Upper Bound for Abundancy Index

Theorem

Let $n \in \N_{>0}$ be a non-zero natural number.

Let $\map h n$ be the abundancy index of $n$.

Then:

$\ds \map h n \le \prod_{p \mathop \divides n} \dfrac p {p - 1}$

such that equality holds if and only if $n = 1$.

Proof

First we note that $\map h 1 = 1$ demonstrating that equality holds if $n = 1$.

Let $n$ be expressed as its prime decomposition:

\(\ds n\)

\(=\)

\(\ds \prod_{p_k \mathop \divides n} {p_k}^{a_k}\)

\(\ds \)

\(=\)

\(\ds {p_1}^{a_1} {p_2}^{a_2} \cdots {p_k}^{a_k}\)

where $p_1, p_2, \ldots, p_k$ are distinct primes.

By definition of abundancy index:

$\map h n = \dfrac {\map {\sigma_1} n} n$

where $\map {\sigma_1} n$ denotes the divisor sum function of $n$.

We have:

\(\ds \dfrac {\map {\sigma_1} n} n\)

\(=\)

\(\ds \dfrac {\ds \map {\sigma_1} {\prod_{p_k \mathop \divides n} {p_k}^{a_k} } } {\ds \prod_{p_k \mathop \divides n} {p_k}^{a_k} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {\ds \prod_{p_k \mathop \divides n} \map {\sigma_1} { {p_k}^{a_k} } } {\ds \prod_{p_k \mathop \divides n} {p_k}^{a_k} }\)

Divisor Sum Function is Multiplicative

\(\ds \)

\(=\)

\(\ds \prod_{p_k \mathop \divides n} \dfrac {\map {\sigma_1} { {p_k}^{a_k} } } { {p_k}^{a_k} }\)

\(\ds \)

\(=\)

\(\ds \prod_{p_k \mathop \divides n} \map h { {p_k}^{a_k} }\)

Definition of Abundancy Index

It remains to be shown that:

$\map h {p^a} < \dfrac p {p - 1}$

for prime $p$.

We have that:

\(\ds m\)

\(=\)

\(\ds p^a\)

\(\ds \leadsto \ \ \)

\(\ds \map h m\)

\(=\)

\(\ds \dfrac {\map {\sigma_1} m} m\)

\(\ds \)

\(=\)

\(\ds \dfrac {p^{a + 1} - 1} {p - 1} / p^a\)

Divisor Sum of Power of Prime

\(\ds \)

\(=\)

\(\ds \dfrac {p^{a + 1} - 1} {p^a \paren {p - 1} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {p^{a + 1} } {p^a \paren {p - 1} } - \dfrac 1 {p^a \paren {p - 1} }\)

\(\ds \)

\(<\)

\(\ds \dfrac p {p - 1}\)

$\blacksquare$