Prove that, for every natural number \(n\), there exists some natural number \(m > n\) such that \(\dfrac{p_m \#}{p_n \#}\) is *abundant*.

**Clarification**:

\(p_i\) is the \(i^\text{th}\) prime number.

\(p_i \# = 2 \times 3 \times \cdots \times p_i\) is the \(i^\text{th}\) primorial.

An abundant number is a number for which the sum of its proper divisors is greater than the number itself. For example, the proper divisors of 12 are 1, 2, 3, 4, 6, and the sum of these proper divisor exceed the number 12, thus 12 is an abundant number.

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TopNewestThe sum of divisors \(\sigma(n)\) of a natural number \(n = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k}\) is given by

\[\displaystyle \prod_{i=1}^k (1 + p_i + p_i^2 + \ldots + p_i^{e_i}) .\]

Thus, \(\frac{\sigma(n)}{n}\) is given by

\[\displaystyle\begin{align*} \frac{\sigma(n)}{n} &= \frac{\prod_{i=1}^k (1 + p_i + p_i^2 + \ldots + p_i^{e_i})}{\prod_{i=1}^k p_i^{e_i}} \\ &= \prod_{i=1}^k \frac{1 + p_i + p_i^2 + \ldots + p_i^{e_i}}{p_i^{e_i}} \\ &= \prod_{i=1}^k \left( 1 + \frac{1}{p_i} + \frac{1}{p_i^2} + \ldots + \frac{1}{p_i^{e_i}} \right) \end{align*}\]

An abundant number has \(\frac{\sigma(n)}{n} > 2\).

Note that \(\frac{p_m \#}{p_n \#} = p_{n+1} p_{n+2} \ldots p_m\), thus

\[\displaystyle \sigma \left( \frac{p_m \#}{p_n \#} \right) = \prod_{i=n+1}^m \left( 1 + \frac{1}{p_i} \right)\]

We know that \(\displaystyle\prod_{i=1}^\infty \left( 1 + \frac{1}{p_i} \right)\) diverges to infinity, so given any \(n\), we can pick \(m\) large enough so that the result is as large as desired (otherwise the product doesn't diverge to infinity). To be more precise, we can pick some \(m\) so that

\[\displaystyle \prod_{i=1}^m \left( 1 + \frac{1}{p_i} \right) > \prod_{i=1}^n \left( 1 + \frac{1}{p_i} \right) \cdot 2\]

because the right hand side is a constant, while the left hand side grows to infinity. If we couldn't pick one, that would imply that \(\displaystyle \prod_{i=1}^n \left( 1 + \frac{1}{p_i} \right) \cdot 2\) is an upper bound to \(\displaystyle\prod_{i=1}^\infty \left( 1 + \frac{1}{p_i} \right)\), contradiction.

This is similar reasoning to the following: given \(n\), we can pick \(m\) so that \(\frac{1}{n+1} + \frac{1}{n+2} + \ldots + \frac{1}{m} > 1\), because \(\sum \frac{1}{i}\) diverges to infinity. – Ivan Koswara · 1 year, 1 month ago

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– Jake Lai · 1 year, 1 month ago

Exactly. Neat proof.Log in to reply