LimSup and LimInf

Motivation

If a sequence in $\mathbb{R}$ fails to converge, we would like to be able to measure the extent of the failure. To do so, we will define LimSup and LimInf in the extended real numbers (allowing them to possibly be infinity) for every sequence of real numbers. These will always exist.

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Preliminaries

To allow limits to be infinity, we must define what it means to converge to $\pm \infty$.

Definition 1(Definition).

We write $P_n\to +\infty$ if $\forall M\in \mathbb{R}$ there exists an integer $N$ such that $\forall n\ge N,p_n>M$.

Ditto with a swap for $-\infty$.

It should be noted that we now use the symbol $\to$ for certain types of divergent sequences, as well as for convergent sequences, but that the definitions of convergence and of limit, given here, are unchanged.

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Definition

À la Kulkarni

Definition 2(Definition).

Given a sequence $(p_n)$ in $\mathbb{R}$, define

$$\underset{n\to \infty }{\mathrm{limsup}}{p}_n:=\underset{n\to \infty }{\lim }(\sup \underset{\text{called a tail of }(p_n)}{\underbrace{\{p_n,p_{n+1},\dots \}}}),$$ $$\underset{n\to \infty }{\mathrm{liminf}}{p}_n:=\underset{n\to \infty }{\lim }(\inf \{p_n,p_{n+1},\dots \}).$$

Both values may be infinite.

À la Rudin

Definition 3(Definition).

Let $(s_n)$ be a sequence of real numbers. Let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{n_k}\to x$ for some subsequence $(s_{n_k})$. This set $E$ contains all subsequential limits of $(s_n)$, plus possibly $+\infty$ and $-\infty$. Define

$$\begin{array}{rl}{s}^{\ast }& :=\sup E,\\ s_{\ast }& :=\inf E.\end{array}$$

The numbers $s^{\ast }$ and $s_{\ast }$ are called the upper and lower limits of $(s_n)$. We use the notation

$$\begin{array}{r}\underset{n\to \infty }{\mathrm{limsup}}{s}_n:=s^{\ast },\\ \underset{n\to \infty }{\mathrm{liminf}}{s}_n:=s_{\ast }.\end{array}$$

Reconciling the two

We have to show that, for a sequence $(p_n)$,

$$\underset{n\to \infty }{\lim }(\sup \{p_n,p_{n+1},\dots \})=\sup \{x|p_{n_k}\to x\}$$

Proof

FTSOC, assume not. Let $(a_n):=(\sup \{p_n,p_{n+1},\dots \})$ and let $(a_n)\to A$. Let $\sup \{x|p_{n_k}\to x\}$ be $B$. We have two cases:

Case 1: ${\lim }_{n\to \infty }(\sup \{p_n,p_{n+1},\dots \})<\sup \{x|p_{n_k}\to x\}$.

Let $ϵ=\frac{B-A}{2}$. There exists $N$ such that for all $n\ge N$, $|a_n-A|<ϵ$. Thus, for all $n\ge N$, $p_n<A+ϵ$. However, this implies no subsequence of $(p_n)$ can converge to any point greater than $A+ϵ$. Thus, $A+ϵ$ is an upper bound for $\{x|p_{n_k}\to x\}$. This contradicts our hypothesis that $B$ is the supremum of all subsequential limits.

Case 2: ${\lim }_{n\to \infty }(\sup \{p_n,p_{n+1},\dots \})>\sup \{x|p_{n_k}\to x\}$.

Let $ϵ>0$. Note that $(a_n)$ is a monotone decreasing sequence. Since $(a_n)\to A$, there exists $a_k$ such that $A<a_k<A+ϵ$, and since $a_k$ is the supremum of the tail $\{p_k,p_{k+1},\dots \}$, there exists $p_i$, $i\ge k$ such that $A<p_i\le a_k<A+ϵ$. Thus, we can construct a subsequence $(p_{n_k})$ of $(p_n)$ which converges to $A$. However, this contradicts our hypothesis that $B$ is an upper bound of all subsequential limits. ❏

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Some observations and theorems

These will be using Kulkarni’s definitions.

