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\newtheorem{theorem}{Theorem}
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\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
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\lhead{\color{blue} \Large Lecture Notes}
\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2012}
\cfoot{}
\chead{\LARGE Sequences - Part 2}
\rfoot{\small Last revised:  March 20, 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.\medskip

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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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Notice that $\left\vert a_{n}-L\right\vert <\varepsilon $ and \ $%
L-\varepsilon <a_{n}<L+\varepsilon $ \ are equivalent statements.\medskip

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Theorem 1. \ Convergent sequences have unique limits: if $\left\{
a_{n}\right\} $ is a sequence with $\lim\limits_{n\rightarrow \infty
}a_{n}=A $ and $\lim\limits_{n\rightarrow \infty }a_{n}=B$, then $A=B$%
.\medskip 
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\medskip

Proof: \ Suppose for a contradiction that a sequence $a_{n}$ converges to
two different numbers $A$ and $B$. \ The basic idea here is that if $%
\varepsilon $ is selected to be small enough, then the $\varepsilon $
neighborhood of $A$ will be disjoint of the $\varepsilon $ neighborhood of $%
B $ and so $a_{n}$ can not be in both intervals.\FRAME{dtbpF}{3.5431in}{%
0.6581in}{0pt}{}{}{pic1.bmp}{\special{language "Scientific Word";type
"GRAPHIC";display "USEDEF";valid_file "F";width 3.5431in;height
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"0";croptop "1";cropright "1";cropbottom "0";filename
'pic1.bmp';file-properties "XNPEU";}}Suppose that $A\not=B$. \ We may assume
that $A<B$. \ \ (Otherwise just re-label them so that the larger number is
denoted by $B$.) \ Define $\varepsilon =\dfrac{B-A}{2}$. \ Since $\left\{
a_{n}\right\} $ converges to $A$, there exists $N_{A}$ so that for all $%
n>N_{A}$, 
\begin{equation*}
A-\varepsilon <a_{n}<A+\varepsilon
\end{equation*}%
Similarly, since $\left\{ a_{n}\right\} $ converges to $B$, there exists $%
N_{B}$ so that for all $n>N_{B}$, 
\begin{equation*}
B-\varepsilon <a_{n}<B+\varepsilon
\end{equation*}%
Now let $n>\max \left( N_{A},N_{B}\right) $, so both conditions hold. \ Then%
\begin{equation*}
A-\varepsilon <a_{n}<A+\varepsilon \text{ \ \ and \ }B-\varepsilon
<a_{n}<B+\varepsilon
\end{equation*}%
We will only need the right-hand side of the first inequality and the
left-hand side of the other:%
\begin{eqnarray*}
a_{n} &<&A+\varepsilon \text{ \ \ and \ }B-\varepsilon <a_{n}\text{ \ \ \
recall that }\varepsilon =\dfrac{B-A}{2} \\
a_{n} &<&A+\dfrac{B-A}{2}\text{ \ \ and \ }B-\dfrac{B-A}{2}<a_{n} \\
a_{n} &<&\dfrac{2A}{2}+\dfrac{B-A}{2}\text{ \ \ and \ }\dfrac{2B}{2}-\dfrac{%
B-A}{2}<a_{n} \\
a_{n} &<&\dfrac{2A+B-A}{2}\text{ \ \ and \ }\dfrac{2B-B+A}{2}<a_{n} \\
a_{n} &<&\dfrac{A+B}{2}\text{ \ \ and \ }\dfrac{A+B}{2}<a_{n}
\end{eqnarray*}%
These two can not be true at the same time. \ This is a contradiction, so $%
A\not=B$ is impossible. \ This completes our proof.\bigskip

In the following, we will prove properties of limits that enable us to
compute limits based on other limits.

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Theorem 2. \ (Sum Rule) \ Let $\left\{ a_{n}\right\} $ and $\left\{
b_{n}\right\} $ be sequences of real numbers. \ Suppose that $A$ and $B$ are

\qquad real numbers such that $\lim\limits_{n\rightarrow \infty }a_{n}=A$
and $\lim\limits_{n\rightarrow \infty }b_{n}=B$. \ Then $\
\lim\limits_{n\rightarrow \infty }\left( a_{n}+b_{n}\right) =A+B$.

