$x_n = \frac {(n-1)(-1)^n}{n}$, find $\lim \sup x_n$ and $\lim \inf x_n$.
Let $a_n = \sup\{x_k : k\ge n\}$ and $b_n = \inf\{x_k : k\ge n\}$.
$a_n = \sup \{\frac {n-1}{n}, \frac {-n}{n+1}, \frac {n+1}{n+2}, \frac {-(n+2)}{n+3}, ...\}$ when $n=even$.
$a_n = \sup \{\frac {-(n-1)}{n}, \frac {n}{n+1}, \frac {-(n+1)}{n+2}, \frac {n+2}{n+3}, ...\}$ when $n=odd$
In both cases, $a_n = \lim_{k \to \infty} \frac {n+k}{n+k+1} = 1$. (*)
Similarly, I found $b_n = -1$.
Therefore, since $\lim \sup x_n= \lim a_n$ and $\lim \inf x_n = \lim b_n$, $\lim \sup x_n = 1$ and $\lim \inf x_n = -1$ .
Could you check this proof is okay? Is (*) correct?