% This is lecture no. 1 of MH1101/: Calculus II.

\chapter{Sequences}%
\label{cha:sequences}

\section{Notation \& Representations}%
\label{sec:notation_&_representations}

\subsection{Notation}%
\label{sub:notation}

The sequence:
\[
	\{a_1,a_2,a_3,\dotsc \}
.\]
is usually denoted by one of the following notations:
\[
	\{a_n\} \quad \{a_n \}^{\infty}_{n=1} \quad (a_n) \quad (a_n )^{\infty}_{n=1}
.\]

Often times, the index of a sequence may start from an integer other than 0 or 1.

\begin{note}
	A sequence of real numbers can be considered as a \textbf{function} from a subset of $\mathbb{N}$ to $\mathbb{R}$.
\end{note}

\begin{exmp}
	Sequence defined by a formula for the $n$-th term:
	\[
		\left\{\frac{n}{n+1}\right\}^{\infty}_{n=1}
		\implies
		a_n = \frac{n}{n+1}
	.\]
	Therefore the sequence's \textbf{listing} is of the following form:
	\[
		\left\{\underbrace{\frac{1}{2}}_{n=1}, \underbrace{\frac{2}{3}}_{n=2}, \underbrace{\frac{3}{4}}_{n=3}, \dotsc, \frac{n}{n+1}, \dotsc \right\}
	.\]
\end{exmp}

\subsection{Missing start index}%
\label{sub:missing_start_index}

If the starting index of a sequence is not stated, then it will be the smallest integer that the formula defining $a_n$ makes sense and is a real number.
\begin{exmp}
	Determine the starting index for the sequence:
	\[
		a_n = \sqrt{n-3}
	.\]
	The term $a_n = \sqrt{n-3}$ does not make sense for $n-3 < 0$.
	It is only a real number when $n-3 \geqslant 0$.
	Thus we have:
	\[
		\left\{\sqrt{n-3}\right\}^{\infty}_{n=3}
	.\]
\end{exmp}

\begin{exmp}
	Determine the starting index for the sequence:
	\[
		b_n = \ln (n^2 -5)
	.\]
	The term is defined when $n^2 -5 > 0$, and therefore:
	\[
		n > \sqrt{5}, \quad n \in \mathbb{N}
		\implies
		n \geqslant 3
	.\]
	Thus we have:
	\[
		\left\{\ln (n^2 -5)\right\}^{\infty}_{n=3}
	.\]
\end{exmp}

\subsection{Recursively defined sequences}%
\label{sub:recursively_defined_sequences}

The Fibonacci sequence is defined recursively by the following conditions:
\[
f_1 = 1,
\quad
f_2 = 1,
\quad
f_n = f_{n-1} + f_{n-3}\ (n\geqslant 3)
.\]
The formula for $f_n$ involves the preceding terms $f_{n-1}$ and $f_{n-2}$, making it recursive.

\section{Convergent Sequences \& their Limits}%
\label{sec:convergent_sequences_&_their_limits}

\subsection{Convergent \& Divergent Sequences}%
\label{sub:convergent_&_divergent_sequences}

The sequence:
\[
	\left\{\frac{n}{n+1}\right\}
.\]
approaches the number 1 as $n$ becomes larger.
Therefore we say the sequence \textbf{converges} to 1, and we write:
\[
	\lim_{n \to \infty}\frac{n}{n+1}=1
.\]
The sequence $\{-1,1,-1,1,\dotsc\}$ does not approach a number as $n$ becomes larger.
Therefore we say the sequence \textbf{diverges}.
The sequence:
\[
	\left\{\frac{n^3+1}{2}\right\}
.\]
becomes larger and larger \textbf{uboundedly} as $n$ becomes larger.
Therefore we say the sequence \textbf{diverges}, and we write:
\[
\lim_{n \to \infty}\frac{n^3+1}{2}=\infty
.\]
\subsection{Formal Definition of the Limit of a Sequence}%
\label{sub:formal_definition_of_the_limit_of_a_sequence}

\begin{definition}
	A sequence $\{a_n\}$ has the limit $L$ and we write:
	\[
	\lim_{n \to \infty}a_n = L
	\quad
	\text{or} 
	\quad
	a_n \to L
	\text{ as }
	n\to \infty
	.\]
	if for every $\epsilon > 0$ there is a corresponding integer $N$ such that:
	\[
	|a_n - L|<\epsilon
	\quad
	\text{for every} 
	\quad
	n > N
	.\]
\end{definition}

