Is there an analytical solution to the inequality $\frac{1 + k({p}^{(k+1)/2)})}{p^k} < \frac{p}{p - 1}$ if one were to bound $k$ in terms of $p$?

Question

My question is as is in the title:

Is there an analytical solution to the inequality $$\frac{1 + k({p}^{(k+1)/2)})}{p^k} < \frac{p}{p - 1}$$ if one were to bound $k$ in terms of $p$?

Here, $p \equiv 1 \pmod 4$ is a prime number and $k \equiv 1 \pmod 4$ is a positive integer.

MOTIVATION

This inquiry arises as an offshoot of this MSE question.

In particular, by using the Arithmetic Mean-Geometric Mean Inequality, then we obtain $$\sigma_1(p^k) \geq 1 + k (\sqrt{p})^{1+k} \tag{1}$$ so that $$\frac{\sigma_1(p^k)}{p^k} \geq \frac{1 + k (\sqrt{p})^{1+k}}{p^k} \tag{2},$$ where $\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.

But $\sigma_1(p^k)/{p^k}$ is equal to $$\frac{\sigma_1(p^k)}{p^k} = \frac{p^{k+1} - 1}{p^k (p - 1)} \tag{3},$$ which is bounded from above by $$\frac{\sigma_1(p^k)}{p^k} = \frac{p^{k+1} - 1}{p^k (p - 1)} < \frac{p^{k+1}}{p^k (p - 1)} = \frac{p}{p - 1} \tag{4}.$$

Using $(2)$ and $(4)$, we get $$\frac{1 + k({p}^{(k+1)/2)})}{p^k} < \frac{p}{p - 1} \tag{5}.$$

MY ATTEMPT

I tried asking WolframAlpha for the solutions to Inequality $(5)$, it is giving me the following Inequality Plot:

I also noticed that Inequality $(5)$ is true for all $p$, when $k=1$.

Furthermore, I tried to rewrite Inequality $(5)$ as $$k p^{(k+1)/2} (p - 1) < p^{k+1} - p + 1$$ $$k^2 p^{k+1} (p - 1)^2 < \left(p^{k+1} - p + 1\right)^2$$ but admittedly, I was not getting anywhere near to an analytical solution.

Basically, I would like to know whether a bound better than $$k \geq \log_{5}\left(\frac{p}{p-4}\right) \tag{6}$$ could be obtained from assuming Inequality $(5)$, as detailed in this MSE question. Hence, my question whether there was an analytical solution to Inequality $(5)$.

Alas, this is where I get stuck!

Answer 1

You got $(5)$ from $\dfrac{1+kp^{(k+1)/2}}{p^k}\leqslant \dfrac{\sigma_1(p^k)}{p^k}<\dfrac{p}{p−1}$. This means that $(5)$ is weaker than $(4)$. (So, I think that, from $(5)$, you cannot get a better result than $(4)$.)

The inequality $(1)$ holds iff $k\geqslant 1$ since the equality of $p+p^2+\cdots +p^k\geqslant kp^{(k+1)/2}$ is attained only when $p=p^2=\cdots =p^k$, i.e. $k=1$, and if $k\gt 1$, then $p+p^2+\cdots +p^k\color{red}>kp^{(k+1)/2}$ holds.

I think that $(6)$ is weaker than an obvious inequality $k\geqslant 1$ since for $p\geqslant 5$, we have $k\geqslant 1\geqslant\log_{5}\left(\dfrac{p}{p-4}\right)$.

Answer 2

This is only a partial answer. (I merely wanted to collect some new thoughts that recently occurred to me, which are too long to be included in the Comments section.)

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As in the original question, Inequality $(5)$ $$\frac{1 + k({p}^{(k+1)/2)})}{p^k} < \frac{p}{p - 1} \tag{5}$$ is equivalent to $$k({p}^{(k+1)/2)}) (p - 1) < p^{k+1} - p + 1 \tag{7}$$ which is equivalent to $$k({p}^{(k+1)/2)}) < p\cdot\left(\frac{p^k - 1}{p - 1}\right) + \frac{1}{p - 1} \tag{8}.$$

Rewriting $s(p^k) = (p^k - 1)/(p - 1)$ where $s(x)=\sigma(x)-x$ is the aliquot sum of $x$, then we have, from Inequality $(8)$: $$k({p}^{(k+1)/2)}) < ps(p^k) + \frac{1}{p - 1} \tag{9}.$$

In other words, we have the lower bound $$s(p^k) > \frac{1}{p}\cdot\left(k({p}^{(k+1)/2)}) - \frac{1}{p - 1}\right).$$

Define the two-variable function $$f(k,p) = \frac{1}{p}\cdot\left(k({p}^{(k+1)/2)}) - \frac{1}{p - 1}\right).$$

It can be shown that $$\frac{\partial f(k,p)}{\partial k} = \frac{1}{2}\cdot {p^{(k-1)/2}} \left(k \log p + 2\right).$$ Since $p$ is a prime number satisfying $p \equiv 1 \pmod 4$, then we know that $p \geq 5$. In particular, this means that $$\frac{\partial f(k,p)}{\partial k} = \frac{1}{2}\cdot {p^{(k-1)/2}} \left(k \log p + 2\right) > 0$$ so that $f(k,p)$ is an increasing function of $k$. Since $k$ is a positive integer satisfying $k \equiv 1 \pmod 4$, then we know that $k \geq 1$. We infer that $$f(k,p) \geq f(1,p) = 1 - \frac{1}{p(p-1)}$$ because $f(k,p)$ is an increasing function of $k$.

But we also know that $$\frac{\partial f(1,p)}{\partial p} = \frac{2p - 1}{\left(p(p-1)\right)^2}$$ which is positive since $p \geq 5$. This means that $f(1,p)$ is an increasing function of $p$, so that we obtain $$f(1,p) = 1 - \frac{1}{p(p-1)} \geq f(1,5) = \frac{19}{20}.$$

Thus, we conclude that $$s(p^k) > f(k,p) \geq f(1,p) \geq f(1,5) = \frac{19}{20} \tag{10}.$$

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I now believe that Inequality $(10)$ $$\frac{p^k - 1}{p - 1} = s(p^k) > \frac{19}{20}$$ will result to a bound better than $(6)$ $$k \geq \log_{5}\left(\frac{p}{p-4}\right)$$ but I am not really 100% sure, as I have no proof.