# upper lower bounds – On an inequality involving the abundancy index of the Eulerian component \$p^k\$ of an odd perfect number \$p^k m^2\$

The topic of odd perfect numbers likely needs no introduction.

Denote the classical sum of divisors of the positive integer $$x$$ by $$sigma(x)=sigma_1(x)$$, the abundancy index of $$x$$ by $$I(x)=sigma(x)/x$$, and the deficiency of $$x$$ by $$D(x)=2x-sigma(x)$$.

Euler proved that a hypothetical odd perfect number must take the form
$$n = p^k m^2$$
where the Eulerian component $$p^k$$ satisfies the constraints $$p equiv k equiv 1 pmod 4$$ and $$gcd(p,m)=1$$.

Since $$p$$ is (the special) prime, we have the formula
$$I(p^k)=frac{sigma(p^k)}{p^k}=frac{p^{k+1} – 1}{p^k (p – 1)}$$
and corresponding (sharp?) upper bound
$$I(p^k)
We also have the formula
$$frac{D(p^k)}{p^k}=2-I(p^k),$$
and corresponding (sharp?) upper bound
$$frac{D(p^k)}{p^k}=2-I(p^k) leq 2-frac{p+1}{p}=frac{p-1}{p}.$$

We obtain
$$I(p^k)bigg(2 – I(p^k)bigg) < frac{p}{p-1}cdotfrac{p-1}{p}=1.$$

Here are my questions:

(1) Can one substantially improve on the bound
$$I(p^k)bigg(2 – I(p^k)bigg) < 1?$$

(2) If the answer to Question (1) is YES, my next question is “How?”.

(3) If the answer to Question (1) is NO, can you explain/show why the bound cannot be substantially improved?

MY ATTEMPT

I do am aware of the fact that
$$f(k):=g(p):=I(p^k)bigg(2 – I(p^k)bigg)=frac{sigma(p^k) D(p^k)}{p^{2k}}=frac{(p^{k+1} – 1)(p^{k+1} – 2p^k + 1)}{Bigg(p^k (p – 1)Bigg)^2}$$
and that
$$frac{partial f}{partial k} = -frac{2(p^k – 1)log(p)}{Bigg(p^k (p – 1)Bigg)^2} < 0$$
while
$$frac{partial g}{partial p} = frac{2(p^k – 1)(p^{k+1} – (k+1)p + k)}{p^{2k+1} (p – 1)^3} > 0.$$

This means that
$$g(5) leq g(q) = f(k) leq f(1)$$
since the computations above show that $$f(k)$$ is decreasing while $$g(q)$$ is increasing.

In particular, the quantity $$f(1)$$ in the inequality $$f(k) leq f(1)$$ simplifies to:
$$f(k) = I(p^k)bigg(2 – I(p^k)bigg) leq f(1) = I(p)bigg(2 – I(p)bigg) = frac{p+1}{p}bigg(frac{2p – (p+1)}{p}bigg) = frac{p^2 – 1}{p^2},$$
which somehow improves on the upper bound of $$1$$.

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