If such a real number $L$ exists, we say that the sequence \textbf{converges} to $L$.
Otherwise, we say that the sequence \textbf{diverges} and is \textbf{divergent}.
To prove that the limit exists, we have to find (or show that it exists) a positive integer $N$ such that:
\[
n > N \implies |a_n-L| < \epsilon
.\]
\begin{exmp}
	Use the definition to prove that the sequence
	$\left\{\frac{1}{n}\right\}$
	\textbf{converges} to 0.
	\begin{proof}
		Let $\epsilon >0$ be an arbitrarily chosen positive number.
		Our aim is to find an integer $N$ such that for all $n> N$,
		\[
		\left|\frac{1}{n}-0\right|<\epsilon
		\implies
		\left|\frac{1}{n}\right| < \epsilon
		\implies
		\frac{1}{n}< \epsilon
		.\]
		Thus we have $n> \frac{1}{\epsilon}$.
		We choose $N$ to be an integer such that $N>\frac{1}{\epsilon}$.
		For example, we let $N = \left\lceil 1+ \frac{1}{\epsilon} \right\rceil = 1 + \left\lceil \frac{1}{\epsilon} \right\rceil$.
		For $n>N$, we have:
		\[
		\left| \frac{1}{n}-0\right| = \frac{1}{n} < \frac{1}{N} < \epsilon
		.\]
		By definition, we have proved that $\{\frac{1}{n}\}$ converges to 0:
		\[
		\lim_{n \to \infty}\frac{1}{n} = 0
		.\]
	\end{proof}
\end{exmp}
\begin{exmp}
	Suppose $0<|r|<1$.
	Prove that $\lim_{n \to \infty}r^n = 0$.
	\begin{proof}
		Let $\epsilon > 0$ be an arbitrary positive real number.
		Our aim is to find an integer $N$ such that $\forall n > N$,
		\[
			|r^n - 0| = |r|^n < \epsilon
		.\]
		which is equivalent to:
		\[
		n \ln |r| < \ln \epsilon
		.\]
		\begin{note}
			We can only do this because $\ln x$ is a \textbf{strictly increasing} function.
		\end{note}
		\noindent
		Since $0<|r|<1$, the real number $\ln |r| < 0$.
		Thus we have:
		\[
			n > \frac{\ln \epsilon}{\ln |r|}
		.\]
		We choose:
		\[
			N = \left\lceil 1 + \frac{\ln \epsilon}{\ln |r|} \right\rceil
		.\]
		so that:
		\[
			N > \frac{\ln \epsilon}{\ln |r|}
		.\]
		Thus, $\forall n > N$ we have:
		\[
			n > \frac{\ln \epsilon}{\ln |r|}
			\implies
			|r^n -0| < \epsilon
		.\]
		Proving that:
		\[
		\lim_{n \to \infty}r^n = 0
		.\]
	\end{proof}
\end{exmp}
\begin{exmp}
	Use the definition to prove that:
	\[
	\lim_{n \to \infty}\frac{n^2}{n^2+n-3}=1
	.\]
	\begin{proof}
		Let $\epsilon > 0$ be an arbitrary positive real number.
		Our aim is to find an integer $N$ such that $\forall n > N$,
		\[
		\left|\frac{n^2}{n^2+n-3}-1\right| < \epsilon
		.\]
		Note that:
		\[
			\left|\frac{n^2}{n^2+n-3}-1\right|
			=
			\left| \frac{-(n-3)}{n^2 + n -3}\right|
			=
			\frac{n-3}{n^2 + n - 3}, \text{ for every }
			n > 3
		.\]
		and that:
		\[
		n > 3
		\implies
		n -3 < n
		.\]
		and that:
		\[
		n > 3 \implies n^2 + n -3 > n^2 \implies \frac{1}{n^2+n-3} < \frac{1}{n^2}
		.\]
		and therefore:
		\[
		n>3\implies \frac{n-3}{n^2+n-3}<\frac{n}{n^2}=\frac{1}{n}
		.\]
		Now, if both $n>3$ and $n>\frac{1}{\epsilon}$, then we have:
		\[
		\frac{n-3}{n^2+n-3}<\frac{1}{n}<\epsilon
		.\]
		Therefore we choose $N$ to be an integer such that $N>\max\left(3,\frac{1}{\epsilon}\right)$ :
		\[
			N = 1 + \left\lceil \max\left(3,\frac{1}{\epsilon}\right) \right\rceil 
		.\]
		Thus, $\forall n > N$ we have:
		\[
		\left|\frac{n^2}{n^2+n-3}-1\right| < \epsilon
		.\]
		This proves that:
		\[
		\lim_{n \to \infty}\frac{n^2}{n^2+n-3}=1
		.\]
	\end{proof}
\end{exmp}
\begin{exmp}
	Use the definition of the limit of a sequence to prove that:
	\[
	\lim_{n \to \infty}\frac{1}{n}=1
	.\]
	is \textbf{not} valid.
	\begin{proof}
		Note that:
		\[
			\left|\frac{1}{n}-1\right| > \frac{1}{2}, \text{ if } n > 2
		.\]
		Therefore, let $\epsilon = 0.5$, then there is \textbf{no} positive integer $N$ such that:
		\[
		n > N \implies \left|\frac{1}{n}-1\right| < 0.5
		.\]
		Which concludes that:
		\[
		\lim_{n \to \infty}\frac{1}{n}=1
		.\]
		is not valid.
	\end{proof}
\end{exmp}

