Can $k$ be arbitrarily large in the following equations?

Question

(Note: This has been cross-posted to MO.)

Let $N = q^k n^2$ be an odd perfect number in Eulerian form. (That is, $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Note that $q \geq 5$.) The Descartes-Frenicle-Sorli conjecture for odd perfect numbers predicts that $k=1$ always holds.

I believe it is currently unknown whether $k$ can be bounded from above, but the following appears to be a viable approach for this problem:

Consider the expression $$N - (q^k + n^2) + 1 = (q^k - 1)(n^2 - 1) = (q - 1)\sigma(q^{k-1})(n + 1)(n - 1).$$

The reason for the $\sigma(q^{k-1})$ term is because of the following expression in this preprint, which gives: $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$ which would then be equal to $$\frac{(q - 1)(n + 1)(n - 1)(2n^2 - \sigma(n^2))}{N - (q^k + n^2) + 1},$$ from which it follows that $$\frac{\sigma(n^2)}{q^k}\cdot\bigg[N - (q^k + n^2) + 1\bigg]=\frac{2n^2}{\sigma(q^k)}\cdot\bigg[N - (q^k + n^2) + 1\bigg]$$ $$=\gcd(n^2,\sigma(n^2))\cdot\bigg[N - (q^k + n^2) + 1\bigg]=(q-1)(n+1)(n-1)(2n^2 - \sigma(n^2)).$$

Now, it is conjectured that $q^k < n$ (see this M. Sc. thesis and this paper). This will then imply that $$\frac{2n^2}{\sigma(q^k)}>\frac{8n^2}{5q^k}>\frac{8n}{5}>\frac{8}{5}\cdot{\sqrt[3]{N}}$$ since $\sigma(q^k)/q^k < 5/4$ holds for all $k$. Hence, by using Ochem and Rao's lower bound for the magnitude of an odd perfect number, we get that $N>{10}^{1500}$, so that $$\frac{2n^2}{\sigma(q^k)}>\frac{8}{5}\cdot{{10}^{500}}$$

Notice that in the equation $$\frac{2n^2}{\sigma(q^k)}\cdot\bigg[N - (q^k + n^2) + 1\bigg]=(q-1)(n+1)(n-1)(2n^2 - \sigma(n^2))$$ the RHS does not involve $k$. Holding $q$ and $n$ constant (i.e. considering a specific Euler prime $q$ and a specific square root of non-Euler part $n$), and allowing $k$ to vary, then we have $$\frac{2n^2}{\sigma(q^k)}\cdot\bigg[N - (q^k + n^2) + 1\bigg]=(q-1)(n+1)(n-1)(2n^2 - \sigma(n^2))$$ $$> \frac{8}{5}\cdot{\sqrt[3]{N}}\cdot\bigg[N - (q^k + n^2) + 1\bigg]$$ $$\lim_{k \to \infty}{\bigg(\frac{8}{5}\cdot{\sqrt[3]{N}}\cdot\bigg[N - (q^k + n^2) + 1\bigg]\bigg)} \leq (q-1)(n+1)(n-1)(2n^2 - \sigma(n^2)),$$ from which we get the "contradiction" $$(q-1)(n+1)(n-1)(2n^2 - \sigma(n^2)) \geq \lim_{k \to \infty}{\bigg(\frac{8}{5}\cdot{\sqrt[3]{N}}\cdot\bigg[N - (q^k + n^2) + 1\bigg]\bigg)} \to \infty.$$

Here are my questions:

(1) Is there indeed a contradiction in the last step of this "proof" that $k$ is bounded from above?

(2) Is it possible to make the argument in this "proof" more rigorous?

Answer 1

This claim is flawed; you have claimed:

$$q^k\lt n$$

which means $k$ and $n$ are dependent.

Answer 2

(Note: This answer was copied verbatim from this blog post.)

If $k$ is arbitrarily large in the equations $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))=\frac{2n^2-\sigma(n^2)}{\sigma(q^{k-1})}$$ then $$\lim_{k \to \infty}{\frac{\sigma(n^2)}{q^k}}=\lim_{k \to \infty}{\frac{2n^2}{\sigma(q^k)}}=\lim_{k \to \infty}{\bigg(\frac{2n^2-\sigma(n^2)}{\sigma(q^{k-1})}\bigg)}=0$$ while $$\gcd(n^2,\sigma(n^2))>1$$ for a fixed Euler part $n^2$.

Does this imply that $k$ is bounded?

UPDATE - June 4 2017

Unconditionally, we know that $q^k < n^2$ [Dris, 2012]. Like in JonMark Perry's answer, this means that $k$ and $n$ are dependent.