What is the relative density of the abundant numbers in the positive integers?

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The Art and Craft of Problem Solving by Paul Zeitz has the following problem. Now, I have been able to solve parts (a) and (b), part (a) by showing that it can get arbitrarily large, and part (b) by first finding the sum and then taking the limit. However, how do you guys think I should approach part (c)?

9.4.37 Analogous to the concept of perfect numbers (see Problems 7.5.33-7.5.35) are the abundant numbers. The natural number $n$ is considered abundant if $\sigma(n) > 2n,$ where $\sigma(n)$ is the sum of the positive divisors of $n$ (e.g. $\sigma(12)=1+2+3+4+6+12$).

(a) How abundant can a number get? In other words. what is the largest possible value for the ratio $\sigma(n)/n?$

(b) What is the expected value of this "abundancy quotient" $\sigma(n)/n?$ In other words if you pick an integer $n$ at random and compute the value of $\sigma(n)/n$ what limiting average value do we get if we repeat this experiment indefinitely?

(c) What relative fraction of positive integers is abundant?

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(c) To my knowledge the exact value is an open problem. We do know, however, reasonably sharp bounds for the natural density of abundant numbers (that is, for the limit $$p := \lim_{n \to \infty} \frac{a(n)}{n},$$ where $a(n)$ is the counting function for the abundant numbers, $$a(n) := \# \{m \in \Bbb N : m \leq n, m \text{ abundant}\}).$$ Specifically, Deléglise (1998) gave the bounds $$0.2474 < p < 0.2480,$$ which improves much weaker (and rather older) bounds due to Wall.

Deléglise, Marc. Bounds for the density of abundant integers, Experiment. Math., Volume 7, Issue 2 (1998), 137-143.

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(a) there is no upper bound. For a number that is the product of the first power of primes, $n=pqr\dots z$, we have $a(n)=(1+p)(1+q)(1+r)\dots (1+z)$, so $\frac {a(n)}n=(1+\frac 1p)(1+\frac 1q)(1+\frac 1r) \dots (1+\frac 1z) \gt 1+\frac 1p+\frac 1q+\frac 1r \dots \frac 1z$. As the sum of the inverses of the primes diverges, this grows without bound.