The definition I am given is as follows: Let $(x_n)$ be a real valued sequence. For each positive integer $n$, let $s_n:=\sup\{x_m:m\geq n\}$. If $(s_n)$ converges, we denote its limit by $\overline{\lim_{n\rightarrow\infty}}x_n$ and call it the limit superior of $(x_n)$.
Now I am to prove that the limit superior of a real valued sequence exists if and only if the sequence is bounded above.
I understand that if the sequence is $\textit{not}$ bounded above the the limit superior will not exist, since there is no supremum of the tail of the sequence.
But as far as the other direction goes, I assume that the sequence is bounded above. But what if it is not bounded below? For example, what if $(x_n)=(-1,-2,-3,-4,...)$. Then, if my understanding is right, we should have $s_1=-1,s_2=-2,s_3=-3$, and in general, $s_n=-n$. But then $(s_n)$ does not converge, and by the definition given to me I have to conclude that the limit superior does not exist.
Am I confused somewhere?
The sequence $s_n:=\sup\{x_m:m\geq n\}$ is non-increasing.
To see this just notice that if $n\le k$, then $\{x_m \colon m\geq n\}\supseteq\{x_m \colon m\geq k\}$ and thus $s_n=\sup \{x_m \colon m\geq n\} \ge x_k = \{x_m \colon m\geq k\}$.
Every bounded monotone sequence is convergent. See, for example, Wikipedia.
If the sequence is not bounded below, then so is the sequence $(s_n)$. So in this case we get $\limsup x_n = \lim s_n = -\infty$.If we allow limit also to be $\pm\infty$, then very monotone sequence has either finite limit or limit equal to $+\infty$ or $-\infty$. So we can say that every sequence (without any condition on boundedness) has limit superior, if we allow also $\pm\infty$.
You are correct in saying that your example $x_n=-n$ does not have a finite limit superior. (And, as you write in your comment, the same is true whenever $\lim x_n=-\infty$.