Fastest way to find $\sup$ and $\inf$ of this set

Question

I was asked to find the $\sup$ and $\inf$ of the set $X= \left \{\frac{n^2+(-1)^nn}{n^2+n+1} : n\in \mathbb{N^+}\right\}$. Since I wanted to work without that $(-1)^n$ term, I wrote X as $X=A \cup B$, where $A= \left\{ \frac{n^2-n}{n^2+n+1}:n \not\in \mathbb{2N}, n \geq1 \right\}$ and $B=\left\{ \frac{n^2+n}{n^2+n+1}: n\in \mathbb{2N}, \ n\geq2 \right\}$. Letting $a_n$ and $b_n$ being the elements of A and B respectively, I proved that $a_n<a_{n+1}$ and that $b_n<b_{n+1}$, moreover, both sequences converge to 1 so $\sup(a_n)=\sup(b_n)=1$ and so $\sup(X)=1$. Also, since the sequences are increasing, $\min(a_n)=a_2=6/7$ and $\min(b_n)=b_1=0$, so we deduce that $\inf(X)=\min(X)=0$. Is my reasoning correct? I would like to know whether there's a faster way do the job, thanks

Answer 1

What you did looks correct.

If $c_n=\frac{n^2+(-1)^nn}{n^2+n+1}$, we always have $c_n\geqslant0$. Besides, $c_1=0$, and therefore $\inf\{c_n\mid n\in\Bbb N\}=0$. On the other hand, we always have $c_n\leqslant\frac{n^2+n}{n^2+n+1}\leqslant1$, and therefore $\sup\{c_n\mid n\in\Bbb N\}\leqslant1$. But$$\lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{1+\frac{(-1)^n}n}{1+\frac1n+\frac1{n^2}}=1,$$and so $\sup\{c_n\mid n\in\Bbb N\}=1$.