In what follows, we denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, the aliquot sum of $x$ by $s(x)=\sigma(x)-x$, and the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Let $m = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $\gcd(q^k,\sigma(q^k))=1$, we get
$$\frac{\sigma(n^2)}{q^k}=\frac{D(n^2)}{s(q^k)}=\frac{n^2}{\sigma(q^k)/2}.$$
But
$$\frac{\sigma(q^k)}{2} \geq \frac{q^k + 1}{2} \geq \frac{5 + 1}{2} = 3,$$
so that
$$\frac{D(n^2)}{s(q^k)} \leq \frac{n^2}{3}.$$
However, $D(n^2) = 2n^2 – \sigma(n^2)$, so that dividing both sides of the last inequality by $n^2 > 0$ yields
$$\frac{2 – I(n^2)}{s(q^k)} \leq \frac{1}{3}.$$
Multiplying both sides of the last inequality by $s(q^k) > 0$ gives
$$2 – I(n^2) \leq \frac{s(q^k)}{3}.$$
We finally obtain
$$I(n^2) \geq 2 – \frac{s(q^k)}{3} = 2 – \frac{q^k – 1}{3(q – 1)} = \frac{6(q – 1) – (q^k – 1)}{3(q – 1)} = \frac{-q^k + 6q – 5}{3(q – 1)} := f(q, k).$$
Note that
$$\frac{\partial f}{\partial q} = -\frac{(k-1)q^{k+1} – kq^k + q}{3q(q-1)^2} < 0$$
and that
$$\frac{\partial f}{\partial k} = -\frac{q^k \ln q}{3(q – 1)} < 0,$$
which means that
$$f(q, k) = \frac{-q^k + 6q – 5}{3(q – 1)}$$
is a decreasing function of both $q$ and $k$.
This implies that
$$I(n^2) \geq \lim_{q \rightarrow \infty}{f(q,k)},$$
which does not exist when $k > 1$, and
$$I(n^2) \geq \lim_{k \rightarrow \infty}{f(q,k)},$$
which likewise does not exist.
Here are my:
QUESTIONS: Does this argument prove that $k=1$? If it does not, how can the proof be mended to make it valid?
Best Answer
When $k\gt 1$, one has $$\lim_{q\to\infty}f(q,k)=\lim_{k\to\infty}f(q,k)=-\infty.$$
So, I think that you cannot prove that $k=1$ from the two inequalities $$I(n^2) \geq \lim_{q \rightarrow \infty}{f(q,k)}$$ and $$I(n^2) \geq \lim_{k \rightarrow \infty}{f(q,k)}$$