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%\newtheorem{example}[theorem]{Example}
%\newtheorem{exercise}[theorem]{Exercise}
%\newtheorem{lemma}[theorem]{Lemma}
%\newtheorem{proposition}[theorem]{Proposition}
%\newtheorem{remark}[theorem]{Remark}


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\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
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\begin{document}


\section{\protect\vspace{1pt}Ma 116 Lecture 2/16/00}


\subsection{The Comparison Test\protect\vspace{1pt}}

\emph{Comparison Test:} \ Suppose that $\sum a_{n}$ and $\sum b_{n}$ are
series with \emph{positive} terms and $a_{n}\leq b_{n}$.

\vspace{1pt}

\qquad (a) \ If $\sum b_{n}$\ is convergent; then $\sum a_{n}$ is also
convergent.

\qquad (b) \ If $\sum a_{n}$ is divergent; then $\sum b_{n}$ is also
divergent.

\paragraph{\protect\vspace{1pt}}

\begin{example}
Determine whether 
\begin{equation*}
\dsum\limits_{n=1}^{\infty }\dfrac{3}{n(n+3)}
\end{equation*}
converges. \ 
\end{example}

\emph{Solution: }Now $n^{2}+3n>n^{2}$ so 
\begin{equation*}
\dfrac{3}{n^{2}+3n}<\dfrac{3}{n^{2}}.
\end{equation*}

\vspace{1pt}

But $\sum \dfrac{3}{n^{2}}$ converges, since it is a \emph{p-series} with $%
p=2>1$. \ Hence, the given series converges by comparison.

\vspace{1pt}

\begin{example}
Determine whether 
\begin{equation*}
\dsum_{n=3}^{\infty }\dfrac{1}{n^{\frac{1}{2}}-2}
\end{equation*}
converges.
\end{example}

Since $n^{\frac{1}{2}}-2<n^{\frac{1}{2}}$ for $n\geq 3,$ then $\dfrac{1}{n^{%
\frac{1}{2}}-2}>\dfrac{1}{n^{\frac{1}{2}}}.$ However, $\dsum \dfrac{1}{n^{%
\frac{1}{2}}}$ diverges (why? 
\CustomNote{AnswerNote}{%
It is a \emph{p-series }with $p=\frac{1}{2}<1.$}). Hence the original series
diverges.

\vspace{1pt}

\begin{example}
Show that 
\begin{equation*}
\dsum_{n=1}^{\infty }\dfrac{1}{n\left( n+1\right) \left( n+2\right) }
\end{equation*}
converges.
\end{example}

\begin{equation*}
\dfrac{1}{n\left( n+1\right) \left( n+2\right) }<\dfrac{1}{n^{3}}
\end{equation*}

$\sum \frac{1}{n^{3}}$ is a $p-$series with $p=3,$ and hence converges. This
implies by the Comparison Test that the original series converges.

\vspace{1pt}

\begin{example}
Test the series%
\begin{equation*}
\sum_{n=0}^{\infty }\dfrac{1}{n!}
\end{equation*}
\end{example}

\vspace{1pt}

for convergence.

\begin{eqnarray*}
n! &=&n\left( n-1\right) \left( n-2\right) \cdots 3\cdot 2\cdot 1 \\
&\geq &2\cdot 2\cdot 2\cdots 2\cdot 2\cdot 1 \\
&=&2^{n-1}
\end{eqnarray*}

Thus

\begin{equation*}
\dfrac{1}{n!}\leq \dfrac{1}{2^{n-1}}\text{ \ \ \ \ for \ }n\geq 1
\end{equation*}

Hence 
\begin{equation*}
\sum_{n=0}^{\infty }\dfrac{1}{n!}\leq 1+\sum_{n=1}^{\infty }\dfrac{1}{2^{n-1}%
}=1+\sum_{n=0}^{\infty }\dfrac{1}{2^{n}}
\end{equation*}

But this last series is a convergent geometric series. Since the original
series and this series are positive, the original series converges by the
Comparison Test.

