I'm aware that there are multiple questions on this topic and I have read most of them, but I still don't really understand the definition and would really appreciate some quick help with this, because I have been struggling with this concept for quiet some time.
At least after having read through these questions I understand now that intuitively the limit superior is the supremum of the set containing the limits of all subsequences for a given sequence $(x_n)_{n \in \mathbb{N}}$. However, what I don't understand is how that intuitive definition relates to this formal definition:
$$ \limsup_{k \rightarrow \infty} (x_k) = \lim_{k \rightarrow \infty} \left( \sup \left(\bigcup_{i=k}^\infty \{x_i\}\right) \right)$$
From what I read into this equation, you first take the union starting from a given k, let's say $k=1$. Then you basically have all the terms of the sequence in this union. For simplicity let's assume that the supremum is included in this union. Then the supremum of this union is the largest element in the whole sequence. As you increase k and eventually as $k \rightarrow \infty$ the supremum should not change, since you already have had the largest element which is the supremum in the case $k=1$. So why would you take the limit of k?
So basically what I don't understand is that from looking at the equation (my interpretation is very likely wrong), the limit superior is the largest term in the sequence, but the largest element does not necessarily need to be a limit of some subsequence, does it? So have can I relate this equation to the actually meaning of limit superior?
Let $(x_{n})_{1}^{\infty}$ be a sequence of reals. Remove $x_{1}$ and take the supremum of the remaining sequence $(x_{n})_{2}^{\infty}$. Remove $x_{2}$ and take the supremum of the remaining sequence $(x_{n})_{3}^{\infty}$. By repeating this procedure we will form a new sequence $(\sup_{n \geq k}x_{n})_{2}^{\infty}$ of suprema. Now the limit superior of the sequence $(x_{n})_{1}^{\infty}$ is simply the limit of the sequence $(\sup_{n \geq k}x_{n})_{2}^{\infty}$ thus formed.
Note that the limit inferior is constructed in the same fashion, with taking the infimum instead.