Definition

Let $(a_{n})_{n\in\mathbb{N}}\subseteq \bar{\mathbb{R}}$. Define a new sequence $(b_{n})_{n\in\mathbb{N}}$ as $$b_{n}=\sup_{k\ge n}a_{k}$$where we observe $b_{1}\ge b_{2}\ge\dots$ (i.e. monotonically decreasing). Then we define the limit superior of our sequence as $$\limsup_{n\ge 1}a_{n}:=\inf_{n\ge 1}b_{n}=\inf_{n\ge 1}\sup_{k\ge n}a_{k}$$Let $T=\{ t\in \bar{\mathbb{R}}: \exists(a_{n_{k}})_{k\in\mathbb{N}}\subset(a_{n})_{n\in\mathbb{N}}:\lim_{ k \to \infty }a_{n_{k}}=t \}$. Then $$\limsup_{ n \ge 1 }a_{n}=\sup_{t\in T}t$$

---

The limit superior of $(a_n)$, $$\limsup_{n\to\infty}a_n = \lim_{n\to\infty}(\sup_{m\ge n}a_n)$$ is defined as follows: 1. If $(a_n)_{n\in\mathbb{N}}\subset T$ is bounded above and $T\not=\emptyset$ then $\limsup_{n\to\infty}a_n$ is the largest number in $T$ or $$\limsup_{n\to\infty}a_n=\sup T$$ 2. If $(a_n)$ is not bounded above then $$\limsup_{n\to\infty}a_n=\infty$$ 3. If $(a_n)$ is bounded above and $T=\emptyset$ then $$\limsup_{n\to\infty}a_n=-\infty$$ ## Remark To provide intuition for this you can think of the $\limsup$ as giving “the largest value that the sequence approaches infinitely often”. While the supremum gives you the static LUB of a sequence the $\limsup$ focuses on the “tail behaviour” of a sequence and what values the sequence “settles down to” as n approaches infinity. This is important for understanding the long-term behaviour of a sequence.

Intuition