What is the least upper bound of the sequence defined, for $n\in\mathbb{N}$, by
$$a_{n}=\begin{cases} 3/n, & \text{if }n\text{ is odd;}\\ 1/n, & \text{if }n\text{is even.} \end{cases}$$
I know that the least upper bound is $3$, but how do I determine the least upper bound?
Notice the behavior of the sequence as $n \to \infty$ on even and odd $n$ separately. In each case, for every even $n$, the sequence is monotonic decreasing, and the same for every odd $n$ the same is also true.
Logically, then, this suggests that the sequences' highest points are at the start of the sequence, i.e. $n=1$ for odd $n$ and $n=2$ for even $n$. Well, in those cases, you get the values $a_1 = 3$ and $a_2 = 1/2$.
The former is visibly higher, so we say that the least upper bound is $3$ - it is the least element such that each subsequent element is less than it.