Q: Improving the lower bound $I(n^2) > \frac{2(q-1)}{q}$ when $q^k n^2$ is an odd perfect number Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the abundancy index
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical sum of divisors of $x$.
Since $q$ is prime, we have the bounds
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$
which implies, since $N$ is perfect, that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
We now prove the following claim:

CLAIM: $$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$

PROOF:  We know that
$$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k}=\gcd(n^2,\sigma(n^2)),$$
(since $\gcd(q^k,\sigma(q^k))=1$).
However, we have
$$\sigma(q^k) - q^k = 1 + q + \ldots + q^{k-1} = \frac{q^k - 1}{q - 1},$$
so that we obtain
$$\frac{\sigma(n^2)}{q^k}=\frac{\bigg(q - 1\bigg)\bigg(2n^2 - \sigma(n^2)\bigg)}{q^k - 1}=\sigma(n^2) - \bigg(q - 1\bigg)\bigg(2n^2 - \sigma(n^2)\bigg)$$
$$=q\sigma(n^2) - 2(q - 1)n^2.$$
Dividing both sides by $qn^2$, we get
$$I(n^2) - \frac{2(q-1)}{q} = \frac{I(n^2)}{q^{k+1}} > \frac{1}{q^{k+1}}\cdot\frac{2(q-1)}{q},$$
which implies that
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg).$$
QED.
To illustrate the improved bound:
(1) Unconditionally, we have
$$I(n^2) > \frac{2(q-1)}{q} \geq \frac{8}{5} = 1.6.$$
(2) Under the assumption that $k=1$:
$$I(n^2) > 2\bigg(1 - \frac{1}{q}\bigg)\bigg(1 + \left(\frac{1}{q}\right)^2\bigg) \geq \frac{208}{125} = 1.664.$$
(3) However, it is known that under the assumption $k=1$, we actually have
$$I(q^k) = 1 + \frac{1}{q} \leq \frac{6}{5} \implies I(n^2) = \frac{2}{I(q^k)} \geq \frac{5}{3} = 1.\overline{666}.$$
Here are my questions:


(A) Is it possible to improve further on the unconditional lower bound for $I(n^2)$?




(B) If the answer to Question (A) is YES, my next question is "How"?


I did notice that
$$\frac{2(q-1)}{q}+\frac{2}{q(q+1)}=I(n^2)=\frac{2q}{q+1}$$
when $k=1$.

A: Here is a quick way to further refine the improved lower bound for $I(n^2)$:
Write
$$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg).$$
Now use, for instance,
$$q^{k+1} - \frac{1}{q^2} > q^{k+1} - 1$$
to obtain
$$I(n^2)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1+\frac{1}{q^{k+1}-1}\bigg)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - \frac{1}{q^2}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\Bigg(1 + \frac{q^2}{q^{k+3} - 1}\Bigg).$$
Note that
$$\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{q^2}{q^{k+3} - 1}\bigg) - \bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\frac{2(q-1)}{q^{k+2} (q^{k+3} - 1)}>0$$
since $q$ is a prime satisfying $q \equiv k \equiv 1 \pmod 4$.
In fact, this method shows that there are infinitely many ways to refine the improved lower bound
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}+1}{q^{k+1}}\bigg).$$
It remains to be seen whether there is a refined (improved) lower bound that is independent of $k$ (and therefore expressed entirely in terms of $q$).

A: Here is a way to come up with an improved lower bound for $I(n^2)$, albeit in terms of $q$ and $n$:
We write
$$I(n^2) - \frac{2(q - 1)}{q} = \frac{I(n^2)}{q^{k+1}} = \frac{\sigma(n^2)}{q^k}\cdot\frac{1}{qn^2} > \frac{1}{qn^2},$$
from which it follows that
$$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{qn^2}.$$
This improved lower bound for $I(n^2)$, which does not contain $k$, can then be used to produce an upper bound for $k$.
