Find the inf and sup of a set A

Question

Let $X=\{x_n=(-1)^n\frac{n^2}{n^3}: n\in\mathbb{N}, n\geq 1\}$. My idea to find the inf and sup is the following:
let the sequence $x_n$ for $n$ odd: $-\frac{n^2}{n^3}$. It is increasing and so $\lim_{n\to\infty}x_n=0=\sup$. The inf instead is -1.
Now let the sequence for $n$ even. It is decreasing and so $\lim_{n\to\infty}x_n=0=\inf$. The sup is instead $\frac{1}{2}$.
Now we conclude that $\sup X=\max\{0,\frac{1}{2}\}=\frac{1}{2}$ and $\inf X=\min\{0,-1\}=-1$. To do: check if my idea is correct, please.