The relation between the supremum and the limit superior of a sequence

Question

If $(a_n)_{n\ge 1} \subset \mathbb{R}$ is a bounded sequence, what is the relation between $\limsup\limits_{n\to \infty} a_n$ and $\sup\limits_{n\in \mathbb{N}}a_n$? My intuition tells me that they should be equal, but when I looked this up on Wikipedia I only found that $\limsup\limits_{n\to \infty}a_n\le \sup\limits_{n\in \mathbb{N}}a_n$. I can't produce an example where the inequality is strict and I believe that since $\sup\limits_{n\in \mathbb{N}}a_n$ is the limit of some subsequence of $(a_n)_{n\ge 1}$ and $\limsup\limits_{n\to \infty}a_n$ is the maximum of the limit points we should always have equality. So, please show me an example where this isn't true or prove that it is always true.

Answer 1

Take any strictly decreasing and convergent sequence $(a_n)_{n\in\mathbb{N}}\subseteq\mathbb{R}$ and let $a$ be the limit of $(a_n)_{n\in\mathbb{N}}$. Then $a$ is the only accumulation point of $(a_n)_{n\in\mathbb{N}}$, hence $\limsup_{n\to \infty} a_n = \lim_{n\to\infty} a_n= a$. As $(a_n)_{n\in\mathbb{N}}$ is strictly decreasing by assumption, we have $a_1>a_2>...>a$, so that $\sup_{n\in \mathbb{N}} a_n =a_1>a=\limsup_{n\to \infty} a_n$.

For instance, the sequence $a_n:=\frac{1}{n}$, $n\in \mathbb{N}$ (with $a:=0$) has such properties.