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?
Best Answer
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 $a\ge x$.