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\newtheorem{case}[theorem]{Case}
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\newtheorem{conclusion}[theorem]{Conclusion}
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\lhead{\color{blue} \Large Lecture Notes}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2012}
\cfoot{}
\chead{\LARGE Sequences - Part 1}
\rfoot{\small Last revised: March 15, 2014}
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\begin{document}


The real numbers has the \textbf{completeness property}: If a non-empty set
of real numbers is bounded above, then there exists a least upper bound; if
a non-empty set of real numbers is bounded below, then there exists 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

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\textbf{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 $%
a_{k}$.%
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\medskip

An infinite sequence is a function with domain $%
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\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\} \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ }\left\{ c_{n}\right\} =\left\{ 0,\dfrac{1}{2},\dfrac{2}{3},%
\dfrac{3}{4},\dfrac{4}{5},...,\dfrac{n-1}{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\} \text{ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ d_{n}\right\} =\left\{
1,-1,1-1,.....,\left( -1\right) ,...\right\}
\end{eqnarray*}%
\bigskip

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Definition: The sequence $\left\{ a_{n}\right\} $ \textbf{converges} to the
number $L$ if for every positive number $\varepsilon $ 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 <\varepsilon \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.\medskip 
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\bigskip

Notice that $\left\vert a_{n}-L\right\vert <\varepsilon $ and \ $%
L-\varepsilon <a_{n}<L+\varepsilon $ \ are equivalent statements.%
\begin{eqnarray*}
L-\varepsilon &<&a_{n}<L+\varepsilon \\
-\varepsilon &<&a_{n}-L<\varepsilon \\
\left\vert a_{n}-L\right\vert &<&\varepsilon
\end{eqnarray*}

This definition is fundamental to our material. \ \ We can think of a
convergent sequences as one whose elements are eventually arbitrarily close
to its limit. \ For any positive value of $\varepsilon $ (think of $%
\varepsilon $ as the error), all but finitely many elements (think of them
as the first few) of the sequence are inside an $\varepsilon -$neighborhood
of $L$.\FRAME{dtbpF}{4.9018in}{1.0542in}{0pt}{}{}{convergent1.bmp}{\special%
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If the sequence $\left\{ a_{n}\right\} $ converges to $L$, then no matter
how small $\varepsilon $ is, all but a finitely many terms of the sequence
will fall inside the $\varepsilon $-neightborhood of $L$. \ For every value
of $\varepsilon $, there is generally a different value of $N$.\pagebreak

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Definition: \ the symbols $\left\lceil \text{ and }\right\rceil $ denote the 
\textbf{ceiling function}: for any real number $x$, $\left\lceil
x\right\rceil $ denotes the smallest

\qquad integer greater than or equal to $x$. \ \ For example, $\left\lceil
5\right\rceil =5$, $\left\lceil 17.34\right\rceil =18$, and $\left\lceil
-2.37\right\rceil =-2$. \ \medskip

\qquad We similarly define the \textbf{floor function}: \ $\lfloor x\rfloor $
denotes the greatest integer less than or equal to $x$. \ 

\qquad For example, $\lfloor 5\rfloor =5,$ $\lfloor 17.34\rfloor =17$, and $%
\lfloor -2.37\rfloor =-3$.\medskip 
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\bigskip

The floor and ceiling functions are useful when we need to start with a real
number such as $\dfrac{1}{\varepsilon }$ and turn it into an integer such as 
$N$.\bigskip

Before we see some examples, let us establish a fact that we will use very
often. \ 

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\bigskip

Theorem: \ If $A$ and $B$ are real numbers such that they are either both
positive or both negative, then 
\begin{equation*}
A<B\text{ implies }\dfrac{1}{A}>\dfrac{1}{B}
\end{equation*}

\medskip 
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\bigskip

This theorem allows us to (carefully) take the reciprocal of both sides in
an inequality. \ If both sides are of the same sign, we can take the
reciprocal of both sides but must reverse the inequality sign. \ (We will
prove this theorem in the exercises.)\bigskip

Example 1. \ The constant sequence $\left\{ a_{n}\right\} =\left\{
c,c,c,....\right\} $. \ Clearly $\lim\limits_{n\rightarrow \infty }a_{n}=c$.

proof: \ Let $\varepsilon >0$ be given. \ Then $N=1$ will do, because for
all $n>1$ we will have that $\left\vert c-c\right\vert =0<\varepsilon $.

