As in the title, I have to find supremum and infimum of the set $$S=\left\\mid n \in \mathbb N,\, k \in \mathbb N, 2|k\right\}$$
I know that $k^6$ should "grow faster" than $n^2$ if both are natural numbers. Then I can transform that expression into: $$$$ For $n \implies \infty$ I get: $$\frac{n^2}{8n^2}=\frac{1}{8}$$ That would be my supremum. BUT there is a question of infimum and the information that $2|k$. I don't know how to use that information in the process of finding the infimum. Any help would be much appreciated.
Important note: in that case, we take $\mathbb N as \{1, 2, 3, 4, 5, 6, 7...\}$ (without $0$).
Denote for $n,k \in \mathbb N$
$$s(n,k)= \frac{n^2k^6}{n^2+8n^2k^6+9}.$$
All the elements of $S$ are less than $1/8$. Now taking $n=1$ and $k=2l$ we have
$$\lim\limits_{l \to \infty} s(1,2l) \to 1/8$$
and $s(1,2l) \in S$ for all $l \in \mathbb N$, proving that $\sup S=1/8$.
Regarding the $\inf$, consider the map
$f(x) = \frac{n^2 x}{n^2 + 8n^2x + 9}$ defined for $x \in [2^6, \infty)$. You have $$f^\prime(x) = \frac{n^2(n^2 + 8n^2x + 9) - 8n^4x}{(n^2 + 8n^2x + 9)^2}= n^2\frac{n^2+9}{(n^2 + 8n^2x + 9)^2}$$ which is positive, proving that f achieve its minimum when $x=2^6$, i.e. $k=2$. We get
$$s(n, k) \ge \frac{64n^2}{513n^2+9}$$ for $n,k \in \mathbb N$ when $k$ is even. Analyzing the map $g(x) = \frac{64x}{513x+9}$ for $x \in [1,\infty)$, you'll find that it is increasing on that interval. Therefore its minimim is achieved for $x=1$ and the minimum of $S$ is $s(1,2)=\frac{32}{261}$.