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Preamble: This post is an offshoot of this earlier question, which was not so well-received in MathOverflow.

Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

If $m$ is odd and $\sigma(m)=2m$, then $m$ is called an odd perfect number. Euler proved that a hypothetical odd perfect number must necessarily have the form $$m = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

As proved in MO, we have the following equation: $$I(n^2) - \frac{2(q - 1)}{q} = \frac{1}{q^{k+1}}\cdot{I(n^2)},$$ whereupon, starting from the lower bound $$I(n^2) > \frac{2(q - 1)}{q}$$ we get the recursive estimates $$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\cdot\frac{2(q - 1)}{q}$$ $$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\bigg(\frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\cdot\frac{2(q - 1)}{q}\bigg)$$ $$= \frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2\Bigg)$$ $$I(n^2) > \frac{2(q - 1)}{q} + \frac{1}{q^{k+1}}\Bigg(\frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2\Bigg)\Bigg)$$ $$= \frac{2(q - 1)}{q}\cdot\Bigg(1 + \frac{1}{q^{k+1}} + \bigg(\frac{1}{q^{k+1}}\bigg)^2 + \bigg(\frac{1}{q^{k+1}}\bigg)^3\Bigg)$$ $$\ldots$$ $$\ldots$$ $$\ldots$$ Repeating the process ad infinitum, we get: $$I(n^2) > \frac{2(q - 1)}{q}\cdot\Bigg(\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i}\Bigg).$$ But $$\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i}$$ is an infinite geometric series, with sum $$\frac{a_0}{1 - r}$$ where the first term $a_0 = 1$ and the common ratio $$r = \frac{1}{q^{k+1}}.$$ Hence, we obtain $$\sum_{i=0}^{\infty}{\bigg(\frac{1}{q^{k+1}}\bigg)^i} = \frac{1}{1 - \frac{1}{q^{k+1}}},$$ from which we finally get $$I(n^2) > \frac{2(q - 1)}{q}\cdot\frac{q^{k+1}}{q^{k+1} - 1}.$$

But we can simplify the RHS of the last inequality as follows: $$\frac{2(q - 1)}{q}\cdot\frac{q^{k+1}}{q^{k+1} - 1} = \frac{2q^k (q - 1)}{q^{k+1} - 1} = \frac{2q^k}{\sigma(q^k)} = \frac{2}{I(q^k)} = I(n^2).$$

We have therefore finally arrived at the contradiction $$I(n^2) > I(n^2).$$

We therefore conclude that there cannot be any odd perfect numbers.

Here is my:

QUESTION: Does this proof hold water? If it does not, where is the error in the argument, and can it be mended so as to produce a valid proof?

Jose Arnaldo Bebita Dris
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    If $I(n^2)$ is greater than all the individual recursive estimates, this does not prove it is greater than the limit of the recursive estimates. – Peter Jul 31 '21 at 05:37
  • From the point "As proved in MO" on, you do not use at all that $I(n^2)$ is the anything special, i.e., you can replace $I(n^2)$ with $y$ throughout. In other words, you believe to arrive at a contradiction $y>y$ from $y-\frac{2(q-1)}q=\frac1{q^{k+1}}y$, when it is straightforward so solve this equation for $y$. – Hagen von Eitzen Jul 31 '21 at 05:42
  • @Peter: I think your point would be debatable, depending on one's point of view regarding limits and infinity. Hence, my reason for posting the question in the first place. – Jose Arnaldo Bebita Dris Jul 31 '21 at 05:43
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    @ArnieBebita-Dris There is no debate. $\frac1n>0$ does not imply $0=\lim_{n\to\infty}\frac1n>0$ – Hagen von Eitzen Jul 31 '21 at 05:44
  • I have submitted this comment as an answer. – Peter Jul 31 '21 at 05:45
  • @HagenvonEitzen: If one solves $$y - \frac{2(q - 1)}{q} = \frac{1}{q^{k+1}}y$$ for $y$, then one gets $$y = \frac{2q^k (q - 1)}{q^{k+1} - 1} = \frac{2q^k}{\sigma(q^k)} = \frac{2}{I(q^k)} = I(n^2),$$ which I am sure you are well-aware that I know of. – Jose Arnaldo Bebita Dris Jul 31 '21 at 05:46

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Assuming your statements about the recursive estimates for $I(n^2)$ are correct, $I(n^2)$ may still not be greater than the limit of the recursive estimates. If we know $a>x_n$ for all $n$, and $x_n\to x$ as $n\to\infty$, we can only deduce that $a\ge x$.

Jose Arnaldo Bebita Dris
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Peter
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In your mathoverflow post, you write $$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \gcd(n^2, \sigma(n^2))$$ because $\gcd(\sigma(q^k), q^k) = 1.$ Why is this true? I do not see any reason it should hold.

user951252
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