n = \prod_p \paren {1 + \map f p + \map f {p^2} + \cdots}$