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\medskip

Proof: \ Suppose that $A$ and $B$ are real numbers\ such that $%
\lim\limits_{n\rightarrow \infty }a_{n}=A$ and $\lim\limits_{n\rightarrow
\infty }b_{n}=B$. \ Let $\varepsilon >0$ be given. \ There exist $N_{a}$ and 
$N_{b}$ natural numbers such that for all $n>N_{a},$%
\begin{equation*}
A-\dfrac{\varepsilon }{2}<a_{n}<A+\dfrac{\varepsilon }{2}
\end{equation*}

\qquad and for all $k>N_{b},$%
\begin{equation*}
B-\dfrac{\varepsilon }{2}<b_{k}<B+\dfrac{\varepsilon }{2}
\end{equation*}

\qquad Let $N=\max \left( N_{a},N_{b}\right) .$ \ If $n>N,$ then 
\begin{equation*}
A-\dfrac{\varepsilon }{2}<a_{n}<A+\dfrac{\varepsilon }{2}\text{ and }B-%
\dfrac{\varepsilon }{2}<b_{n}<B+\dfrac{\varepsilon }{2}
\end{equation*}

\qquad Adding these two inequalities we obtain%
\begin{eqnarray*}
A-\dfrac{\varepsilon }{2}+B-\dfrac{\varepsilon }{2} &<&a_{n}+b_{n}<A+\dfrac{%
\varepsilon }{2}+B+\dfrac{\varepsilon }{2} \\
A+B-\varepsilon &<&a_{n}+b_{n}<A+B+\varepsilon
\end{eqnarray*}

\qquad Thus $a_{n}+b_{n}$ converges to \ $A+B$.\bigskip

Example 1. \bigskip

\qquad a) $\ \lim\limits_{n\rightarrow \infty }\left( 2+\dfrac{1}{n}\right)
=\lim\limits_{n\rightarrow \infty }2+\lim\limits_{n\rightarrow \infty }%
\dfrac{1}{n}=2+0=2$\bigskip

\qquad b) $\ \lim\limits_{n\rightarrow \infty }\dfrac{3n+1}{n}%
=\lim\limits_{n\rightarrow \infty }\left( \dfrac{3n}{n}+\dfrac{1}{n}\right)
=\lim\limits_{n\rightarrow \infty }\left( 3+\dfrac{1}{n}\right)
=\lim\limits_{n\rightarrow \infty }3+\lim\limits_{n\rightarrow \infty }%
\dfrac{1}{n}=3+0=3$\bigskip

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Theorem 3. \ (Difference Rule) \ Let $\left\{ a_{n}\right\} $ and $\left\{
b_{n}\right\} $ be sequences of real numbers. \ Suppose that $A$ and $B$ are

\qquad real numbers such that $\lim\limits_{n\rightarrow \infty }a_{n}=A$
and $\lim\limits_{n\rightarrow \infty }b_{n}=B$. \ Then $\
\lim\limits_{n\rightarrow \infty }\left( a_{n}-b_{n}\right) =A-B$.

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\medskip \bigskip

Proof: \ Suppose that $A$ and $B$ are real numbers\ such that $%
\lim\limits_{n\rightarrow \infty }a_{n}=A$ and $\lim\limits_{n\rightarrow
\infty }b_{n}=B$. \ Let $\varepsilon >0$ be given. \ There exist $N_{a}$ and 
$N_{b}$ natural numbers such that for all $n>N_{a},$%
\begin{equation*}
A-\dfrac{\varepsilon }{2}<a_{n}<A+\dfrac{\varepsilon }{2}
\end{equation*}

\qquad and for all $k>N_{b},$%
\begin{equation*}
B-\dfrac{\varepsilon }{2}<b_{k}<B+\dfrac{\varepsilon }{2}
\end{equation*}

\qquad Multiply all sides by $-1$.%
\begin{equation*}
-B+\dfrac{\varepsilon }{2}>-b_{k}>-B-\dfrac{\varepsilon }{2}
\end{equation*}

\qquad We turn the inequality around:%
\begin{equation*}
-B-\dfrac{\varepsilon }{2}<-b_{k}<-B+\dfrac{\varepsilon }{2}
\end{equation*}

\qquad Let $N=\max \left( N_{a},N_{b}\right) .$ \ If $n>N,$ then 
\begin{equation*}
A-\dfrac{\varepsilon }{2}<a_{n}<A+\dfrac{\varepsilon }{2}\text{ and \ }-B-%
\dfrac{\varepsilon }{2}<-b_{n}<-B+\dfrac{\varepsilon }{2}
\end{equation*}

