Find, if there exist, infimum and supremum of the set $$S:=\{[\cos((m^2+n)\pi)+\cos((m^2-n)\pi)]\frac{2-3m+2n-3mn}{2n+2mn};m,n\in \mathbb N\}$$ $$m,n \in \mathbb N \implies (m^2+n),(m^2-n)\in \mathbb Z$$ The sum of cosines transformed into product: $$\cos((m^2+n)\pi)+\cos((m^2-n)\pi)=2\cos(m^2\pi)\cos(n\pi)=\pm2,$$ which depends if $m,n\in \mathbb N$ are odd or even, so there are $4$ cases, $2$ $distinct$. I simplified the initial expression and got: $$\cos(m^2\pi)\cos(n\pi)\frac{(2-3m)(1+n)}{n(1+m)}$$ As mentioned before: $\cos(m^2\pi)\cos(n\pi)=\pm 1.$
Can you suggest how to continue and finish the task? Should I ramify the expression I got into more cases or $fix$ $the$ $values$ $1$ first for $n$, then for $m$ (as I was already suggested in a similar task today if you've seen the post)?
As Paul suggests, you can write each element $s$ of $S$ as $$ s=(-1)^m\left(-3 +\frac 5{1+m}\right)\times (-1)^n\left(1 + \frac 1n\right)\text{ for some }m,n\in \Bbb N.$$$$\implies |s|=\bigg|(-1)^m\left(-3 +\frac 5{1+m}\right)\times (-1)^n\left(1 + \frac 1n\right)\bigg|=\bigg|-3 +\frac 5{1+m}\bigg|\times \left(1 + \frac 1n\right)\leq |-3|\times\left(1 + \frac 1n\right)=3\left(1 + \frac 1n\right) \leq 3\times\bigg(1+\frac 11\bigg)=6$$$$\implies -6\leq s\leq 6,\forall s\in S.$$ Now the sequence $\{a_k\}$ defined by $$a_k=(-1)^{2k}\left(-3 +\frac 5{1+(2k)}\right)\times (-1)^1\left(1 + \frac 11\right)\in S$$ converges to $6$.
Also the sequence $\{b_k\}$ defined by $$b_k=(-1)^{2k+1}\left(-3 +\frac 5{1+(2k+1)}\right)\times (-1)^{1}\left(1 + \frac {1}{1}\right)\in S$$ converges to $-6$.
So $\text{inf}(S)=-6$ and $\sup(S)=6$.