\section{Properties of Convergent Sequences}%
\label{sec:properties_of_convergent_sequences}

\subsection{Constant Sequences}%
\label{sub:constant_sequences}

The sequence $\{a_n=C\}$, where $a_n=C$ for every $n$ and $C$ is a real number is known as a constant sequence, and its limit is $C$ :
\[
\lim_{n \to \infty}a_n = C
.\]

\subsection{Uniqueness of the Limit}%
\label{sub:uniqueness_of_the_limit}

\begin{thrm}
	Uniqueness.
	If $\lim_{n \to \infty}a_n = L$ and $\lim_{n \to \infty}a_n = L^\prime$, then:
	\[
	L = L^\prime
	.\]
\end{thrm}

\begin{proof}
	Uniqueness of a Limit.
	Suppose $L=L^\prime$ is not true:
	\[
	L=L^\prime \implies L \neq L^\prime \implies L-L^\prime \neq 0
	.\]
	Consider the case where in the formal definition of the limit, epsilon has the following value:
	\[
	\epsilon = \frac{|L-L^\prime|}{2}
	.\]
	Which can be used since:
	\[
	L-L^\prime \neq 0 \implies \epsilon > 0
	.\]
	We know by the definition of the limit that:
	\[
	\lim_{n \to \infty}a_n = L
	\iff
	\exists N_1 \in \mathbb{N}
	\text{ such that }
	n > N_1 \implies |a_n -L| < \epsilon 
	.\]
	\[
	\lim_{n \to \infty}a_n = L^\prime
	\iff
	\exists N_2 \in \mathbb{N}
	\text{ such that }
	n > N_2 \implies |a_n -L^\prime| < \epsilon 
	.\]
	Which therefore implies that:
	\begin{align}
		\label{ineq1}
		|a_n-L| & < \frac{|L-L^\prime|}{2}\\
		\label{ineq2}
		|a_n-L^\prime| & < \frac{|L-L^\prime|}{2}
	\end{align}
	\bigbreak
	\noindent
	Note that:
	\[
	|L-L^\prime| = |L - a_n + a_n - L^\prime|
	.\]
	And using the triangle inequality:
	\[
	|L - a_n + a_n - L^\prime| \leqslant |L - a_n| + |a_n - L^\prime|
	.\]
	Therefore:
	\begin{equation}
		\label{ineq3}
	|L-L^\prime| \leqslant |L - a_n| + |a_n - L^\prime|
	.\end{equation}
if $(n > N_1) \land (n > N_2)$ then by \crefrange{ineq1}{ineq2}:
	\begin{align*}
		|a_n - L| + |a_n - L^\prime| < & \frac{|L-L^\prime|}{2} + \frac{|L-L^\prime|}{2}\\
		|a_n - L| + |a_n - L^\prime| < & |L-L^\prime|
	\end{align*}
	Then using \cref{ineq3}:
	\[
		|L-L^\prime| \leqslant |a_n - L| + |a_n - L^\prime| < |L-L^\prime|
	.\]
	Therefore:
	\[
		|L-L^\prime| < |L-L^\prime|
	.\]
	A strict inequality, which contradicts our assumptions, therefore the theorem is true.
\end{proof}

