Infimum and supremum of the set $Z = \left\{ \frac{m}{n}: m, n \in \mathbb{N} \space \land 4^n n^m < m^m < 27^n n^m \right\}$

Question

The problem is to find supremum and infimum of set $$Z = \left\{ \frac{m}{n}: m, n \in \mathbb{N} \space \land \space4^n n^m < m^m < 27^n n^m \right\}$$

I have an answer to this question, infimum is $2$ and supremum is $3$ but I don't see how to actually do the problem, so I would like to ask for some help with that. Thank you.

Answer 1

We have

\begin{align}4^nn^m<m^m&<27^nn^m \implies 4^{n/m}n<m<27^{n/m}n\\[1ex] &\implies 4^{n/m}<m/n<27^{n/m}\\[1ex] &\implies 4<(m/n)^{m/n}<27\\[1ex] &\implies 2^2<q^q<3^3 \end{align}

where $\;q=m/n$.

Now, it is easy to see that the function $\;q\longmapsto q^q\;$ is strictly increasing for $q\geq 1/e$, and it is $\;\geq 1\;$ iff $q\geq1$. Hence, from the previous calculation we deduce that $\;2<q<3$. The bounds $2$ and $3$ are respectively $\;\it inf\;$ and $\;\it sup\;$ because

$$ \lim_{q\to2^+}q^q=2^2\qquad \text{and}\qquad \lim_{q\to3^-}q^q = 3^3.$$

Answer 2

Suppose $n,m \ne 0$, let's manipulate the second condition of the elements of that set: $$\begin{align} 4^n n^m < m^m < 27^n n^m & \iff 4^n < \left(\frac{m}{n}\right)^m < 27^n \\[8pt] & \iff 4^{\frac{n}{m}} < \frac{m}{n} < 27^{\frac{n}{m}} \end{align}$$

Now, if you call for simplicity $a := \frac{m}{n}$ the last equation gives $$4^{\frac{1}{a}} < a < 27^{\frac{1}{a}}$$ that holds if and only if $$2 < a < 3$$

With this conclusion we arrived to the following set: $$Z := \left\{a \in \mathbb{Q} \ : \ 2 < a < 3 \right\}$$