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A Proof That There Are No Odd Perfect Numbers

Recall that we have from [Dris, Dagal (2021) - discussion of the proof of Theorem 2.1 from pp. 13 to 15] that $$I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}$$ which is equivalent to $$I(m^2) \leq \left(I(m)\right)^{\ln(13/9)/\ln(4/3)},$$ from which we get $$\frac{I(m^2)}{I(m)} \leq \left(I(m)\right)^{\ln(13/12)/\ln(4/3)}.$$ This last inequality implies that $$1 > \frac{I(m)}{I(m^2)} \geq \left(I(m)\right)^{\ln(4/3)/\ln(13/12)}$$ $$> \Bigg(\left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)}\Bigg)^{\ln(4/3)/\ln(13/12)}= \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}.$$ But in general, we know that $I(m^2) > 8/5$, so that $$1>\frac{I(m)}{I(m^2)}> \left(I(m^2)\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}$$ $$>\left(\frac{8}{5}\right)^{{\left(\ln(4/3)\right)^2}/(\ln(13/9)\cdot\ln(13/12))}> 3$$ which is a contradiction. This seems to prove that there are, in fact, no odd perfect numbers. We will stop here for the time be
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Step Two: Almost Perfect Numbers... (#OPNResearch #May2023)

Let $N = p^k m^2$ be an odd perfect number with special prime $p$.  Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x) = \sigma_1(x)$ is the classical sum of divisors of $x$.  Additionally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$. I claim that the estimate $m < p$ holds if and only if $D(m) = 1$. Recall the following material facts: $$\left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)} \approx 1.4912789897463723558019369442648428792691131693617462298521097731$$ $$k = 1 \Rightarrow I(m) \geq \left(I(m^2)\right)^{\ln(4/3)/\ln(13/9)} \geq \left(\frac{5}{3}\right)^{\ln(4/3)/\ln(13/9)}$$ [ Dagal, Dris (NNTDM, 06/2021)  -  The Abundancy Index of Divisors of Odd Perfect Numbers - Part II ] We have the (generic/sharp [?]) upper bound $$\frac{I(m^2)}{I(m)} \leq \frac{\zeta(2)}{\zeta(3)}.$$ Reference: Comment by Erick Wong  underneath an answer by Will Jagy  to MSE question https://math.stackexchange.com/questions/1097803 . If $I(m) < 2$
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Step One - Odd Perfect Numbers Circa 2013

In what follows, we will denote the classical sum of divisors of the positive integer $z$ by $$\sigma(z)=\sigma_1(z).$$ We will also denote the abundancy index  of $z$ by $I(z)=\sigma(z)/z=\sigma_{-1}(z)$, the deficiency of $z$ by $D(z)=2z-\sigma(z)$, and the aliquot sum of $z$ by $s(z)=\sigma(z)-z$. (Note that, in general, $D(z)+s(z)=z$ holds.)   Notice that $$I(z) = \sum_{d \mid z}{\frac{1}{d}} \leq \sum_{i=1}^{z}{\frac{1}{i}} = H(z).$$   Dagal showed in 2015 that we actually have the sharper upper bound $$I(z) \leq H(d(z)) \hspace{2.50in} (1)$$ where $d(z)$ is nothing but $\tau(z)$ or $\sigma_{0}(z)$.  Equality holds in $(1)$ if and only if $z=1$, since unity is a fixed point of $\sigma_{-1}(z), \sigma_{0}(z), \sigma(z),$ and $H(z)$. We start with a minor technical lemma: Lemma 1: If $N = p^k m^2$ is an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$, then $I(m^2) < \zeta(2)$ if and only if $p = 5$ and $k \neq 1$. Proof: S
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