Some Questions about $\limsup$

Question

The following theorem appears (unproven) in Rudin's RCA:

If $\{a_n\}$ converges, then evidently $\displaystyle \limsup_{n \to \infty} a_n = \liminf_{n \to \infty} a_n = \lim_{n \to \infty} a_n $

Note that $\displaystyle \limsup_{n \to \infty} = \inf \bigg\{ \sup \{a_k \mid k \ge n \} \mid n \in \Bbb{N} \bigg\}$, and since $\displaystyle \liminf_{n \to \infty} a_n = - \limsup_{n \to \infty} (-a_n)$, we need only prove the one equality. Since $\sup \{a_k \mid k \ge n \}$ is a decreasing sequence, if it converges its limit is $\inf \bigg\{ \sup \{a_k \mid k \ge n \} \mid n \in \Bbb{N} \bigg\}$, and so the goal is to show that $\sup \{a_k \mid k \ge n \}$ converges to $\ell = \lim_{n \to \infty} a_n$ to prove that $\displaystyle \limsup_{n \to \infty} a_n = \lim_{n \to \infty} a_n$. Observe that

$$|\sup \{a_k \mid k \ge n \} - \ell| \le |\sup \{a_k \mid k \ge n \} - a_n| + |a_n - \ell|,$$

so I just need to prove that $|\sup \{a_k \mid k \ge n \} - a_n| \to 0$ as $n \to \infty$, which I am having trouble doing...

I am also wondering whether $\displaystyle \lim_{n \to \infty} a_n \le \limsup_{n \to \infty} a_n$ holds whether $a_n$ converges (either to a finite or infinte value) or diverges. Obviously once the above theorem is proven, then if $x_n$ converges, the inequality holds (in fact, equality to be precise). But I am not sure about the case when $x_n$ doesn't converge to a finite value.

Answer 1

$\sup \{a_n\}\ge a_n \ge \inf \{a_n\}$

$\limsup_\limits{n\to \infty} a_n, \limsup_\limits{n\to \infty} a_n$ is restricting ourselves to the special case when $n>N$

For example, if $a_n = \cos n$ then $\{a_n\}$ does not converge: $\limsup_\limits{n\to \infty} a_n =1$ and $\liminf_\limits{n\to \infty} a_n = -1$

If $\{a_n\}$ converges, then $n>N \implies |a_n - a| < \epsilon$ for any epsilon and $\limsup_\limits{n\to \infty} a_n = \lim_\limits{n\to \infty} a_n = \liminf_\limits{n\to \infty} a_n = a$

Answer 2

Suppose $\lim_{n\to\infty}a_n=l$. Take $\varepsilon>0$. There is then some $p\in\mathbb N$ such that $n\geqslant p\implies|a_n-l|<\varepsilon$. Therefore, if $n\geqslant p$, $\sup\{a_k\,|\,k\geqslant n\}\leqslant l+\varepsilon$ and $\inf\{a_k\,|\,k\geqslant n\}\geqslant l-\varepsilon$ and so$$\limsup_{n\to\infty}a_n=\inf_{n\in\mathbb N}\left\{\sup\{a_k\,|\,k\geqslant n\}\right\}\leqslant l+\varepsilon\text{ and }\liminf_{n\to\infty}a_n=\sup_{n\in\mathbb N}\left\{\inf\{a_k\,|\,k\geqslant n\}\right\}\geqslant l-\varepsilon.$$Since $\limsup_{n\to\infty}a_n\geqslant\liminf_{n\to\infty}a_n$, it follows from this that$$\limsup_{n\to\infty}a_n,\liminf_{n\to\infty}a_n\in[l-\varepsilon,l+\varepsilon]$$and, since this occurs for each $\varepsilon>0$, $\limsup_{n\to\infty}a_n,\liminf_{n\to\infty}a_n=l=\lim_{n\to\infty}a_n$.