\vspace{1pt}

For more on the Comparison Test hold down the Ctrl key and click on 
\hyperref{Comparison Test}{}{}{%
http://archives.math.utk.edu/visual.calculus/6/series.8/index.html}

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\subsection{The Alternating Series Test}

\begin{definition}
A series $\dsum\limits_{n=n_{o}}^{\infty }a_{n}$ is \emph{alternating} if $%
a_{n}=\left( -1\right) ^{n}b_{n}$ (or $a_{n}=\left( -1\right) ^{n+1}b_{n}$)
where $b_{n}$ is positive for all $n$. \ Notice that $b_{n}=\left|
a_{n}\right| ,$ so $b_{n}$ is positive..
\end{definition}

\vspace{1pt}

\emph{The Alternating Series Test:} \ Given the series $\sum\limits_{n=1}^{%
\infty }(-1)^{n-1}b_{n}$, if 
\begin{equation*}
(i)b_{n+1}\leq b_{n}\text{ \ for all \ }n
\end{equation*}%
\emph{and} 
\begin{equation*}
(ii)\lim\limits_{n\rightarrow \infty }b_{n}=0;
\end{equation*}
then the series is convergent.

\paragraph{\protect\vspace{1pt}}

The alternating harmonic series 
\begin{equation*}
\sum\limits_{n=1}^{\infty }\dfrac{(-1)^{n-1}}{n}
\end{equation*}
has $b_{n}=\dfrac{1}{n}$ and therefore satisfies (i) $b_{n+1}<b_{n}$ because

\begin{center}
\begin{equation*}
\dfrac{1}{n+1}<\dfrac{1}{n}
\end{equation*}
\end{center}

\emph{and} (ii) $\lim\limits_{n\rightarrow \infty
}b_{n}=\lim\limits_{n\rightarrow \infty }\dfrac{1}{n}$ $=0$.

Hence, the Alternating Harmonic Series converges.

\begin{example}
The series
\end{example}

\begin{equation*}
\sum_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n+1}n}{2n-1}=1-\dfrac{2}{3}+%
\dfrac{3}{5}-\dfrac{4}{7}+\dfrac{5}{9}-\cdots
\end{equation*}

\vspace{1pt}

is an alternating series. Let $b_{n}=\dfrac{n}{2n-1}.$ Then

\vspace{1pt}

\begin{equation*}
b_{n}=\dfrac{n}{2n-1}>\dfrac{n+1}{2n+1}=b_{n+1}
\end{equation*}

for $n\geq 1.$ But

\vspace{1pt}

\begin{equation*}
\lim_{n\rightarrow \infty }b_{2}=\dfrac{1}{2}\neq 0
\end{equation*}

so the Alternating Series Test does not apply. We see that the series
diverges because the $n$th term does not go to zero.

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\emph{Alternating Series Estimation Series:} \ If $s=\sum (-1)^{n-1}b_{n}$
is the sum of an alternating series that satisfies 
\begin{equation*}
(i)0<b_{n+1}\leq b_{n}\text{ \ }\emph{and}\text{ \ }(ii)\lim\limits_{n%
\rightarrow \infty }b_{n}=0;
\end{equation*}
then 
\begin{equation*}
|R_{n}|=|S-S_{n}|\leq b_{n+1}
\end{equation*}
where $R_{n}$ is the remainder if we use the partial sum $S_{n}$ as an
approximation to the total sum $S$.

\vspace{1pt}For more on alternating series hold down the Ctrl key and click
on \hyperref{Alternating Series}{}{}{%
http://archives.math.utk.edu/visual.calculus/6/series.10/index.html}.

