I tasked myself with proving the equivalence of these two definitions of limit superior for a bounded sequence:
$ (1) \ \overline{\lim} a_n = \sup \{ x \in \mathbb{R} : x < a_n \ \text{ for infinitely many } n\in \mathbb{N} \} $
$ (2) \ \displaystyle \overline{\lim} a_n = \lim_{n \to \infty} \sup \{ a_n, a_{n+1}, \ldots \} $
I would love some feedback on my attempt below. I'm curious to know whether I could make my arguments more concise, or if there is perhaps a shorter approach.
Proof:
Denote $\lambda_n = \sup\{a_n, a_{n+1}, \ldots \}$. Then $\lambda_n \geq \lambda_{n+1}$ for each natural $n$.
The sequence $(\lambda_n)$ is monotonely decreasing and bounded, hence convergent. The limit is $\inf \{\lambda_1, \lambda_2, \ldots \}$.
The task therefore is to show $ \inf \{\lambda_1, \lambda_2, \ldots \} = \sup \{ x \in \mathbb{R} : x < a_n \ \text{ for infinitely many } n\in \mathbb{N} \} $.
In interest of brevity, I will denote $A = \{ x \in \mathbb{R} : x < a_n \ \text{ for infinitely many } n\in \mathbb{N} \} $.
Notice that each $\lambda_n = \sup \{ a_n, a_{n+1}, \ldots \}$ is greater than or equal to all terms after (and including) $n$. This means that $\lambda_n$ at most can have $n-1$ terms above it, namely a finite number. Therefore, $\lambda_n \not \in A$ for each $n$.
We will prove the equivalence by cases.
Suppose first $\inf \{\lambda_1, \lambda_2, \ldots \} < \sup A$. Then there is an $x \in A$ such that $ \inf \{\lambda_1, \lambda_2, \ldots \} < x \leq \sup A$. But since $\inf \{ \lambda_1, \lambda_2, \ldots \} < x $, there must exist some $\lambda_k$ such that $ \lambda_k < x $. But then $\lambda_k$ will have infinitely many terms above it (due to $x \in A$), which will imply $\lambda_k \in A$ and this is false.
We conclude that $\inf \{\lambda_1, \lambda_2, \ldots \} < \sup A$ cannot hold.
Suppose next $\inf \{\lambda_1, \lambda_2, \ldots \} > \sup A$. Then there is a real $x$ such that $\inf \{ \lambda_1, \lambda_2, \ldots \} > x > \sup A$. This means $x < \lambda_n = \sup \{a_n, a_{n+1}, \ldots \}$ for each natural $n$. Consequently $x$ will have infinitely many terms above it, which will imply $x \in A$. But this cannot be possible, since also $x > \sup A$. Therefore there cannot exist a real $x$ between $\inf \{ \lambda_1, \lambda_2, \ldots \}$ and $ \sup A$.
We conclude that $\inf \{\lambda_1, \lambda_2, \ldots \} > \sup A$ also cannot hold.
This leaves us with $\inf \{ \lambda_1, \lambda_2, \ldots \} = \sup A$. Done. $\blacksquare$.
Thoughts? I'm especially curious about the latter case $\inf \{ \lambda_1, \lambda_2, \ldots \} > \sup A$; I couldn't do it without having to assert the existence of $x \in \mathbb{R}$ and insert it between the two real numbers.