Let $\sigma(x)$ be the sum of the divisors of the positive integer $x$, and let the abundancy index of $x$ be defined as
$$I(x) = \frac{\sigma(x)}{x}.$$
My question is this: What proportion of the positive integers satisfy
$$\frac{\zeta(3)}{\zeta(2)}\cdot\frac{2n}{n + 1} < I(n) < \frac{2n}{n + 1}?$$
Note that we have the rational approximation $\zeta(3)/\zeta(2) \approx 0.7307629694$.
I'll copy part of my (not really mine, but I found it) answer from here: Does this inequality hold true, in general?
I did a Google search for "density of euler phi function". The second link is http://www.ams.org/journals/proc/2007-135-09/S0002-9939-07-08771-0/S0002-9939-07-08771-0.pdf. This paper, by ANDREAS WEINGARTNER, is titled "THE DISTRIBUTION FUNCTIONS OF σ(n)/n AND n/ϕ(n)". Here is its abstract:
"Let σ(n) be the sum of the positive divisors of n. We show that the natural density of the set of integers n satisfying σ(n)/n ≥ t is given by $\exp\big(−e^{t e^{−γ}(1 + O(t^{−2}))}\big)$ , where γ denotes Euler’s constant. The same result holds when σ(n)/n is replaced by n/ϕ(n), where ϕ is Euler’s totient function."
This paper has clearly done the heavy lifting. If we put $t = 2$, and use $\gamma \approx 0.5772156649$ (I show each stage in the computation for checkability), $te^{-\gamma} = 1.1229189671$, $e^{te^{-\gamma}}=3.0738134815$, $\exp(-e^{te^{-\gamma}}) =0.0462444658 $.
This is the density of $n$ for which $\dfrac{n}{\phi(n)} > 2 $. The density for which $\dfrac{n}{\phi(n)} < 2 $ is one minus this or $0.9537555342$.
Do this for your bounds.