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%TCIDATA{<META NAME="Title" CONTENT="Sequences and Series">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Thu Feb 13 17:07:15 1997}
%TCIDATA{Language=American English}
%TCIDATA{CSTFile=MathHelp.cst}
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%TCIDATA{<LINK REL="contents" HREF="DM0-0.tex">}
%TCIDATA{<LINK REL="end" HREF="DM7-9.tex">}
%TCIDATA{<LINK REL="index" HREF="DMIndex.tex">}
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\input{tcilatex}
\begin{document}


\section{Sequences and Series\label{Sequences}}

A sequence can be thought of as an infinite list, and a series as a sum of
the terms of a sequence.

\subsection{\ Sequences}

A \emph{sequence} 
\index{Sequence} $\left\{ a_{n}\right\} _{n=1}^{\infty }$ is a function
whose domain is the set of positive integers. Calculate limits of sequences
in \textsl{Scientific Notebook} by selecting an expression such as $%
\lim_{n\rightarrow \infty }\left( 1+%
\frac{1}{n}\right) ^{n}$ and choosing \textsf{Evaluate}, or by defining $%
a_{n}$, writing $\lim_{n\rightarrow \infty }a_{n}$, and choosing \textsf{%
Evaluate}.

\label{Subscripted function}To define $a_{n}$, with the insertion point in
the equation $a_{n}=\left( 1+\frac{1}{n}\right) ^{n}$, choose \textsf{New
Definition} from the \textsf{Define} submenu. You will get a dialog box that
asks you how you wish to interpret the subscript. Choose \textsf{A Function
Argument}.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\lim\limits_{n\rightarrow \infty }a_{n}=\allowbreak e$

$\lim\limits_{n\rightarrow \infty }\left( 1+\frac{1}{n}\right)
^{n}=\allowbreak e$
\end{quotation}

\vspace{1pt}

\label{Plotting sequences}A sequence such as $\left\{ \left( 1+\frac{1}{n}%
\right) ^{n}\right\} _{n=1}^{\infty }$ can be visualized graphically by
plotting the expression $\left( 1+\frac{1}{n}\right) ^{n}$ at integer values
of $n.$ To get the following plot, set the \textsf{Domain Interval} to $%
1\leq n\leq 49$, as \textsf{Plot Style} choose \textsf{Point}, and for 
\textsf{Point Symbol} choose \textsf{Circle}.

\begin{quote}
$\blacktriangleright $ \textsf{Plot + Rectangular}
\end{quote}

\begin{quotation}
$\left( 1+\frac{1}{n}\right) ^{n}$
\end{quotation}

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"1";xmax "49";xviewmin "0.04";xviewmax "49.98";yviewmin "1.986";yviewmax
"2.705";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle
"normal";xis \TEXUX{n};var1name \TEXUX{$x$};}}

You can generate this figure by plotting $\left( 1+\frac{1}{n}\right) ^{n}$,
then revising the \textsf{Plot Components} page so that the \textsf{Plot
Style} is \textsf{Point}, the \textsf{Point Symbol} is \textsf{Circle}, the 
\textsf{Domain Interval} is 1 to 49, and the \textsf{Sample Size} is 49.

This plot indicates that $\lim_{n\rightarrow \infty }\left( 1+\frac{1}{n}%
\right) ^{n}\approx 2.7$. Indeed, \textsf{Evaluate} yields $e$ and \textsf{%
Evaluate Numerically} produces $e=\allowbreak 2.718281828$.\medskip

\subsection{\ Series\label{Series}}

To sum a series 
\index{Series}, place the insertion point in the series and choose \textsf{%
Evaluate}.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\dsum\limits_{n=1}^{\infty }%
\dfrac{1}{n^{2}}=\allowbreak \frac{1}{6}\pi ^{2}$

$\dsum\limits_{n=1}^{\infty }\dfrac{(-1)^{n}}{n}=\allowbreak -\ln 2$

$\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{3}}=\allowbreak \zeta \left(
3\right) $

$\dsum\limits_{n=1}^{\infty }\dfrac{20^{n}}{n!}=\allowbreak e^{20}-1$

$\dsum\limits_{n=1}^{\infty }\left( 0.99\right) ^{n}=\allowbreak 99.0$

$\dsum\limits_{n=1}^{\infty }\sin n\pi =\allowbreak
\dsum\limits_{n=1}^{\infty }\sin n\pi $

