Using the standard order on $\mathbb{Q}$, I am trying to find the supremum of $A = \{1/n + (-1)^n \mid n \in \mathbb{Z}_{+}\}$. It made the most sense to me to try to break apart $A$ by the parity of $n$. Let's write $A = E \cup \mathcal{O}$, with $E$ corresponding to even $n$ and $\mathcal{O}$ corresponding to odd $n$.
\begin{align*}
A &= \{1/n + (-1)^n \mid \text{$n$ is even}\} \cup \{1/n + (-1)^n \mid \text{$n$ is odd}\} \\
&= \{1/n + (-1)^{2n} \mid n \in \mathbb{Z}_{+}\} \cup \{1/n + (-1)^{2n-1} \mid n \in \mathbb{Z}_{+}\} \\
&= \{1/n + 1 \mid n \in \mathbb{Z}_{+} \} \cup \{1/n - 1 \mid n \in \mathbb{Z}_{+} \} \\
&= \left\{2, 3/2, \right\} \cup \left\{0, -1/2, -2/3, \ldots \right\}
\end{align*}
Both $E$ and $\mathcal{O}$ are bounded above and below, so
$$
\sup A = \max\{\sup E, \sup \mathcal{O} \}, \qquad \inf A = \min\{\inf E, \inf \mathcal{O} \}.
$$
Both $A$ and $B$ are decreasing in $n$, so $\sup E = 2$, $\sup \mathcal{O} = 0$, and $\sup A = \max\{2,0\} = 2$.
I'm far less certain on finding the infimum. As $n \to \infty$, $\frac{1}{n} \to 0$, so I believe the elements of $E$ tend (from above) to $1$. Similarly, the elements of $\mathcal{O}$ tend to $-1$ from above. So $\inf E = 1$, $\inf \mathcal{O} = -1$, and $\inf A = \min\{-1,1\} = -1$.
How does this look?
You have changed the exponents from $n$ to $2n$ and $2n-1$. You should change the denominators as well.