# Improving the bound for $\sigma(q^k)/q^k$ where $q^k n^2$ is an odd perfect number given in Eulerian form

Let $$x$$ be a positive integer. (That is, let $$x \in \mathbb{N}$$.)

We denote the sum of divisors of $$x$$ as $$\sigma(x) = \sum_{d \mid x}{d}.$$

We also denote the abundancy index of $$x$$ as $$I(x)=\sigma(x)/x$$.

If $$N$$ is odd and $$\sigma(N)=2N$$, then $$N$$ is called an odd perfect number. Euler proved that an odd perfect number, if one exists, must have the so-called Eulerian form $$N = q^k n^2$$, i.e. $$q$$ is the special prime satisfying $$q \equiv k \equiv 1 \pmod 4$$ and $$\gcd(q,n)=1$$.

Since $$k \equiv 1 \pmod 4$$, then we have $$\frac{q+1}{q} = I(q) \leq I(q^k) = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$

Since $$q$$ is prime and satisfies $$q \equiv 1 \pmod 4$$, we have $$q \geq 5$$, from which we obtain $$\frac{1}{q} \leq \frac{1}{5} \implies -\frac{1}{q} \geq - \frac{1}{5} \implies \frac{q-1}{q} = 1 - \frac{1}{q} \geq 1 - \frac{1}{5} = \frac{4}{5}.$$ We get that $$I(q^k) < \frac{q}{q - 1} \leq \frac{5}{4}$$ from which we conclude that $$I(q^k) < \frac{5}{4}.$$

We also obtain $$I(q^k) < \frac{5}{4} < \sqrt{\frac{8}{5}} < \sqrt{I(n^2)} = \sqrt{\frac{2}{I(q^k)}},$$ from which we get $$\bigg(I(q^k)\bigg)^2 < \frac{2}{I(q^k)} \implies \bigg(I(q^k)\bigg)^3 < 2 \implies I(q^k) < \sqrt{2}.$$

Here is my question:

Is it possible to improve on the inequality $$I(q^k) < \sqrt{\frac{2}{I(q^k)}},$$ to something like (say) $$I(q^k) < \sqrt{\frac{2}{I(q^k)}}?$$

Note that $$1/3 = 0.\overline{333}$$.

MOTIVATION

Here is the reason why I think it might be possible to improve on the bound for $$I(q^k)$$.

Let $$N = q^k n^2$$ be an odd perfect number given in Eulerian form.

The Descartes-Frenicle-Sorli Conjecture predicts that $$k=1$$. Suppose that this Conjecture holds.

Then $$I(q^k) = I(q) = \frac{q+1}{q} = 1+\frac{1}{q} \leq 1+\frac{1}{5}=\frac{6}{5}$$ and $$I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} \geq \frac{2}{\frac{6}{5}} = \frac{5}{3}.$$

Let $$\frac{6}{5} = \bigg(\frac{5}{3}\bigg)^y.$$

Note that we then have that $$I(q) \leq \frac{6}{5} = \bigg(\frac{5}{3}\bigg)^y \leq \bigg(I(n^2)\bigg)^y = \bigg(\frac{2}{I(q)}\bigg)^y$$ where $$y = \frac{\log\bigg(\frac{6}{5}\bigg)}{\log\bigg(\frac{5}{3}\bigg)} \approx 0.356915448856724.$$

WLOG, if we assume that $$k>1$$ and let $$\frac{5}{4} = \bigg(\frac{8}{5}\bigg)^z,$$ then we have that $$I(q^k) < \frac{5}{4} = \bigg(\frac{8}{5}\bigg)^z < \bigg(I(n^2)\bigg)^z = \bigg(\frac{2}{I(q^k)}\bigg)^z$$ where $$z = \frac{\log\bigg(\frac{5}{4}\bigg)}{\log\bigg(\frac{8}{5}\bigg)} \approx 0.474769847356948651282146696312271.$$

It turns out that $$I(q^k) < \sqrt{\frac{2}{I(q^k)}}$$ implies $$1 + \frac{1}{q} = I(q) \leq I(q^k) < \sqrt{2}$$ from which we obtain $$q > \bigg(\sqrt{2} - 1\bigg)^{-1} \approx 5.2852135$$ thereby giving $$q \geq 13$$ since $$q$$ is a prime satisfying $$q \equiv 1 \pmod 4$$. Thus, the implication $$I(q^k) < \sqrt{\frac{2}{I(q^k)}} \implies q \geq 13$$ holds.
If the reverse inequality $$I(q^k) > \sqrt{\frac{2}{I(q^k)}}$$ holds, then we get $$\frac{q}{q - 1} > I(q^k) > \sqrt{2}$$ which implies that $$1 < q < \frac{\sqrt{2}}{\sqrt{2} - 1} \approx 6.28521$$ from which we conclude that $$q = 5$$.
Since $$q \geq 5$$ and $$q$$ is a prime satisfying $$q \equiv 1 \pmod 4$$, by the contrapositive of the last implication, we get the implication $$q \geq 13 \implies I(q^k) < \sqrt{\frac{2}{I(q^k)}}.$$
We therefore have the biconditional $$I(q^k) < \sqrt{\frac{2}{I(q^k)}} \iff q \geq 13.$$
(Note that we cannot have $$I(q^k) = \sqrt{\frac{2}{I(q^k)}}$$ as equality implies that $$I(q^k) = \sqrt{2}$$, contradicting the fact that $$I(q^k) = \frac{q^{k+1} - 1}{q^k (q - 1)}$$ is rational.)
Thus, to prove the inequality $$I(q^k) < \sqrt{\frac{2}{I(q^k)}}$$ we need to rule out $$q=5$$. This is currently open, and is also unknown even if we assume the Descartes-Frenicle-Sorli Conjecture that $$k=1$$.