\qquad Adding these two inequalities we obtain%
\begin{eqnarray*}
A-\dfrac{\varepsilon }{2}+\left( -B\right) -\dfrac{\varepsilon }{2}
&<&a_{n}+\left( -b_{n}\right) <A+\dfrac{\varepsilon }{2}+\left( -B\right) +%
\dfrac{\varepsilon }{2} \\
A-B-\varepsilon &<&a_{n}-b_{n}<A-B+\varepsilon
\end{eqnarray*}

\qquad Thus $a_{n}-b_{n}$ converges to \ $A-B$.\bigskip

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Theorem 4. \ (Constant Multiple Rule) \ Let $\left\{ a_{n}\right\} $ be a
sequence of real numbers and $c$ a real number. \ Suppose

\qquad that $\lim\limits_{n\rightarrow \infty }a_{n}=A.$ \ Then \ $%
\lim\limits_{n\rightarrow \infty }\left( ca_{n}\right) =cA$.

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\medskip

Proof: \ Case 1. \ Suppose that $c=0$. \ Then $ca_{n}$ is a constant
sequence and its limit is clearly zero.\medskip

\qquad\ \ \ \ Case 2. \ Suppose that $c\not=0$. \ Let $\varepsilon >0$ be
given. \ Since $\lim\limits_{n\rightarrow \infty }a_{n}=A,$ there exists $%
N>0 $ so

\qquad\ \ \ \ \ \ that for all $n>N$, 
\begin{eqnarray*}
\left\vert a_{n}-A\right\vert &<&\dfrac{\varepsilon }{\left\vert
c\right\vert }\text{ \ \ \ \ \ \ \ \ mulitply by }\left\vert c\right\vert \\
\left\vert c\right\vert \left\vert a_{n}-A\right\vert &<&\varepsilon \\
\left\vert ca_{n}-cA\right\vert &<&\varepsilon
\end{eqnarray*}

\qquad\ \ \ \ \ and so $\lim\limits_{n\rightarrow \infty }ca_{n}=cA$.\bigskip

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Theorem 5. \ (Product Rule) \ Let $\left\{ a_{n}\right\} $ and $\left\{
b_{n}\right\} $ be sequences of real numbers. \ Suppose that $A$ and $B$ are

\qquad real numbers such that $\lim\limits_{n\rightarrow \infty }a_{n}=A$
and $\lim\limits_{n\rightarrow \infty }b_{n}=B$. \ Then \ $%
\lim\limits_{n\rightarrow \infty }\left( a_{n}b_{n}\right) =AB$

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Theorem 6. \ (Quotient Rule) \ Let $\left\{ a_{n}\right\} $ and $\left\{
b_{n}\right\} $ be sequences of real numbers. \ Suppose that $A$ and $%
B\not=0 $ are

\qquad real numbers such that $\lim\limits_{n\rightarrow \infty }a_{n}=A$
and $\lim\limits_{n\rightarrow \infty }b_{n}=B$. \ Then \ $%
\lim\limits_{n\rightarrow \infty }\left( \dfrac{a_{n}}{b_{n}}\right) =\dfrac{%
A}{B}$.

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\medskip

Proving these theorems is more difficult. \ We will not cover it in this
course.\bigskip \bigskip

Example 2. \bigskip

\qquad a) $\lim\limits_{n\rightarrow \infty }\left( -\dfrac{3}{n^{2}}\right)
=-3\lim\limits_{n\rightarrow \infty }\left( \dfrac{1}{n}\cdot \dfrac{1}{n}%
\right) =-3\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}\cdot
\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}=-3\cdot 0\cdot 0=0$\bigskip

\qquad b) \ $\lim\limits_{n\rightarrow \infty }\dfrac{3-2n^{4}}{7n^{4}+2}%
=\lim\limits_{n\rightarrow \infty }\dfrac{\dfrac{3}{n^{4}}-2}{7+\dfrac{2}{%
n^{4}}}=\dfrac{\lim\limits_{n\rightarrow \infty }\left( \dfrac{3}{n^{4}}%
-2\right) }{\lim\limits_{n\rightarrow \infty }\left( 7+\dfrac{2}{n^{4}}%
\right) }=\dfrac{-2}{7}=-\dfrac{2}{7}$ \ \bigskip

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Theorem 7. \ (The Sandwich Theorem for Sequences): \ Suppose that $\left\{
a_{n}\right\} ,$ $\left\{ b_{n}\right\} ,$ and $\left\{ c_{n}\right\} $ are
sequences

