What proportion of the positive integers satisfy this number-theoretic inequality? 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 $\zeta(3)/\zeta(2) \approx 0.7307629694$.
 A: 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.
