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

$$I(p^k)<frac{p^{k+1}}{p^k (p – 1)}=frac{p}{p – 1}.$$

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)isYES, my next question is “How?”.

(3)If the answer to Question(1)isNO, 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 $1$.