\qquad with $\lim\limits_{n\rightarrow \infty
}a_{n}=\lim\limits_{n\rightarrow \infty }c_{n}=L$. \ Suppose that there
exists $N$ positive integer such that for all $n>N,$ 
\begin{equation*}
a_{n}\leq b_{n}\leq c_{n}
\end{equation*}

\qquad then $b_{n}$ converges to $L$.%
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\medskip \bigskip

Proof: \ Suppose the conditions hold. \ Let $\varepsilon >0$ be given. \
Since $\left\{ a_{n}\right\} $ and $\left\{ b_{n}\right\} $ are converge to $%
L$, there exist $N_{a}$ and $N_{c}$ such that when $n>N_{a}$, then \ 
\begin{equation*}
L-\varepsilon <a_{n}<L+\varepsilon
\end{equation*}

\qquad and when $n>N_{c}$, then \ 
\begin{equation*}
L-\varepsilon <c_{n}<L+\varepsilon
\end{equation*}

\qquad Let $N=\max \left( N_{a},N_{b}\right) $. \ If $n>N$, then 
\begin{equation*}
L-\varepsilon <a_{n}\leq b_{n}\leq c_{n}<L+\varepsilon
\end{equation*}

\qquad and so $\left\{ b_{n}\right\} $ converges to $L$.\bigskip

Consequence: \ If $\left\vert b_{n}\right\vert \leq c_{n}$ and $%
c_{n}\rightarrow 0$, then $b_{n}\rightarrow 0$.\bigskip

Example 3. \ \bigskip

\qquad a) \ $\dfrac{\sin n}{n}\rightarrow 0$ since $-\dfrac{1}{n}\leq \dfrac{%
\sin n}{n}\leq \dfrac{1}{n}$ and $\dfrac{1}{n}\rightarrow 0$ and $-\dfrac{1}{%
n}\rightarrow 0$\bigskip

\qquad b) \ $\dfrac{\left( -1\right) ^{n}}{n^{2}}\rightarrow 0$ since $-%
\dfrac{1}{n^{2}}\leq \dfrac{\left( -1\right) ^{n}}{n^{2}}\leq \dfrac{1}{n^{2}%
}$\bigskip

Some sequences are defined \textbf{recursively}. \ Recursive definitions
enable us to compute the first, second, third, ... $n$th term, but we cannot
compute the nth term without first computing the first $n-1$. \ 

The \textbf{Fibonacci sequence} is a perfect example for this.\bigskip

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Definition: \ The \textbf{Fibonacci sequence} is defined recursively as 
\begin{equation*}
F_{1}=1,~~F_{2}=1,~~\text{and for all }n\in 
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,~~~F_{n+2}=F_{n}+F_{n+1}
\end{equation*}

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\qquad

The first few terms of the Fibonacci Sequence are $%
1,1,2,3,5,8,13,21,34,55,89,...$ The explicit formula for the nth term of
this sequence is a very interesting formula.\bigskip

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Definition: \ A sequence $\left\{ a_{n}\right\} $ is \textbf{bounded from
above} if there exists a number $M$ such that $a_{n}\leq M$

\qquad for all $n$. \ The number $M$ is called an \textbf{upper bound} for
the sequence $\left\{ a_{n}\right\} $. \ If $M$ is an upper bound

\qquad for $\left\{ a_{n}\right\} $ but no number less than $M$ is an upper
bound for $\left\{ a_{n}\right\} ,$ then $M$ is the \textbf{least upper }

\qquad \textbf{bound} for $\left\{ a_{n}\right\} $.\bigskip

\qquad A sequence $\left\{ a_{n}\right\} $ is \textbf{bounded from below} if
there exists a number $m$ such that $a_{n}\geq m$ for all $n$. \ 

\qquad The number $m$ is called a \textbf{lower bound} for the sequence $%
\left\{ a_{n}\right\} $. \ If $m$ is a lower bound for $\left\{
a_{n}\right\} $ but

\qquad no number greater than $m$ is a lower bound for $\left\{
a_{n}\right\} $, then $m$ is the \textbf{greatest lower bound} for $\left\{
a_{n}\right\} $.\bigskip

\qquad If $\left\{ a_{n}\right\} $ is bounded from above and from below, we
say that $\left\{ a_{n}\right\} $ is \textbf{bounded}. \ If $\left\{
a_{n}\right\} $ is not bounded,

\qquad we say that $\left\{ a_{n}\right\} $ is an \textbf{unbounded}
sequence.%
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\medskip \bigskip

