Consider a sequence $(x_n)$ defined by $x_n=(-1)^n\frac{n+1}{n}$. It is clear that
$$\limsup_{n\to\infty} x_n=\inf_{n\in\mathbb N}\sup_{k\ge n}(-1)^n\frac{n+1}{n}=\inf_{n\in\mathbb N}\begin{cases}\frac{n+1}{n} & \text{$n$ even} \\\ \frac{n+2}{n+1} & \text{$n$ odd}\end{cases}=1$$
I am not sure how to prove why it is true.
I think it is enough to prove that $x_n$ is bounded above by $1$ at sufficiently large $n$, i.e. $x_n\le1$ for all but finite $n$. Is that right? I have trouble proving this result because $x_n>1$ for all even $n$.