\qquad That is the same as $\left\vert a_{n}-c\right\vert <\varepsilon $ and
so the sequence converges to $c$.\bigskip

Example 2. $\ \lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0$

proof: \ Let $\varepsilon >0$ be given. \ Define \ $N=\left\lceil \dfrac{1}{%
\varepsilon }\right\rceil +1$. \ \bigskip

\qquad If $n>N$, then $n>N>\dfrac{1}{\varepsilon }$ and so $\dfrac{1}{n}%
<\varepsilon $. \ Since $\dfrac{1}{n}$ is positive, we have that%
\begin{eqnarray*}
-\varepsilon &<&\dfrac{1}{n}<\varepsilon \\
-\varepsilon &<&\dfrac{1}{n}-0<\varepsilon \text{ \ same as }\left\vert 
\dfrac{1}{n}-0\right\vert <\varepsilon
\end{eqnarray*}

\qquad and so $\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=0$.\bigskip

Example 3. \ The sequence defined $a_{n}=\dfrac{3n^{3}+2}{n^{3}}$ approach $%
3 $. \ We will give an $\varepsilon -N$ proof.\bigskip

proof: \ Since $\dfrac{3n^{3}+2}{n^{3}}=3+\dfrac{2}{n^{3}}$, as $n$ becomes
larger and larger, $\dfrac{2}{n^{3}}$ will approach zero and so $a_{n}$ will
approach $3$. \ 

\qquad We will prove that $3$ is the limit.\bigskip

\qquad Let $\varepsilon >0$ be given. \ Let $N$ be a positive integer with $%
N\geq \dfrac{2}{\varepsilon }$. \ For all $n>N$, we have that%
\begin{eqnarray*}
n &>&\dfrac{2}{\varepsilon } \\
n^{3} &>&n>\dfrac{2}{\varepsilon } \\
n^{3} &>&\dfrac{2}{\varepsilon }\text{ \ \ \ \ \ \ \ both sides are positive}
\\
\dfrac{1}{n^{3}} &<&\dfrac{\varepsilon }{2} \\
\dfrac{2}{n^{3}} &<&\varepsilon
\end{eqnarray*}
\ \ \ \ \ \ \ \ 

\qquad also, $\dfrac{2}{n^{3}}>0$ and since $-\varepsilon <0$, we also have
that%
\begin{eqnarray*}
-\varepsilon &<&\dfrac{2}{n^{3}}<\varepsilon \\
-\varepsilon &<&\dfrac{2}{n^{3}}+3-3<\varepsilon \\
-\varepsilon &<&\dfrac{3n^{3}+2}{n^{3}}-3<\varepsilon \text{ \ \ same as \ }%
\left\vert \dfrac{3n^{3}+2}{n^{3}}-3\right\vert <\varepsilon \\
-\varepsilon &<&a_{n}-3<\varepsilon \text{ \ \ same as \ }\left\vert
a_{n}-3\right\vert <\varepsilon
\end{eqnarray*}

\qquad and so \ the sequence converges to $3$.\bigskip

Example 4. $\ \left\{ 1,~-1,~1,~-1,~1,~-1,~....\right\} $ diverges.

proof: \ Suppose for a contradiction that such a number $L$ exists. \ Let $%
\varepsilon =\dfrac{1}{3}$. \ 

\qquad There exists $N\in 
%TCIMACRO{\U{2115} }%
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%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 $\dfrac{2}{3}<L<\dfrac{4}{3}$ and $-%
\dfrac{4}{3}<L<-\dfrac{2}{3},$ so the sequence diverges.\bigskip \pagebreak

Example 5. \ The sequence $\left\{ \sqrt{n}\right\} $ diverges
differently.\bigskip