\subsection{Bounded Sequences}%
\label{sub:bounded_sequences}

A sequence is bounded if there exists a \textbf{positive} real number $M$ such that:
\[
|a_n| \leqslant M, \quad \forall n
.\]
A sequence is said to be bounded from above if there is a real number $M$ such that:
\[
a_n \leqslant M, \quad \forall n
.\]
and we say that $M$ is an \textbf{upper bound} of $\{a_n\}$.
A sequence is said to be bounded from below if there is a real number $m$ such that:
\[
m \leqslant a_n, \quad \forall n
.\]
and we say that $m$ is a \textbf{lower bound} of $\{a_n\}$.
\begin{thrm}
	Bounded.
	If $\{a_n\}$ is convergent, then $\{a_n\}$ is bounded.
	Consequently, an unbounded sequence is divergent.
\end{thrm}
\begin{proof}
	Every convergent sequence is bounded.
	We know by the definition of the limit of a convergent sequence that:
	\[
	\lim_{n \to \infty}a_n = L
	\iff
	\forall \epsilon > 0,\ \exists N \in \mathbb{N} \text{ s.t } n > N \implies |a_n-L| < \epsilon
	.\]
	Take $\epsilon = 1$ then:
	\[
		\exists N_1 \in \mathbb{N} \text{ s.t } n > N_1 \implies |a_n-L| < 1
	.\]
	Note that because of the triangle inequality:
	\[
	|a_n-L| \geqslant |a_n| - |L|
	.\]
	So for $n>N_1$, we have:
	\[
	|a_n| - |L| \leqslant |a_n-L| < 1
	.\]
	Therefore:
	\[
	|a_n| - |L| < 1
	.\]
	And therefore:
	\[
	|a_n| < |L|+1
	.\]
	Now, let $M_1 = \max(|a_1|,|a_2|,\dotsc,|a_{N_1}|)$.
	This means that:
	\[
	n \leqslant N_1 \implies |a_n| \leqslant M_1
	.\]
	Now take $M = \max(M_1,(|L|+1))$.
	Then:
	\[
		|a_n| \leqslant M, \quad \forall n \in \mathbb{N}
	.\]
\end{proof}

\subsection{Sum \& Difference Laws}%
\label{sub:sum_&_difference_laws}

Suppose $\{a_n\}$ and $\{b_n\}$ are \textbf{convergent}.
Then $\{a_n+b_n\}$ is convergent and:
\[
\lim_{n \to \infty}a_n + b_n = \lim_{n \to \infty}a_n + \lim_{n \to \infty}b_n
.\]
\begin{proof}
	Suppose the following:
	\[
	\lim_{n \to \infty}a_n = A,
	\quad
	\lim_{n \to \infty}b_n = B,
	\quad
	A,B \in \mathbb{R}
	.\]
	Let $\epsilon > 0$ then:
	\begin{align}
	\label{lima}
	\lim_{n \to \infty}a_n = A
	\iff \exists N_1 \in \mathbb{N} \text{ s.t }
	n > N_1 \implies |a_n-A| < \frac{\epsilon}{2}\\
	\label{limb}
	\lim_{n \to \infty}b_n = B
	\iff \exists N_2 \in \mathbb{N} \text{ s.t }
	n > N_2 \implies |b_n-B| < \frac{\epsilon}{2}
	.\end{align}
	Now let $N=\max(N_1,N_2)$ so that:
	\[
		n > N \implies (n>N_1) \land (n>N_2)
	.\]
	and note that:
	\[
		|(a_n + b_n)-(A+B)| = |(a_n - A)+(b_n-B)|
	.\]
	by the triangle inequality:
	\[
	|(a_n - A)+(b_n-B)| \leqslant |a_n -A| + |b_n-B|
	.\]
	And by \crefrange{lima}{limb}, for $n>N$:
	\[
	|(a_n - A)+(B-b_n)| \leqslant |a_n -A| + |b_n-B| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
	.\]
	and finally:
	\[
		n>N \implies |(a_n+b_n)-(A+B)| < \epsilon
	.\]
\end{proof}

\noindent
Suppose $\{a_n\}$ and $\{b_n\}$ are \textbf{convergent}.
Then $\{a_n-b_n\}$ is convergent and:
\[
\lim_{n \to \infty}a_n - b_n = \lim_{n \to \infty}a_n - \lim_{n \to \infty}b_n
.\]

