If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$? Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$.  Denote the abundancy index of $y \in \mathbb{N}$ by $I(y)=\sigma(y)/y$.
If $\sigma(N)=2N$, then $N$ is said to be perfect.
Euler proved that every odd perfect number has the form $q^k n^2$ where $q$ is prime (called the Euler prime) with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
If $k=1$, then since $q \geq 5$, then we have
$$I(q)I(n^2)=2 \Longleftrightarrow I(n^2)=\dfrac{2q}{q+1} \geq \dfrac{5}{3}.$$
If $k>1$, then we can use the bound
$$I(q^k) < \dfrac{q}{q-1} \leq \dfrac{5}{4}$$
so that
$$I(n^2)=\dfrac{2}{I(q^k)} > \dfrac{2(q-1)}{q} \geq \dfrac{8}{5}.$$
Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does $I(n^2) \geq 5/3$ imply $k=1$?

 A: In an answer to a related MSE question, we get that
$$I(n^2) = \dfrac{2{q^k}(q - 1)}{q^{k+1} - 1} = 2 - 2\dfrac{q^k - 1}{q^{k+1} - 1}.$$
Then $I(n^2) \geq 5/3$ implies that
$$2 - 2\dfrac{q^k - 1}{q^{k+1} - 1} \geq \dfrac{5}{3}$$
$$\dfrac{1}{6} \geq \dfrac{q^k - 1}{q^{k+1} - 1}$$
$$q^{k+1} - 6q^k + 5 \geq 0,$$
which does not force $k=1$.
$k=1$ does follow if it is known a priori that $q=5$.  In this case, $I(n^2) \geq 5/3$ and $q=5$ imply that $I(n^2)=5/3$.
UPDATE (September 03 2017 - Manila time)
Suppose that $k > 1$, so that we have $k \geq 5$ (since $k \equiv 1 \pmod 4$).  We are given that
$$\frac{2}{I(q^k)}=I(n^2) \geq \frac{5}{3}.$$
It follows that
$$I(q^k) \leq \frac{6}{5},$$
whereupon we obtain
$$I(q^5) \leq I(q^k) \leq \frac{6}{5}.$$
It follows that
$$\frac{q^6 - 1}{q^6 - q^5} = \frac{q^6 - 1}{{q^5}(q - 1)} \leq \frac{6}{5}.$$
Consequently, since $q \geq 5 > 1$, we have
$$5q^6 - 5 = 5(q^6 - 1) \leq 6(q^6 - q^5) = 6q^6 - 6q^5$$
which implies that
$$q^6 - 6q^5 + 5 \geq 0.$$
When $q=5$, we obtain
$$q^6 - 6q^5 + 5 = {q^5}(q - 6) + 5 = {5^5}\cdot(-1) + 5 < 0.$$
A: We consider three cases:
Case 1 $I(n^2) = 5/3$
This implies that
$$2 - I(n^2) = \frac{2n^2 - \sigma(n^2)}{n^2} = 2 - \frac{5}{3} = \frac{1}{3},$$
which implies that $(2n^2 - \sigma(n^2)) \mid n^2$, since
$$\frac{n^2}{2n^2 - \sigma(n^2)} = 3$$
then holds.  This means that $n^2$ is deficient-perfect, which is true if and only if $k=1$ (Dris [2017], pages 17 to 18).
Case 2 $I(n^2) < 5/3$
Since $k = 1 \implies I(n^2) \geq 5/3$ holds, by the contrapositive, we have
$$I(n^2) < 5/3 \implies k > 1.$$
Thus, $k > 1$.
Case 3 $I(n^2) > 5/3$
Since we have the inequality
$$I(n^2) \leq \frac{2q}{q+1}$$
with equality if and only if $k=1$, then we have $q>5$ under this case.
Next, since $q$ is the special/Euler prime (satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), then $q>5$ implies that $q \geq 13$.
So we now have
$$I(q^k) < \frac{q}{q-1} \leq \frac{13}{12}$$
and therefore that
$$I(n^2) = \frac{2}{I(q^k)} > \frac{24}{13}.$$
From the other answer, we have
$$I(n^2) = \dfrac{2{q^k}(q - 1)}{q^{k+1} - 1} = 2 - 2\dfrac{q^k - 1}{q^{k+1} - 1}.$$
Thus, we obtain
$$\frac{2}{13} > 2 - I(n^2) = 2\dfrac{q^k - 1}{q^{k+1} - 1}$$
$$q^{k+1} - 1 > 13q^k - 13$$
$$q^{k+1} - 13q^k + 12 > 0.$$
This last inequality does not force $k=1$, and neither proves that $k>1$.