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\bigskip

Definition: \ The sequence $\left\{ a_{n}\right\} $ \textbf{diverges to
infinity} if for every real number $M$

\qquad there exists an integer $N$ such that for all $n$, 
\begin{equation*}
\text{if }n>N\text{ then }a_{n}>M\text{.}
\end{equation*}

\qquad We denote this as $\lim\limits_{n\rightarrow \infty }a_{n}=\infty $
or $a_{n}\rightarrow \infty $. \ \bigskip

\qquad Similarly, the sequence $\left\{ a_{n}\right\} $ \textbf{diverges to
negative infinity} if for every real number $m$ there exists

\qquad an integer $N$ such that for all $n$, 
\begin{equation*}
\text{if }n>N\text{ then }a_{n}<m\text{.}
\end{equation*}

\qquad We denote this as $\lim\limits_{n\rightarrow \infty }a_{n}=-\infty $
or $a_{n}\rightarrow -\infty $. \ \bigskip

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\bigskip

The sequence $\left\{ \sqrt{n}\right\} $ diverges to infinity. \ The
sequence $\left\{ 1,0,2,0,3,0,...\right\} $ diverges, but does not diverge
to infinity or negative infinity.\bigskip

\bigskip

\bigskip

\begin{center}
{\LARGE Sample Problems\bigskip }
\end{center}

\begin{enumerate}
\item Recall that if $a_{n}=\dfrac{1}{n}$, then $\lim\limits_{n\rightarrow
\infty }a_{n}=0$.

a) \ Find a value of $N$ for $\varepsilon =0.15$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.15$

b) \ Find a value of $N$ for $\varepsilon =0.01$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.01$

\item Define $a_{n}=\dfrac{3n+2}{5n+1}$. \ 

a) \ Find $\lim\limits_{n\rightarrow \infty }a_{n}$.

b) \ Find a value of $N$ for $\varepsilon =\dfrac{1}{2}$. \ That is, find a
value for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <%
\dfrac{1}{2}$.

c) \ Find a value of $N$ for $\varepsilon =0.05$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.05$.

d) \ Find a value of $N$ for $\varepsilon =0.001$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.001$.

e) \ Find a general expression for $N$ in terms of $\varepsilon $.

\item Find the limit of the sequence $a_{n}=\dfrac{2n-5}{n+3}$ and use an
epsilon-N proof to justify your answer.\pagebreak
\end{enumerate}

\begin{center}
{\LARGE Sample Problems- Solutions\bigskip }
\end{center}

\begin{enumerate}
\item Recall that if $a_{n}=\dfrac{1}{n}$, then $\lim\limits_{n\rightarrow
\infty }a_{n}=0$.

a) \ Find a value of $N$ for $\varepsilon =0.15$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.15$

Solution: \ Let us start with what we need and solve for what that means for 
$n$. \ We want:%
\begin{eqnarray*}
\left\vert a_{n}-0\right\vert &<&0.15 \\
-0.15 &<&a_{n}<0.15 \\
-0.15 &<&\dfrac{1}{n}<0.15
\end{eqnarray*}%
The left-hand side is automatically true for all $n$ since $n$ is positive,
so we just need to focus on the right-hand side.%
\begin{equation*}
\dfrac{1}{n}<0.15
\end{equation*}%
Both sides are positive and so we may take the reciprocal of both sides.
Remember to reverse the inequality sign.%
\begin{equation*}
n>\dfrac{1}{0.15}=6.\overline{6}
\end{equation*}%
So $N=6$ \ will work. \ Indeed, if $n>6$, then 
\begin{equation*}
\dfrac{1}{n}\leq \dfrac{1}{7}\approx 0.142\,857<0.15
\end{equation*}%
and then we also have%
\begin{eqnarray*}
-0.15 &<&\dfrac{1}{n}<0.15 \\
-0.15 &<&\dfrac{1}{n}-0<0.15\text{ \ \ same as \ }\left\vert \dfrac{1}{n}%
-0\right\vert <0.15 \\
-0.15 &<&a_{n}-0<0.15\text{ \ \ same as \ }\left\vert a_{n}-0\right\vert
<0.15
\end{eqnarray*}%
b) \ Find a value of $N$ for $\varepsilon =0.01$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.01$