Example 4. a) \ The sequence $1,4,9,16,....$ \ is bounded from below but not
from above. \ $1$ is the greatest lower

\qquad bound for the sequence.\bigskip

b) \ The sequence $\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},...,\dfrac{n}{n+1}$
is bounded. \ $\dfrac{1}{2}$ is the greatest lower bound and $1$ is the
lowest upper bound

\qquad for this sequence.\bigskip

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Theorem 8. \ If $\left\{ a_{n}\right\} $ is convergent, then $a_{n}$ is
bounded.%
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proof: \ Suppose that $\left\{ a_{n}\right\} $ is a convergent sequence and $%
a_{n}\rightarrow L$. \ Let $\varepsilon =1$. \ There exists a natural number 
$N$

\qquad such that for all $n>N$, 
\begin{equation*}
L-1<a_{n}<L+1
\end{equation*}%
\qquad

\qquad Consider now the set $\left\{ a_{1},a_{2},a_{3},...,a_{N}\right\} $.
\ Since this is a finite set, it has a lowest and greatest element. \ 

\qquad Denote these by $m$ and $M$, \ respectively. \ We claim that $\min
\left( m,L-1\right) $ is a lower bound for the sequence

$\qquad \left\{ a_{n}\right\} $ and $\max \left( M,L+1\right) $ is an upper
bound for the sequence. \ \bigskip

\qquad Let $a_{k}$ be any term of the sequence. \ If $k>N,$ then $%
L-1<a_{k}<L+1$ and so 
\begin{equation*}
\min \left( m,L-1\right) \leq L-1<a_{k}<L+1\leq \max \left( M,L+1\right)
\end{equation*}

\qquad and if $k\leq N$, then $a_{k}$ is in the set $\left\{
a_{1},a_{2},a_{3},...,a_{N}\right\} $ and so%
\begin{equation*}
\min \left( m,L-1\right) \leq m<a_{k}<M\leq \max \left( M,L+1\right)
\end{equation*}

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Defintion: \ A sequence $\left\{ a_{n}\right\} $ is \textbf{nondecreasing}
if $a_{n}\leq a_{n+1}$ for all $n$. \ That is, $a_{1}\leq a_{2}\leq
a_{3}\leq ...$ \ The sequence

\qquad is \textbf{nonincreasing} if $a_{n}\geq a_{n+1}$ for all $n$. \ That
is, $a_{1}\geq a_{2}\geq a_{3}\geq ...$ \ \ \ The sequence $\left\{
a_{n}\right\} $ is \textbf{monotonic} if

\qquad it is either nondecreasing or nonincreasing. 
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Example 5. a) \ The sequence $1,\dfrac{1}{2},\dfrac{1}{3},\dfrac{1}{4},....$
\ is nonincreasing.

\qquad b) \ the constant sequence $2,2,2,....$ is both nonincreasing and
nondecreasing.

\qquad c) \ the sequence $1,-\dfrac{1}{4},\dfrac{1}{9},-\dfrac{1}{16}$,....
is not monotonic.\medskip

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Theorem 9. \ If a sequence $\left\{ a_{n}\right\} $ is bounded from above
and non-decreasing, then it is also convergent. \ 

\qquad (Similarly, if a sequence is bounded from below and non-increasing,
then it is convergent.)%
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proof: \ Suppose that $\left\{ a_{n}\right\} $ is bounded and nondecreasing.
\ Let $L$ be the least upper bound for the sequence. \ Since

$\qquad L$ is an upper bound, $a_{n}\leq L$ for all $n$.\medskip

\qquad Let $\varepsilon >0$ be given. \ Since $L$ is the lowest upper bound, 
$L-\varepsilon $ is NOT an upper bound. \ This means that

\qquad there exists $m$ natural number such that $a_{m}>L-\varepsilon $. \
Since $a_{n}$ is nondecreasing, all subsequent terms will

\qquad have this property, i.e. for all $n>m,$ $a_{n}\geq
a_{m}>L-\varepsilon $. \ thus we have that for all $n>m$%
\begin{equation*}
L-\varepsilon <a_{n}\leq L<L+\varepsilon
\end{equation*}

\qquad and so $L-\varepsilon <a_{n}<L+\varepsilon $ and so $a_{n}$ converges
to $L$. \ The proof for nonincreasing sequences is similar.\medskip \medskip
\medskip \medskip \medskip \medskip \medskip \medskip

{\small 
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\href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{%
For more documents like this, visit our page at\
https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail
questions or comments to mhidegkuti@ccc.edu.}

\end{document}