\begin{proof}
	Suppose the following:
	\[
	\lim_{n \to \infty}a_n = A,
	\quad
	\lim_{n \to \infty}b_n = B,
	\quad
	A,B \in \mathbb{R}
	.\]
	Let $\epsilon > 0$ then:
	\begin{align}
	\label{limc}
	\lim_{n \to \infty}a_n = A
	\iff \exists N_1 \in \mathbb{N} \text{ s.t }
	n > N_1 \implies |a_n-A| < \frac{\epsilon}{2}\\
	\label{limd}
	\lim_{n \to \infty}b_n = B
	\iff \exists N_2 \in \mathbb{N} \text{ s.t }
	n > N_2 \implies |b_n-B| < \frac{\epsilon}{2}
	.\end{align}
	Now let $N=\max(N_1,N_2)$ so that:
	\[
		n > N \implies (n>N_1) \land (n>N_2)
	.\]
	and note that:
	\[
		|(a_n - b_n)-(A-B)| = |(a_n - A)-(b_n-B)|
	.\]
	by the triangle inequality:
	\[
	|(a_n - A)-(b_n-B)| \leqslant |a_n -A| + |b_n-B|
	.\]
	And by \crefrange{limc}{limd}, for $n>N$:
	\[
	|(a_n - A)+(B-b_n)| \leqslant |a_n -A| + |b_n-B| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
	.\]
	and finally:
	\[
		n>N \implies |(a_n+b_n)-(A+B)| < \epsilon
	.\]
\end{proof}
\subsection{Order Properties for Convergent Sequences}%
\label{sub:order_properties_for_convergent_sequences}

\begin{thrm}
	\label{propa}
	If $\{a_n\}$ is \textbf{convergent} and $a_n \geqslant 0$, $\forall n \geqslant N_0$, then:
	\[
		\lim_{n \to \infty}a_n \geqslant 0
	.\]
	Therefore:
	\[
	L\geqslant0
	.\]
\end{thrm}

\begin{proof}
	Suppose $\{a_n\}$ is convergent and $a_n\geqslant0$, $\forall n\geqslant N_0$.
	Now suppose the limit $L$ opposes the property in theorem \ref{propa}:
	\[
		\lnot(L\geqslant0)\equiv L<0
	.\]
	Consider the case where $\epsilon = -\frac{L}{2}$ and since $L<0$, this value of $\epsilon$ is valid.
	This means that:
	\[
		\exists N \in \mathbb{N} \text{ s.t. } n>N
		\implies
		|a_n-L|<-\frac{L}{2}
	.\]
	and therefore:
	\[
		-\left(-\frac{L}{2}\right) < a_n-L < -\frac{L}{2}
	.\]
	The latter part of the inequality can be rearranged as such:
	\[
	a_n-L < -\frac{L}{2} \to a_n<L-\frac{L}{2} \to a_n<\frac{L}{2}
	.\]
	and by the supposition at the start of the proof:
	\[
	L<0\implies \frac{L}{2}<0
	.\]
	therefore:
	\[
	a_n <0
	.\]
	which contradicts the assumptions held at the start of the proof.
\end{proof}
\begin{thrm}
	If $\{a_n\}$ is \textbf{convergent} and $m\leqslant a_n\leqslant M$, $\forall n\geqslant N_0$, then:
	\[
	m\leqslant \lim_{n \to \infty}a_n \leqslant M
	.\]
\end{thrm}
\begin{proof}
	Suppose the following conditions:
	\[
		\lim_{n \to \infty}a_n = L,\ L\in\mathbb{R}
		\qquad
		m\leqslant a_n\leqslant M,\ \forall n>N_0
	.\]
	Now considering the sequence $\{a_n-m\}$ constructed using the laws in \cref{sub:sum_&_difference_laws}, we see the following property:
	\[
		\forall n>N_0,\ a_n-m\geqslant0
	.\]
	therefore by theorem \ref{propa}:
	\begin{align*}
		\lim_{n \to \infty}(a_n-m)\geqslant0 & \equiv \lim_{n \to \infty}a_n - \lim_{n \to \infty}m \geqslant0 && \text{by difference law}\\
						     & \equiv \lim_{n \to \infty}a_n -m\geqslant0 &&\text{by constant sequence}\\
						     & \equiv \lim_{n \to \infty}a_n \geqslant m
	\end{align*}
	Now considering the sequence $\{M-a_n\}$ constructed using the laws in \cref{sub:sum_&_difference_laws}, we see the following property:
	\[
		\forall n>N_0,\ M-a_n\geqslant0
	.\]
	therefore by theorem \ref{propa}:
	\begin{align*}
		\lim_{n \to \infty}(M-a_n)\geqslant0 & \equiv \lim_{n \to \infty}M - \lim_{n \to \infty}a_n \geqslant0 && \text{by difference law}\\
						     & \equiv M-\lim_{n \to \infty}a_n \geqslant0 &&\text{by constant sequence}\\
						     & \equiv M\geqslant \lim_{n \to \infty}a_n
	\end{align*}
	and therefore:
	\[
		m\leqslant \lim_{n \to \infty}a_n \leqslant M
	.\]
\end{proof}