Solution: \ Since $0.01=\dfrac{1}{100}$, we can easily guess that $N=100$
will do fine. \ Indeed, if $n>100$, then 
\begin{equation*}
n>100\text{ \ and both sides being positive implies that \ }\dfrac{1}{n}<%
\dfrac{1}{100}
\end{equation*}%
and so 
\begin{eqnarray*}
-0.01 &<&\dfrac{1}{n}<0.01 \\
-0.01 &<&\dfrac{1}{n}-0<0.01\text{ \ \ \ same as }\left\vert \dfrac{1}{n}%
-0\right\vert <0.01\text{ \ same as }\left\vert a_{n}-0\right\vert <0.01
\end{eqnarray*}

\item Define $a_{n}=\dfrac{3n+2}{5n+1}$. \ 

a) \ Find $\lim\limits_{n\rightarrow \infty }a_{n}$.

Solution: \ $a_{n}=\dfrac{3n\left( 1+\dfrac{2}{3n}\right) }{5n\left( 1+%
\dfrac{1}{5n}\right) }$ As $n$ becomes larger and larger, $1+\dfrac{2}{3n}$
and $1+\dfrac{1}{5n}$ both approach $1$ and so the sequence approaches $%
\dfrac{3}{5}$.

b) \ Find a value of $N$ for $\varepsilon =\dfrac{1}{2}$. \ That is, find a
value for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <%
\dfrac{1}{2}$.

Let us look at what we need. \ We need to find $N$ so that if $n>N$, then%
\begin{equation*}
\left\vert a_{n}-L\right\vert <\varepsilon \text{ \ \ in this case this
means that }\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert <\dfrac{1}{%
2}
\end{equation*}%
Let us simplify the expression $\left\vert a_{n}-L\right\vert =\left\vert 
\dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert $.%
\begin{equation*}
\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert =\left\vert \dfrac{%
5\left( 3n+2\right) }{5\left( 5n+1\right) }-\dfrac{3\left( 5n+1\right) }{%
5\left( 5n+1\right) }\right\vert =\left\vert \dfrac{15n+10-15n-3}{5\left(
5n+1\right) }\right\vert =\left\vert \dfrac{7}{25n+5}\right\vert =\dfrac{7}{%
25n+5}
\end{equation*}%
\begin{equation*}
\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert <\dfrac{1}{2}\text{ \
is the same as \ }\dfrac{7}{25n+5}<\dfrac{1}{2}
\end{equation*}%
We solve this inequality:%
\begin{eqnarray*}
\dfrac{7}{25n+5} &<&\dfrac{1}{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ }2\left( 25n+5\right) \text{ \ is positive} \\
14 &<&25n+5\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5 \\
9 &<&25n\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }25 \\
\dfrac{9}{25} &<&n
\end{eqnarray*}%
So all values of $n$, greater than $\dfrac{9}{25}=0.36$ will work. \ So $N=1$

c) \ Find a value of $N$ for $\varepsilon =0.05$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.05$.

Let us look at what we need. \ We need to find $N$ so that if $n>N$, then%
\begin{equation*}
\left\vert a_{n}-L\right\vert <\varepsilon \text{ \ in this case this means
that }\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert <0.05
\end{equation*}%
We already saw that $\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert =%
\dfrac{7}{25n+5}$, so we need that%
\begin{equation*}
\dfrac{7}{25n+5}<0.05
\end{equation*}%
We solve this inequality: 
\begin{eqnarray*}
\dfrac{7}{25n+5} &<&0.05\text{ \ \ \ \ \ \ \ \ \ \ \ \ both sides are
positive - take reciprocal} \\
\dfrac{25n+5}{7} &>&\dfrac{1}{0.05}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }%
\dfrac{1}{0.05}=20 \\
\dfrac{25n+5}{7} &>&20\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ multiply by }7
\\
25n+5 &>&140\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5 \\
25n &>&135\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }25 \\
n &>&\dfrac{135}{25}=5.4
\end{eqnarray*}%
So all values of $n$, greater than $5.4$ will work. \ So $N=5$.\pagebreak

d) \ Find a value of $N$ for $\varepsilon =0.001$. \ That is, find a value
for $N$ so that for all $n>N$, \ $\left\vert a_{n}-L\right\vert <0.001$.

