Supremum, infimum, limsup and liminf

1. For even $n$: $u_n=\frac n{n+1}$ increases to $1$ without reaching it. For odd $n$: $u_n=-\frac n{n+1}$ decreases to $-1$ without reaching it. Hence
$$\sup u_n=1,\ \inf u_n=-1\ \text{(not attained)},\qquad\limsup u_n=1,\ \liminf u_n=-1.$$

2. For even $n$: $v_n=1+\frac1n$ decreases to $1$, with maximum $v_2=\frac32$. For odd $n$: $v_n=-1+\frac1n$ is $0$ at $n=1$ and then decreases to $-1$ without reaching it. Hence
$$\sup v_n=\frac32\ \text{(attained)},\quad\inf v_n=-1\ \text{(not attained)},\qquad\limsup v_n=1,\ \liminf v_n=-1.$$

Remark. The supremum describes the *whole* sequence, whereas $\limsup$ only describes its behaviour at the tail. In example 2 they differ ($\frac32$ versus $1$) because the largest value occurs early. A bounded sequence converges iff $\limsup=\liminf$.