Definition of limit superior Let $x_n$ be a bounded sequence, then the limit superior is the​ infimum of the set $V$ of $v\in\mathbb{R}$, such that $v < x_n$ for at most finite number of $n\in\mathbb{N}$.
How​ can​ the limit superior be the​ infimum? I came across this definition and I am confused
 A: Consider the sequence $y_n = (-1)^n(1+\frac{1}{n})$. Intuitively, we want $\lim \sup y_n = 1$, as this is the largest value that $y_n$ appears to be "approaching" in some sense. The set $V$ you defined contains all of the even terms of this sequence; namely, $1+\frac{1}{2},1+\frac 1 4, 1+\frac 1 6,...$. If we defined the $\lim \sup$ to be the supremum of $V$, then we would have $\lim \sup y_n = \sup V = \frac{3}{2}$. But, rather, $\lim \sup y_n = \inf V = 1$. The idea is that the $\lim \sup$ shouldn't be based on how early terms behave; rather, it should tell us about the long term behavior of the sequence. 
Here's another approach. Let $(x_n)$ be a bounded sequence. For each $n\in \mathbb{N}$, define
$$X_n = \sup \{x_k \, \vert \, k > n\}$$
That is, $X_n$ is the biggest term that appears after $x_n$. Then it's reasonable to define
$$\lim \sup x_n = \lim_{n\to \infty} X_n$$
Note that $X_n$ is a nonincreasing sequence (since the biggest term that appears after $x_{m+1}$ can't be bigger than the biggest term that appears after $x_m$). $X_n$ is also bounded below since $x_n$ is, and so the above limit exists. However, as said, $X_n$ is a nonincreasing sequence, and so
$$\lim_{n\to \infty} X_n = \inf \{X_n\, \vert \, n\in \mathbb{N}\}$$
and so
$$\lim \sup x_n = \inf \{X_n \, \vert \, n\in \mathbb{N}\}$$
