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\pagestyle{fancy}
\lhead{\color{blue} \Large Math 207}
\lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2012}
\cfoot{}
\chead{\LARGE Sequences}
\rfoot{\small Last revised: October 25, 2012}
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\begin{document}
The real numbers has the \textbf{completeness property}: If a set of real
numbers is bounded above, it has a least upper bound; if a set of real
numbers is bounded below, it has a greatest lower bound. \ (This is an axiom
of the real numbers, and this is the one that distinguishes the set of
rational numbers from the set of real numbers.)\bigskip
Definition: \ A \textbf{sequence} is a list of numbers $a_{1},$ $a_{2},$ $%
a_{3},...,a_{n},...$ in a given order. \ The numbers $a_{n}$ are \textbf{%
terms}
\qquad of the sequence. \ The integer $k$ is called the index of the term $%
a_{k}$.\bigskip
An infinite sequence is a function with domain $%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$. \ We may start labeling at a number greater than $1$.\bigskip
We can describe sequences by writing rules%
\begin{equation*}
a_{n}=\sqrt{n}~~~b_{n}=\left( -1\right) ^{n+1}\dfrac{1}{n}~~~~c_{n}=\dfrac{%
n-1}{n}~~~~d_{n}=\left( -1\right) ^{n+1}
\end{equation*}%
or listing the first few terms%
\begin{eqnarray*}
\left\{ a_{n}\right\} &=&\left\{ 1,\sqrt{2},\sqrt{3},...,\sqrt{n}%
,....\right\} \\
\left\{ b_{n}\right\} &=&\left\{ 1,-\dfrac{1}{2},\dfrac{1}{3},-\dfrac{1}{4}%
,...,\left( -1\right) ^{n+1}\dfrac{1}{n},...\right\} \\
\left\{ c_{n}\right\} &=&\left\{ 0,\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},%
\dfrac{4}{5},...,\dfrac{n-1}{n},...\right\} \\
\left\{ d_{n}\right\} &=&\left\{ 1,-1,1-1,.....,\left( -1\right)
,...\right\}
\end{eqnarray*}%
\bigskip
Definition: The sequence $\left\{ a_{n}\right\} $ \textbf{converges} to the
number $L$ if for every positive number $\epsilon $ there exists
\qquad an integer $N$ such that for all $n$,
\begin{equation*}
\text{if }n>N\text{ then }\left\vert a_{n}-L\right\vert <\epsilon \text{.}
\end{equation*}
\qquad If no such number $L$ exists, we say $\left\{ a_{n}\right\} $ \textbf{%
diverges}. \
\qquad If $\left\{ a_{n}\right\} $ converges to $L,$ we write $%
\lim\limits_{n\rightarrow \infty }a_{n}=L$ or $a_{n}\rightarrow L$ and call $%
L$ the \textbf{limit} of the sequence.\bigskip
Example 1. \ a) $\ \lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0$
proof: \ Let $\epsilon >0$ be given. \ Define \ $N=\left\lceil \dfrac{1}{%
\epsilon }\right\rceil +1$.
\qquad If $n>N$, then $n>N>\dfrac{1}{\epsilon }$ \ and so $\dfrac{1}{n}%
<\epsilon $.\bigskip
b) \ the constant sequnece $\left\{ a_{n}\right\} =\left\{
c,c,c,....\right\} .$ Clearly $a_{n}\rightarrow c$.
proof: \ Let $\epsilon >0$ be given. \ Then $N=1$ will do, because for all $%
n>1$ we will have that $\left\vert c-c\right\vert =0<\epsilon $.\bigskip
Example 2. $\ \left\{ 1,-1,1-,1,1,-1....\right\} $ diverges.
proof: \ Suppose for a contradiction that such a number $L$ exists. \ Let $%
\epsilon =\dfrac{1}{3}$. \
\qquad There exists $N\in
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
$ such that for all $n>N$, $\left\vert a_{n}-L\right\vert <\dfrac{1}{3}$. \
Since $1$ occurs in the sequence at arbitrarily high
\qquad index, it must be that
\begin{eqnarray*}
\left\vert 1-L\right\vert &<&\dfrac{1}{3} \\
-\dfrac{1}{3} &<&L-1<\dfrac{1}{3} \\
\dfrac{2}{3} &<&L<\dfrac{4}{3}
\end{eqnarray*}
\qquad Since $-1$ occurs in the sequence at arbitrarily high index, it also
must be that
\begin{eqnarray*}
\left\vert -1-L\right\vert &<&\dfrac{1}{3} \\
-\dfrac{1}{3} &<&L+1<\dfrac{1}{3} \\
-\dfrac{4}{3} &<&L<-\dfrac{2}{3}
\end{eqnarray*}
\qquad There is no number $L$ with $\dfrac{2}{3}