Let $N = q^k n^2$ 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 $\gcd(q,n)=1$).
Inspired by mathlove's answer to this closely related MSE question, I reconsider $$\frac{n^2}{D(n^2)}=\frac{q^{k+1}-1}{2(q^k - 1)}$$ as $$f(k):=\frac{q^{k+1}-1}{2(q^k - 1)}$$ and $$g(q):=\frac{q^{k+1}-1}{2(q^k - 1)},$$ where $D(x)=2x-\sigma(x)$ is the deficiency of the positive integer $x$ and $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.
As noted by mathlove, $$f'(k) = -\frac{(q-1){q^k}\log(q)}{2(q^k - 1)^2} < 0$$ which means $f(k)$ is a decreasing function of $k$, and this implies that $$\frac{q}{2} = \lim_{k \rightarrow \infty}{f(k)} < f(k) \leq f(1) = \frac{q+1}{2}.$$
Also, I noted that $$g'(q) = \frac{q^{k-1}(q(q^k - 1) - k(q - 1))}{2(q^k - 1)^2} > 0$$ which means $g(q)$ is an increasing function of $q$, and this implies that $$\frac{5^{k+1} - 1}{2(5^k - 1)} = g(5) \leq g(q) = f(k) \leq f(1) = \frac{q+1}{2}.$$ We obtain $$5 + \frac{5}{5^k - 1} = \frac{5^{k+1} - 5}{5^k - 1} + \frac{5}{5^k - 1} \leq q + 1.$$ Suppose that $k=1$. Then we get $$5 + \frac{5}{4} \leq q + 1 \implies q \geq 4 + \frac{5}{4} = \frac{21}{4}.$$
Here are my:
QUESTIONS: Is this a logically valid proof for the implication $k=1 \implies q > 5$? If the proof is invalid, would it be possible to mend it so as to produce a logically valid proof?
Blimey!
Found the error: The expression for $$\frac{5^{k+1} - 1}{5^k - 1}$$ should have been $$\frac{5^{k+1} - 1}{5^k - 1}=\frac{5^{k+1} - 5}{5^k - 1}+\frac{5 - 1}{5^k - 1}=5+\frac{4}{5^k - 1}.$$