Definition 1 : Let $A=\{a_n\}$ be a sequence of real numbers not necessarily bounded . Then we define :
$\lim \sup ~ a_n = \inf ~\{\sup ~a_n,~\sup ~a_{n+1}~\sup ~a_{n+2}, \cdots\} $
Definition 2 : If $A =\{a_n\}$ is a sequence of real numbers which is bounded above, then the limit superior of $A$ denoted by $S$ is the infimum of those real numbers $v$ with the property that there are only a finite number of natural numbers $n$ such that $v < a_n$
Please consider this image : 
Let the sequence $A$ denoted on the number line be bounded above by $b$ and may or may not be bounded below. ( In this case, we have taken $A$ to be bounded below by $a$ as well)
Now, $\{\sup ~a_n,~\sup ~a_{n+1}~\sup ~a_{n+2}, \cdots\}$ is a monotonically decreasing and bounded sequence which converges to the point $S . S_o =~ \sup ~a_n , S_1 = ~\sup ~a_{n+1}$ and so on.
Question 1: As per definition $2$, there can only be a finite numbers of terms to the right of $S$. But how can this be possible here when we have a whole sequence of supremums itself convering towards $S$.
Question 2: What could happen when $A$ is not bounded below? I fail to understand how can only a finite number of terms be there on the right side of $S$.
Thank you for help.
I hope that this makes things clear for you.
Let $A$ denote a sequence $\left(a_{n}\right)$.
Definition 1a) $\limsup A:=\lim_{n\rightarrow\infty}s_{n}$ where $s_{n}:=\sup\left\{ a_{k}\mid k\geq n\right\} $
The fact that sequence $\left(s_{n}\right)$ is non-increasing makes it possible to define equivalently:
Definition 1b) $\limsup A:=\inf\left\{ s_{n}\mid n\in\mathbb{N}\right\} $.
Now define $V:=\left\{ v\in\mathbb{R}\mid\text{the set }\left\{ n\in\mathbb{N}\mid v<a_{n}\right\} \text{ is finite}\right\} $ and note:
$$v\in V\iff\exists n\; s_{n}\leq v$$
This shows that $V=\cup_{n=1}\left[s_{n},\infty\right)$ and consequently $\inf V=\inf\left\{ s_{n}\mid n\in\mathbb{N}\right\} =\limsup A$
In special case where $A$ is not bounded above we have $s_n=+\infty$ for each $n$ so that $\limsup A=+\infty$. Secondly we have $V=\emptyset$ and $\inf\emptyset:=+\infty$. So everything remains valid in that case.