Let us look at what we need. \ We need to find $N$ so that if $n>N$, then%
\begin{equation*}
\left\vert a_{n}-L\right\vert <\varepsilon \text{ \ in this case this means
that }\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert <0.001
\end{equation*}%
we have shown already that \ $\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}%
\right\vert =\dfrac{7}{25n+5}$ 
\begin{equation*}
\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert <0.001\text{ \ \ \
same as }\dfrac{7}{25n+5}<0.001\text{ }
\end{equation*}%
We solve this inequality:%
\begin{eqnarray*}
\dfrac{7}{25n+5} &<&0.001\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ both sides are
positive- take reciprocal} \\
\dfrac{25n+5}{7} &>&\dfrac{1}{0.001}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ }\dfrac{1}{0.001}=1000 \\
\dfrac{25n+5}{7} &>&1000\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
multiply by }7 \\
25n+5 &>&7000\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5
\\
25n &>&6995\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }5
\\
n &>&\dfrac{6995}{25}=279.\,8
\end{eqnarray*}%
So all values of $n$, greater than $279.8$ will work. \ So $N=279$.

e) \ Find a general expression for $N$ in terms of $\varepsilon $.

Solution: \ $\left\vert a_{n}-L\right\vert <\varepsilon $ \ in this case
this means that $\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}\right\vert
<\varepsilon $. \ The expression $\left\vert \dfrac{3n+2}{5n+1}-\dfrac{3}{5}%
\right\vert $ can be simplified as $\dfrac{7}{25n+5}$.%
\begin{eqnarray*}
\left\vert a_{n}-L\right\vert &<&\varepsilon \\
\dfrac{7}{25n+5} &<&\varepsilon \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ both sides
are positive - take reciprocal} \\
\dfrac{25n+5}{7} &>&\dfrac{1}{\varepsilon }\text{ \ \ \ \ \ \ \ \ \ \ \ \
multiply by }7 \\
25n+5 &>&\dfrac{7}{\varepsilon }\text{ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5 \\
25n &>&\dfrac{7}{\varepsilon }-5\text{ \ \ \ \ \ \ divide by }25 \\
n &>&\dfrac{\dfrac{7}{\varepsilon }-5}{25}=\dfrac{7}{25\varepsilon }-\dfrac{5%
}{25}=\dfrac{7}{25\varepsilon }-\dfrac{1}{5}
\end{eqnarray*}%
So $N=\lceil \dfrac{7}{25\varepsilon }-\dfrac{1}{5}\rceil $ is a good value
for $N$. \ \ However, once we find a value for $N$ that works, any greater
value would also work. \ So, we migh present a simpler value for $N,$ namely 
$\lceil \dfrac{7}{25\varepsilon }\rceil $. \ As long as we are "growing"
this value, we are OK.

Let us prove that $N=\lceil \dfrac{7}{25\varepsilon }\rceil $ indeed works.

If $n>N$, then $n>\dfrac{7}{25\varepsilon }$. \ Then $n>\dfrac{7}{%
25\varepsilon }-\dfrac{1}{5}$ is also true.%
\begin{eqnarray*}
n &>&\dfrac{7}{25\varepsilon }-\dfrac{1}{5}\text{ \ \ \ \ \ \ \ \ \ \ \ \
add }\dfrac{1}{5} \\
n+\dfrac{1}{5} &>&\dfrac{7}{25\varepsilon } \\
\dfrac{5n+1}{5} &>&\dfrac{7}{25\varepsilon }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ both sides are positive - take reciprocal} \\
\dfrac{5}{5n+1} &<&\dfrac{25\varepsilon }{7}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ multiply by }\dfrac{7}{25} \\
\dfrac{7\cdot 5}{25\left( 5n+1\right) } &<&\varepsilon \\
\dfrac{7}{5\left( 5n+1\right) } &<&\varepsilon
\end{eqnarray*}

and we already know that $\dfrac{7}{5\left( 5n+1\right) }$ is $\left\vert
a_{n}-\dfrac{3}{5}\right\vert $ and so we have that $\left\vert a_{n}-\dfrac{%
3}{5}\right\vert <\varepsilon $. \ This completes our proof.

