Difference between limit superior & supremum of a sequence

The supremum is in fact the least upper bound of the set or sequence. That is, of all upper bounds, it is the smallest. All others are greater than it. Think about the open interval, say A = (0,1), in $\mathbb{R}$. 3, 75, 1.4 are all upper bounds of A, since $\forall x \in$ A, $x \le$ any of these. Of all the upper bounds of A, however, none can be smaller than 1. Thus 1 is the least upper bound, or supremum, of A. Similarly, 0 is the greatest lower bound, or infimum, of A.

Consider the sequence $\{ a_n\}$, where $a_n = (-1)^n$. You can probably determine that this does not converge. There are subsequences, $\{ a_{n_k}\}$ and $\{ a_{n_i}\}$, $n_k$ = 1,3,5... and $n_i$ = 2,4,6..., that do, to -1 and 1, respectively. So E = {-1,1} is the set of subsequential limits of $\{ a_n\}$. The supremum, or maximum of this finite set, is 1. This is the lim sup of $\{ a_n\}$. Similarly, lim inf = -1.

This also works, if the subsequences do not attain the limit either, as in the sequence $\{ b_n\}$, where $b_n$ = $\frac{1}{n} + (-1)^n$. Here, lim sup = 1, but sup = $\frac{3}{2}$, lim inf = inf = -1

Let $(a_{n})$ be a sequence of real numbers. Then the supremum of $(a_{n})$ is by definition the number $\sup_{n}a_{n} := \sup \{ a_{n} | n \in \mathbb{N} \}$. The infimum of $(a_{n})$ is defined likewise.

Now let $b_{n} := \sup_{k \geq n}a_{k}$ for all $n$, which intuitively means that we discard $a_{1},\dots, a_{n}$ and take the supremum of the remaining sequence for every $n$. Then the limit superior of $a_{n}$ is defined as the number $\lim_{n \to \infty}b_{n} = \inf_{n}b_{n}$ (in order that you get a better understanding of this, you may want to try proving the preceding equality and the existence of the numbers involved). The limit inferior of $(a_{n})$ is defined likewise.