Actually, if $\lim_{n\to\infty}a_n=\pm\infty$, then it is still true that $\limsup_{n\to\infty}a_n,\liminf_{n\to\infty}a_n=\lim_{n\to\infty}a_n$.

Answer 3

Goal: $|\sup \{a_k \mid k \ge n \} - a_n| \to 0$ as $n\to0$.

Note that $a_n \in \{a_k \mid k \ge n \}$, so we can drop the absolute value sign and write $$|\sup \{a_k \mid k \ge n \} - a_n|=\sup \{a_k-a_n \mid k \ge n \} \quad \forall\,n \in \Bbb{N}.\tag{*}\label{*}$$

Let $\epsilon>0$.

Method 1: Cauchy sequence

Use the definition of Cauchy to get $N$ responding to the $\epsilon$-challenge (i.e. $\forall\,n,k\ge N, |a_n-a_k|\le\epsilon$.) Choose $n \ge k$ and recall that $b \le |b|\,\forall\,b\in\Bbb{R}$. Take $\sup$ over $k \ge n$.) $$\forall\,n\ge N,\sup \{a_k\mid k \ge n \}-a_n =\sup \{a_k-a_n \mid k \ge n \}\le \epsilon$$

---

Method 2: Think from the tail

Translate the $\sup$ on RHS of \eqref{*} into qualitifers. $$\sup \{a_k-a_n \mid k \ge n \}\le \epsilon \iff \forall\,k\ge n, a_k-a_n \le \epsilon.\tag#\label#$$

We don't need $a_k-a_n \le \epsilon$ in \eqref{#} to hold for all $k\in\Bbb{N}$. We just need this to be "eventually true", because convergence is a "tail fact". (The rigorous definition of "eventually $(P)$" can be defined as $\exists N_0\in\Bbb{N}$ such that $\forall\,n\ge N_0,$ property $P(n)$ holds. I use "tail fact" as convergence doesn't depend on its initial value (i.e. independent of the "head"). I invent the word "tail fact" to capture "tail" as in "tail event" and "tail $\sigma$-algebra" in probability-theory.)

In common terms, if $(x_n)_n$ converges to $x$, its "tail fluctuation" $|x_n-x|$ can be "eventually controlled" by arbittrary $\epsilon$. $$i.e. \exists N_0\in\Bbb{N}\text{ s.t. } \forall\,n\ge N, |x_n-x|\le\epsilon$$

The advantage of talking in eventual truths is that if we know something is eventually $(\heartsuit)$ and eventually $(\clubsuit)$, then it's eventually $(\heartsuit$ and $\clubsuit)$, since one of the tails must contain another ($\{n\mid n\ge N_1\} \subseteq \{n \mid n\ge N_2\}$). The process can be carried $m(\in\Bbb{N})$ times. The typical argument "choose $N'''=\max\{N,N_0,N',\dots,N_0''\}$" can be capsulated.

Writing in "tails", you assume $a_n\to\ell$ as $n\to0$, so the tail fluctuation $|a_n-\ell|$ is eventually controlled by $\epsilon$. In other words, $$\ell-\epsilon<a_n<\ell+\epsilon\tag{@}\label@$$ eventually on a certain tail. For any $n$ on the tail (a set $\{m\in\Bbb{N} \mid m\ge N\}$), the index $k\ge n$ runs over a sub-tail, which is included in the tail, so we are safe to take $\sup$ over $k\ge n$ on \eqref{@}. \begin{align} & \ell-\epsilon<a_n<\ell+\epsilon \\ & \ell-\epsilon<\sup\{a_k\mid k\ge n\}<\ell+\epsilon \\ & \sup\{a_k\mid k\ge n\} - a_n < \ell+\epsilon-(\ell-\epsilon) = 2\epsilon \end{align} Therefore, the tail of $(\sup \{a_k \mid k \ge n \} - a_n)_n$ is eventually controlled by $2\epsilon$. From this, we deduce the desired convergence.