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\subsection{Absolute Convergence}

\begin{definition}
Given a series $\dsum\limits_{n=n_{o}}^{\infty }a_{n}$, its \emph{related
absolute series} is $\dsum\limits_{n=n_{o}}^{\infty }\left| a_{n}\right| $.
\end{definition}

\begin{proposition}
\textsl{The Absolute Convergence Test}

If the related absolute series $\dsum\limits_{n=n_{o}}^{\infty }\left|
a_{n}\right| $ is convergent, then the original series $\dsum%
\limits_{n=n_{o}}^{\infty }a_{n}$\ is convergent.\bigskip
\end{proposition}

\begin{remark}
The Absolute Convergence Test says nothing about $\dsum\limits_{n=n_{o}}^{%
\infty }a_{n}$\ if the related absolute series $\dsum\limits_{n=n_{o}}^{%
\infty }\left| a_{n}\right| $ is divergent.\bigskip
\end{remark}

\begin{remark}
Note that $\sum \left| a_{n}\right| =\left| a_{1}\right| +\left|
a_{2}\right| +\cdots $. \emph{If a series is absolutely convergent, then the
proposition tells us that it is convergent.} The converse is not true,
namely $\sum a_{n}$ can converge, but $\sum \left| a_{n}\right| $ may
diverge. Can you give an example of this case? 
\CustomNote{AnswerNote}{%
Consider $\sum\limits_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n}}{n}.$%
\par
By the Alternating Series Test, this series converges. However, $%
\sum\limits_{n=1}^{\infty }\left| \dfrac{\left( -1\right) ^{n}}{n}\right|
=\sum\limits_{n=1}^{\infty }\dfrac{1}{n}$ and this series diverges by the
Integral Test.}
\end{remark}

\begin{definition}
If $\dsum\limits_{n=n_{o}}^{\infty }\left| a_{n}\right| $ is convergent,
then $\dsum\limits_{n=n_{o}}^{\infty }a_{n}$ is called \emph{absolutely
convergent}. \ ( So the proposition says ``An absolutely convergent series
is convergent.'')

If $\dsum\limits_{n=n_{o}}^{\infty }\left| a_{n}\right| $ is divergent but $%
\dsum\limits_{n=n_{o}}^{\infty }a_{n}$\ is convergent, then $%
\dsum\limits_{n=n_{o}}^{\infty }a_{n}$ is called \emph{conditionally
convergent}.

If $\dsum\limits_{n=n_{o}}^{\infty }\left| a_{n}\right| $ is divergent and $%
\dsum\limits_{n=n_{o}}^{\infty }a_{n}$\ is divergent, then $%
\dsum\limits_{n=n_{o}}^{\infty }a_{n}$ is just called \emph{divergent}.
\end{definition}

\begin{example}
Thus 
\begin{equation*}
\sum_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n}}{n}
\end{equation*}
\end{example}

is conditionally convergent.

\vspace{1pt}

\begin{example}
Test the series
\end{example}

\begin{equation*}
\sum_{n=1}^{\infty }\dfrac{\sin n}{n^{2}}
\end{equation*}

for convergence.

Note that series has both positive and negative terms but it is not
alternating. We shall look at the series of absolute values

\begin{equation*}
\sum_{n=1}^{\infty }\left| \dfrac{\sin n}{n^{2}}\right|
\end{equation*}%
Now

\begin{equation*}
\left| \dfrac{\sin n}{n^{2}}\right| \leq \dfrac{1}{n^{2}}
\end{equation*}

But $\sum \dfrac{1}{n^{2}}$ converges. (Why?). So $\sum_{n=1}^{\infty
}\left| \dfrac{\sin n}{n^{2}}\right| $ is convergent by the Comparision
Test. Since absolute convergence implies convergence, we have the $%
\sum_{n=1}^{\infty }\dfrac{\sin n}{n^{2}}$ is convergent.

For more on absolute Convergence hold down the Ctrl key and click on 
\hyperref{Absolute Convergence}{}{}{%
http://archives.math.utk.edu/visual.calculus/6/series.12/index.html}.