\begin{thrm}
	If $\{a_n\}$ and $\{b_n\}$ are \textbf{convergent} and $a_n\leqslant b_n$, $\forall n\geqslant N_0$, then:
	\[
		\lim_{n \to \infty}a_n \leqslant \lim_{n \to \infty}b_n
	.\]
\end{thrm}
\begin{proof}
	Suppose the following conditions:
	\[
		\{a_n\},\ \{b_n\}
		\text{ are convergent}
		\qquad
		a_n\leqslant b_n,\ \forall n\geqslant N_0
	.\]
	for $n\geqslant N_0$, we see that:
	\[
	a_n\leqslant b_n \to b_n-a_n\geqslant0
	.\]
	Now considering the sequence $\{b_n-a_n\}$ constructed using the laws in \cref{sub:sum_&_difference_laws}, and by theorem \ref{propa}:
	\begin{align*}
		\lim_{n \to \infty}(b_n-a_n)\geqslant0 & \equiv \lim_{n \to \infty}b_n-\lim_{n \to \infty}a_n \geqslant0\\
						       & \equiv \lim_{n \to \infty}b_n \geqslant \lim_{n \to \infty}a_n
	\end{align*}
\end{proof}

\section{Divergent Sequences}%
\label{sec:divergent_sequences}

To prove that a sequence $\{a_n\}$ is divergent, we have to show that for every arbitrary real number $L \in \mathbb{R}$, we can find a positive real number $\epsilon>0$ such that for every positive integer $N$, there is some integer $n>N$ where $|a_n-L|\geqslant\epsilon$.

\bigbreak
\begin{exmp}
	Prove that the sequence $\{(-1)^{n}\}$ diverges.
	\begin{proof}
		Note that $a_{2n}=1$ and $a_{2n+1}=-1$, so for an arbitrary real number $L$, either of the following conditions hold:
		\[
			|a_{2n}-L|\geqslant\frac{1}{2}
			\text{ or }
			|a_{2n+1}-L|\geqslant\frac{1}{2}
		.\]
		So taking $\epsilon = \frac{1}{2}$ for every positive integer $N$, we have either:
		\[
			|a_{2N}-L|\geqslant\frac{1}{2}
			\text{ or }
			|a_{2N+1}-L|\geqslant\frac{1}{2}
		.\]
		and there is always an $n>N$ such that $|a_n-L|\geqslant\frac{1}{2}$.
		Therefore, we conclude that the sequence diverges.
	\end{proof}
\end{exmp}

\subsection{Divergence to $\infty$}%
\label{sub:divergence_to_infty_}

We say a sequence $\{a_n\}$ diverges to $\infty$, denoted $\lim_{n \to \infty}a_n = \infty$, if:
\[
	\forall M\in \mathbb{R}^+ >0,\ \exists N \in \mathbb{N}\text{ s.t. }
	a_n>M,\ \forall n>N
.\]
We say a sequence $\{a_n\}$ diverges to $-\infty$, denoted $\lim_{n \to \infty}a_n = -\infty$, if:
\[
	\forall M\in \mathbb{R}^+ >0,\ \exists N \in \mathbb{N}\text{ s.t. }
	a_n<-M,\ \forall n>N
.\]
\begin{exmp}
	Prove by definition:
	\[
	\lim_{n \to \infty}\frac{n^2}{n+1}=\infty
	.\]
	\begin{proof}
		For every positive real number $M$, we must find a positive integer $N$ such that:
		\[
		n > N \implies \frac{n^2}{n+1}-M >0
		.\]
		Note that:
		\[
			\frac{n^2}{n+1}=(n-1)+\frac{1}{n+1}
		.\]
		Given any positive real number $M$:
		\[
			\frac{n^2}{n+1}-M = (n-1)+\frac{1}{n+1}-M
		.\]
		we see that:
		\[
		(n-1)+\frac{1}{n+1}-M > n-1-M > 0
		.\]
		so we choose $N=M+1$, which works for any $M$.
		Therefore we conclude that:
		\[
		\lim_{n \to \infty}\frac{n^2}{n+1}=\infty
		.\]
	\end{proof}
\end{exmp}