\item Find the limit of the sequence $a_{n}=\dfrac{2n-5}{n+3}$ and use an
epsilon-N proof to justify your answer.

Solution: \ As $n$ approaches infinity, $a_{n}=\dfrac{2n\left( 1-\dfrac{5}{2n%
}\right) }{n\left( 1+\dfrac{3}{n}\right) }$ will approach $2$. \ Let $%
\varepsilon >0$ be given. \ 

Part 1 - This is how we find a correct value of $N$. \ You do not need to
present this.%
\begin{equation*}
\left\vert a_{n}-L\right\vert <\varepsilon \text{ \ means that }\left\vert 
\dfrac{2n-5}{n+3}-2\right\vert <\varepsilon
\end{equation*}%
We simplify $\left\vert \dfrac{2n-5}{n+3}-2\right\vert $:%
\begin{equation*}
\left\vert \dfrac{2n-5}{n+3}-2\right\vert =\left\vert \dfrac{2n-5}{n+3}-%
\dfrac{2\left( n+3\right) }{n+3}\right\vert =\left\vert \dfrac{2n-5-2n-6}{n+3%
}\right\vert =\left\vert \dfrac{-11}{n+3}\right\vert =\dfrac{11}{n+3}
\end{equation*}%
So we need that%
\begin{eqnarray*}
\dfrac{11}{n+3} &<&\varepsilon \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ both
sides are positive - take reciprocal} \\
\dfrac{n+3}{11} &>&\dfrac{1}{\varepsilon }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \
multiply by }11 \\
n+3 &>&\dfrac{11}{\varepsilon }\text{ \ \ \ \ \ \ \ \ \ \ \ subtract }3 \\
n &>&\dfrac{11}{\varepsilon }-3
\end{eqnarray*}%
So $N=\lceil \dfrac{11}{\varepsilon }-3\rceil $ \ is good, and so is any
greater $N$, for example $\lceil \dfrac{11}{\varepsilon }\rceil $ \bigskip
\pagebreak

Part 2 - This is how we present a correct value of $N$.

Claim: \ $\lim\limits_{n\rightarrow \infty }\dfrac{2n-5}{n+3}=2$

proof: \ Let $\varepsilon >0$ be given. \ Define $N=\lceil \dfrac{11}{%
\varepsilon }\rceil $. \ If $n>N$, then 
\begin{eqnarray*}
n &>&\dfrac{11}{\varepsilon }-3\text{ \ \ \ \ \ \ add }3 \\
n+3 &>&\dfrac{11}{\varepsilon }\text{ \ \ \ \ \ \ \ \ \ \ \ \ both sides are
positive - take reciprocal} \\
\dfrac{1}{n+3} &<&\dfrac{\varepsilon }{11}\text{ \ \ \ \ \ \ \ \ \ \ \ \
multiply by }11 \\
\dfrac{11}{n+3} &<&\varepsilon
\end{eqnarray*}%
Also, 
\begin{equation*}
\left\vert a_{n}-L\right\vert =\left\vert \dfrac{2n-5}{n+3}-2\right\vert
=\left\vert \dfrac{2n-5-2n-6}{n+3}\right\vert =\left\vert \dfrac{-11}{n+3}%
\right\vert =\dfrac{11}{n+3}<\varepsilon
\end{equation*}%
This completes our proof.

\vspace{3.98in}\vspace{1in}\medskip \medskip \medskip \medskip \medskip
\medskip \medskip
\end{enumerate}

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