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\title{The limit and convergence of sequences}
\author{Arif Ali}
\date{December 5, 2013}
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\maketitle
\begin{abstract}
This document is a short introduction to the idea of convergence and divergence in sequences and shows how to prove a sequence is convergent by using the definition of convergence. This document also intrdocues the standard notation used for sequences.
\end{abstract}
\section{Sequences and their notation}
sequecnes can be notated in shorthand with some special notation, let some sequences be defined as:
\begin{equation}\label{my_first_equation}a_{n} =\left(\frac{1}{n}\right)
\end{equation}
$$b_{n} =\left(n^2\right)$$
The terms in the sequence would be :
$$(a_{n})_{n=1}^{\infty} = (1,\quad 1/2,\quad 1/3,\quad 1/4,\quad1/5, \dots)$$
$$(b_{n})_{n=1}^{\infty} = (1,\quad 4,\quad 9,\quad 16,\quad25, \dots)$$
\section{The limit of a sequence and definition of convergence}
Sequences which have an infinite number of terms will either diverge or converge. If the sequence diverges then it does not have a limit and may go to infinity or fluctuate over some numbers throughout the terms in the sequence. When the sequence is convergent it will converge to a limit, e.g the sequence (\ref{my_first_equation}) will have
$\lim_{n \to \infty}(a_{n}) = a $ as a limit.
The formal definition of convergence is\cite[P.4]{reade1986introduction} :
$$\forall \varepsilon >0 ,\varepsilon \in \mathbb{R} \quad \exists N(\varepsilon) \in \mathbb{N} : \forall n \geq N(\varepsilon) : |a-a_{n}| < \varepsilon$$
This definition is used to decide whether a sequence has a limit and if it is convergent. if the statemnt does not hold true for a sequnce e,g $(b_{n})$ then it doesn't have a limit and is therefore diveregent.
\subsection{example of a convergence}
to see how sequence the $(a_{n})$ is convergent we will use the definition and a graph to show things more clearly.
If we choose $ N(\epsilon) > 1/\epsilon $ and note that $(a_{n}) = (1/n)$ then we see that the inequality holds true for all $ N(\epsilon) > 1/\epsilon $ : $$ 1/n \leq1/ N(\epsilon) < \frac{1}{(1/\epsilon)}= \epsilon$$
so for whatever value of $\epsilon$ we choose we can find a natural number $ N(\epsilon)$, such that the inequality remains true for all of the succeding $_{n}th$ terms in the sequence. This means that $\lim_{n \to +\infty}(a_{n}) = a = 0 $
$$|a-a_{n}|= |0-a_{n}| = |a_{n}|<\epsilon$$
\begin{center}
\includegraphics[width= 18cm]{sequence_of_convergence.png}
\\Fig 1. This is a graph showing convergence of the sequence ($a_{n}$), the red line is $\epsilon = 1.$
\end{center}
The graph shows that all the terms of the sequence will be smaller than $\epsilon$ after some $ N(\epsilon)$, also as n heads towards +$\infty$, (1/n) heads towards 0, so: $$\lim_{n \to +\infty}\left(\frac{1}{n}\right)=0$$
\subsection{Example of a divergence}
With this exaple it is more clear to see how this sries is divergent, the graph shows that the terms in the sequence do not approach a limit they increase without bound and tend to $+\infty$.
To show that the sequence $b_{n} =\left(n^2\right)$ is not convergent we shall first assume it is convergent and then show a contradicition. If we assume that $(b_{n})$ is convergent then $\forall \epsilon >0, \exists N(\epsilon): \forall n \geq N(\epsilon): |b-b_{n}| < \epsilon $. \\If we choose $N(\epsilon) <\sqrt{\epsilon} \Rightarrow N(\epsilon) \leq n \leq \sqrt{\epsilon} \Rightarrow (N(\epsilon))^2 \leq n^2 \leq \epsilon $ so the strict inequality $n <\epsilon$ is not true $\forall n$. Furthermore since the definition has to be valid $\forall n \geq N(\epsilon) \implies \forall \epsilon>0, \exists n : n^2 \geq \epsilon $ , this is because the (infinite) set of natural numbers $\mathbb{N}$ is not bounded from above, \cite{Upperandlowerbounds}
so there will always be a bigger $N(\epsilon)$ we can select so the inequality becomes false for all $n \geq N(\epsilon)$. Now we have shown the sequence $(b_{n}) \text{ is unbounded and tends to} +\infty \text{. We may write this as } \lim_{n \to +\infty} (b_{n}) = +\infty \text{, which means it} \textit{ increases without bound} \textbf{ and is not convergent and doesn't have a limit.}$
\begin{center}
\includegraphics[width= 18cm]{sequence_of_divergence.png}
\\Fig 2. This is a graph showing the divergence of the sequence ($b_{n}$)
\end{center}
The graph shows that as $n$ grows larger and tends to $+\infty, (n^2) \text{ also tends to +} \infty$. This means that: $$\lim_{n \to +\infty}(n^2) = +\infty $$
\subsection{Conclusion}
Fom this report I have learnt how to use latex and sage toghether to explain the definition of convergence by using the mathematical notation in Latex. I have learnt to plot graphs in sage and import them to Latex to show a sequence converging, it was interesting to learn how graphs and images can be placed in latex and also how to cite many references clearly in the bibliography.
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