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Notation 1.1 The extended real numbers is the set $\overline{\mathbb{R}}:=\mathbb{R} \cup{ \pm \infty}$, i.e. it is $\mathbb{R}$ with two new points called $\infty$ and $-\infty$. We use the following conventions, $\pm \infty \cdot 0=0, \pm \infty \cdot a= \pm \infty$ if $a \in \mathbb{R}$ with $a>0, \pm \infty \cdot a=\mp \infty$ if $a \in \mathbb{R}$ with $a<0, \pm \infty+a= \pm \infty$ for any $a \in \mathbb{R}, \infty+\infty=\infty$ and $-\infty-\infty=-\infty$ while $\infty-\infty$ is not defined. A sequence $a_n \in \overline{\mathbb{R}}$ is said to converge to $\infty(-\infty)$ if for all $M \in \mathbb{R}$ there exists $m \in \mathbb{N}$ such that $a_n \geq M\left(a_n \leq M\right)$ for all $n \geq m$.

Lemma 1.2. Suppose $\left{a_n\right}_{n=1}^{\infty}$ and $\left{b_n\right}_{n=1}^{\infty}$ are convergent sequences in $\mathbb{R}$, then:

If $a_n \leq b_{\mathrm{n}}$ for $^1$ a. $a$. $n$ then $\lim {n \rightarrow \infty} a{\mathrm{n}} \leq \lim _{n \rightarrow \infty} b_n$.

If $c \in \mathbb{R}, \lim {n \rightarrow \infty}\left(c a_n\right)=c \lim {n \rightarrow \infty} a_n$.

If $\left{a_n+b_n\right}_{n=1}^{\infty}$ is convergent and
$$
\lim {n \rightarrow \infty}\left(a_n+b_n\right)=\lim {n \rightarrow \infty} a_n+\lim _{n \rightarrow \infty} b_n
$$
provided the right side is not of the form $\infty-\infty$.

$\left{a_n b_n\right}_{n=1}^{\infty}$ is convergent and
$$
\lim {n \rightarrow \infty}\left(a_n b_n\right)=\lim {n \rightarrow \infty} a_n, \lim {n \rightarrow \infty} b_n $$ provided the right hand side is not of the for $\pm \infty \cdot 0$ of $0 \cdot( \pm \infty)$. Before going to the proof consider the simple example where $a_n=n$ and $b_n=-\alpha n$ with $\alpha>0$. Then $$ \lim \left(a_n+b_n\right)=\left{\begin{array}{cc} \infty & \text { if } \alpha<1 \\ 0 & \text { if } \alpha=1 \\ -\infty & \text { if } \alpha>1 \end{array}\right. $$ while $$ \lim {n \rightarrow \infty} a_n+\lim _{n \rightarrow \infty} b_n{ }^{i n}=” \infty-\infty
$$
This shows that the requirement that the right side of Eq. (1.1) is not of form $\infty-\infty$ is necessary in Lemma 1.2. Similarly by considering the examples
${ }^1$ Here we use “a.a. $n^n$ as an abreviation for almost all $n$. So $a_n \leq b_n$ a.a. $n$ iff there exists $N<\infty$ such that $a_n \leq b_n$ for all $n \geq N$.

12
1 Limsups, Liminfs and Extended Limits
$a_n=n$ and $b_n=n^{-\alpha}$ with $\alpha>0$ shows the necessity for assuming right hand side of Eq. (1.2) is not of the form $\infty \cdot 0$.
Proof. The proofs of items 1. and 2. are left to the reader.
Proof of Eq. (1.1). Let $a:=\lim {n \rightarrow \infty} a_n$ and $b=\lim {n \rightarrow \infty} b_n$. Case 1., suppose $b=\infty$ in which case we must assume $a>-\infty$. In this case, for every $M>0$, there exists $N$ such that $b_n \geq M$ and $a_n \geq a-1$ for all $n \geq N$ and this implies
$$
a_n+b_n \geq M+a-1 \text { for all } n \geq N \text {. }
$$
Since $M$ is arbitrary it follows that $a_n+b_n \rightarrow \infty$ as $n \rightarrow \infty$. The cases where $b=-\infty$ or $a= \pm \infty$ are handled similarly. Case 2 . If $a, b \in \mathbb{R}$, then for every $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that
$$
\left|a-a_n\right| \leq \varepsilon \text { and }\left|b-b_n\right| \leq \varepsilon \text { for all } n \geq N
$$
Therefore,
$$
\left|a+b-\left(a_n+b_n\right)\right|=\left|a-a_n+b-b_n\right| \leq\left|a-a_n\right|+\left|b-b_n\right| \leq 2 \varepsilon
$$
for all $n \geq N$. Since $n$ is arbitrary, it follows that $\lim {n \rightarrow \infty}\left(a_n+b_n\right)=a+b$. Proof of Eq. (1.2). It will be left to the reader to prove the case where $\lim a_n$ and $\lim b_n$ exist in $\mathbb{R}$. I will only consider the case where $a=\lim {n \rightarrow \infty} a_n \neq 0$ and $\lim {n \rightarrow \infty} b_n=\infty$ here. Let us also suppose that $a>0$ (the case $a<0$ is handled similarly) and let $\alpha:=\min \left(\frac{a}{2}, 1\right)$. Given any $M<\infty$, there exists $N \in \mathbb{N}$ such that $a_n \geq \alpha$ and $b_n \geq M$ for all $n \geq N$ and for this choice of $N, a_n b_n \geq M \alpha$ for all $n \geq N$. Since $\alpha>0$ is fixed and $M$ is arbitrary it follows that $\lim {n \rightarrow \infty}\left(a_n b_n\right)=\infty$ as desired.

For any subset $A \subset \overline{\mathbb{R}}$, let sup $A$ and inf $A$ denote the least upper bound and greatest lower bound of $A$ respectively. The convention being that $\sup A=\infty$ if $\infty \in A$ or $A$ is not bounded from above and $\inf A=-\infty$ if $-\infty \in \Lambda$ or $A$ is not bounded from below. We will also use the conventions that $\sup \emptyset=-\infty$ and $\inf \emptyset=+\infty$

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