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\subsection{The Ratio Test}

\vspace{1pt}Recall

\vspace{1pt}

\textbf{Definition:} \ A series $\sum a_{n}$ is called \emph{absolutely
convergent} if the series of absolute values $\sum |a_{n}|$ is convergent. \ 

\vspace{1pt}

\vspace{1pt}

\emph{Ratio Test:} \ Given a series $\sum a_{n}$, let $L=\lim\limits_{n%
\rightarrow \infty }\left| \dfrac{a_{n+1}}{a_{n}}\right| $

\vspace{1pt}

\qquad (a) \ If $L<1$, then the series converges absolutely.

\qquad (b) \ If $L>1$, then the series is divergent.

\qquad (c) \ If $L=1$, then the test fails (i.e., is inconclusive.)

\vspace{1pt}

\begin{remark}
For an example of (c) consider $\sum \dfrac{1}{n}$ and $\sum \dfrac{1}{n^{2}}%
.$ Both have a value of $1$ for $L$. \ Yet, the former diverges, whereas the
latter converges.
\end{remark}

\vspace{1pt}

\begin{example}
Determine whether 
\begin{equation*}
\sum\limits_{n=1}^{\infty }\dfrac{(n+1)5^{n}}{n3^{2n}}
\end{equation*}
\end{example}

is convergent. \ 

\emph{Solution: }We apply the Ratio Test.

\vspace{1pt}

\begin{equation*}
L=\lim_{n\rightarrow \infty }\left| \dfrac{\dfrac{(n+2)5^{n+1}}{(n+1)3^{2n+2}%
}}{\dfrac{(n+1)5^{n}}{n3^{2n}}}\right| =\lim\limits_{n\rightarrow \infty
}\left| \dfrac{(n+2)5^{n+1}}{(n+1)3^{2n+2}}\cdot \dfrac{n3^{2n}}{(n+1)5^{n}}%
\right| =\lim_{n\rightarrow \infty }\dfrac{5}{3^{2}}\left( \dfrac{n^{2}+2n}{%
n^{2}+2n+1}\right) =\dfrac{5}{9}<1.
\end{equation*}

\vspace{1pt}

Therefore, the given series converges by the Ratio Test.

\begin{example}
Test the series
\end{example}

\begin{equation*}
\sum_{n=1}^{\infty }\dfrac{\left( -1\right) ^{n}4^{n}}{n!}
\end{equation*}

for convergence or divergence.

\emph{Solution: }Using the Ratio Test we have

\begin{equation*}
L=\lim_{n\rightarrow \infty }\left| \dfrac{\dfrac{\left( -1\right)
^{n+1}4^{n+1}}{\left( n+1\right) !}}{\dfrac{\left( -1\right) ^{n}4^{n}}{n!}}%
\right| =\lim_{n\rightarrow \infty }\left| \dfrac{4^{n+1}}{\left( n+1\right)
!}\cdot \dfrac{n!}{4^{n}}\right| =\lim_{n\rightarrow \infty }\dfrac{4}{n+1}=0
\end{equation*}

Thus this series converges absolutely, since $L<1.$

\vspace{1pt}

\begin{example}
Test the series%
\begin{equation*}
\sum_{n=1}^{\infty }\dfrac{7^{n}}{n^{2}}
\end{equation*}
\end{example}

for convergence or divergence.

\vspace{1pt}\emph{Solution: }Again the Ratio Test may be applied. We have

\vspace{1pt}

\begin{equation*}
L=\lim_{n\rightarrow \infty }\left| \dfrac{\dfrac{7^{n+1}}{\left( n+1\right)
^{2}}}{\dfrac{7^{n}}{n^{2}}}\right| =\lim_{n\rightarrow \infty }\left| 
\dfrac{7^{n+1}}{\left( n+1\right) ^{2}}\cdot \dfrac{n^{2}}{7^{n}}\right|
=\lim_{n\rightarrow \infty }7\cdot \left( \dfrac{n^{2}}{n^{2}+2n+1}\right) =7
\end{equation*}

\vspace{1pt}

Since $L>1,$ this series diverges.

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