• $\lim {\sup }_{n\to \infty }{p}_n=\infty ⟺(p_n)\text{ is not bounded above.}$
• Note that $s_n=\sup \{p_n,p_{n+1},\dots \}$ form a monotone decreasing sequence. Thus, ${\mathrm{limsup}}_{n\to \infty }{p}_n={\lim }_{n\to \infty }{s}_n=\inf \{s_n|n\in \mathbb{N}\}$.
• Similarly, $i_n=\inf \{p_n,p_{n+1},\dots \}$ form a monotone increasing sequence. Thus, ${\mathrm{liminf}}_{n\to \infty }{p}_n={\lim }_{n\to \infty }{i}_n=\sup \{i_n|n\in \mathbb{N}\}$.
• Quick exercise: show that $i_k\le s_{k^{\prime }}$ for all $k,k^{\prime }\in \mathbb{N}$.
• Thus, we have $i_1\le i_2\le i_3\le \cdots \le s_3\le s_2\le s_1$.
• Note that $\lim {\sup }_{n\to \infty }{p}_n=\infty$ means that every $s_i$ is $\infty$.

Theorem 4(Theorem).

Let $(p_n)$ be a sequence. Then, ${\mathrm{liminf}}_{n\to \infty }(p_n)\le {\mathrm{limsup}}_{n\to \infty }(p_n)$.

Proof
Pick any $s_k$. $i_m\le s_k$ for all $m$. Thus, $s_k$ is an upper bound for the set of all infima. Thus, $\lim {\inf }_{n\to \infty }{p}_n=\sup \{i_m\}\le s_k$. Since $s_k$ was arbitrary, the preceding statement is valid for all $k$. Thus, $\lim {\inf }_{n\to \infty }{p}_n$ is a lower bound for the set of all suprema. Thus, $\lim {\inf }_{n\to \infty }{p}_n\le \inf \{s_k\}=\lim {\sup }_{n\to \infty }{p}_n$. ❏

Theorem 5(Theorem).

If $s_n\le t_n$ for $n\ge N$, where $N$ is fixed, then

$$\begin{array}{rl}\underset{n\to \infty }{\mathrm{liminf}}{s}_n& \le \underset{n\to \infty }{\mathrm{liminf}}{t}_n,\\ \underset{n\to \infty }{\mathrm{limsup}}{s}_n& \le \underset{n\to \infty }{\mathrm{limsup}}{t}_n.\end{array}$$

The above result is rather intuitive once you internalize what limsup and liminf mean:

Important result

Important

Let $\lim {\sup }_{n\to \infty }{t}_n=p$. Then,

• for all $q>p$, there exists $N$ such for all $n>N$, we have $t_n<q$.
• for all $q<p$, for all $N$, there exists $n>N$ such that $t_n>q$.

Let $\lim {\inf }_{n\to \infty }{t}_n=p$. Then,

• for all $q>p$, for all $N$, there exists $n>N$ such that $t_n<q$.
• for all $q<p$, there exists $N$ such for all $n>N$, we have $t_n>q$.

Limsup = Liminf iff the sequence converges

Theorem 6(Theorem).

$$\underset{n\to \infty }{\lim }{p}_n=p⟺(\underset{n\to \infty }{\mathrm{limsup}}{p}_n=p)\text{ and }(\underset{n\to \infty }{\mathrm{liminf}}{p}_n=p)$$

Proof of $⟹$

Suppose $(p_n)\to p\in \mathbb{R}$. Let $ϵ>0$. There exists an $N$ such that for all $n>N$, $|p_n-p|<ϵ$, i.e, $\{p_n,p_{n+1},\dots \}\subset (p-ϵ,p+ϵ)$. Thus, $s_n$ and and $i_n$ must lie in $[p-ϵ,p+ϵ]$ which implies $\inf s_n$ and $\sup i_n$ must also lie in $[p-ϵ,p+ϵ]$. If we consider a sequence of $ϵ$‘s $({ϵ}_n)\to 0$, it follows from the nested interval property that $\inf s_n=\sup i_n=p$. ❏

Proof of $⟸$

$\lim \sup$ = $\lim \inf$ = $p$, i.e, $(s_n)\to p$ (monotonically from above), and $(i_n)\to p$ (monotonically from below). Let $ϵ>0$. Find $N$ such that for all $n>N$, we have $0\le s_n-p<ϵ$ and $0\le p-i_n<ϵ$. Now,

$$\begin{array}{rl}0\le s_n-p<ϵ& ⟹p_n-p<ϵ\\ 0\le p-i_n<ϵ& ⟹p-p_n<ϵ\\ ⟹|p-p_n|<ϵ.& \end{array}$$

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