$\dsum\limits_{n=1}^{\infty }\dfrac{n+1}{n}=\infty $

$\dsum\limits_{n=1}^{\infty }\dfrac{n^{4}-1}{n^{4}}=\infty $

$\dsum\limits_{n=1}^{\infty }\dfrac{n^{2}-1}{n^{4}}=-\dfrac{1}{90}\pi ^{4}+%
\dfrac{1}{6}\pi ^{2}$
\end{quotation}

Occasionally, a result is \label{Evaluate numerically}obtained that may be
obscure, such as the responses to $\dsum\limits_{n=1}^{\infty }\sin n\pi $
and $\dsum\limits_{n=1}^{\infty }\frac{1}{n^{3}}$. The lack of response to $%
\dsum\limits_{n=1}^{\infty }\sin n\pi $ is the Maple version of ``I give
up'' or ``I do not have enough information.'' The series $%
\dsum\limits_{n=1}^{\infty }\frac{1}{n^{3}}$ and the values of the \label%
{Zeta function} 
\index{Zeta function}zeta function $\zeta \left( \cdot \right) $ can be
estimated numerically.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate Numerically}
\end{quote}

\begin{quotation}
$\zeta \left( 3\right) =\allowbreak 1.202056903$

$\dsum\limits_{n=1}^{\infty }%
\dfrac{1}{n^{3}}=\allowbreak 1.202056903$
\end{quotation}

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate, Evaluate Numerically}
\end{quote}

\begin{quotation}
$\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{5}}=\allowbreak \zeta \left(
5\right) =\allowbreak 1.036927755$
\end{quotation}

To sum a series of a form similar to $\sum_{n=1}^{\infty }a_{n}$, first
enter an equation such as $a_{n}=\frac{n^{2}}{2^{n}}$. Then, from the 
\textsf{Define} submenu, choose \textsf{New Definition}.

\subsubsection{Ratio Test\label{Absolute convergence}\label{Ratio test}}

\begin{quotation}
A series $\sum_{n=1}^{\infty }a_{n}$ \emph{converges absolutely} if $%
\sum_{n=1}^{\infty }\left| a_{n}\right| $ converges, in which case the
series $\sum_{n=1}^{\infty }a_{n}$ also converges. The \emph{ratio test}
states that a series $\sum_{n=1}^{\infty }a_{n}$ converges absolutely (and
therefore converges) if 
\[
\lim_{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| =L<1 
\]
\end{quotation}

\medskip

To verify the convergence of $\sum_{n=1}^{\infty }\frac{n^{2}}{2^{n}}$ using
the ratio test, note the following

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\lim\limits_{n\rightarrow \infty }\dfrac{a_{n+1}}{a_{n}}=\allowbreak \dfrac{%
1}{2}$
\end{quotation}

\strut

Thus, $L=1/2$, which is less than 1, so the series converges absolutely.

\subsubsection{Root Test\label{Root test}}

\begin{quotation}
The \emph{root test} states 
\index{Series!root test} that a series $\sum_{n=1}^{\infty }a_{n}$ converges
absolutely (and therefore converges) if 
\[
\lim_{n\rightarrow \infty }\sqrt[n]{\left| a_{n}\right| }=L<1 
\]
\end{quotation}

\strut

To verify convergence of $\sum_{n=1}^{\infty }%
\frac{n^{2}}{2^{n}}$ using the root test, note the following.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\lim\limits_{n\rightarrow \infty }\sqrt[n]{\left| a_{n}\right| }%
=\allowbreak \dfrac{1}{2}$
\end{quotation}

Thus, $L=1/2$, which is less than 1, again showing that the series converges
absolutely.

\subsubsection{Integral Test\label{Integral test}}

\begin{quotation}
The \emph{integral test} states that a series $\sum_{n=1}^{\infty }a_{n}$
converges absolutely if these exists a positive decreasing function $f$ such
that $f(n)=\left| a_{n}\right| $ for each positive integer $n$ and 
\[
\int_{1}^{\infty }f(x)dx<\infty 
\]
\end{quotation}

\medskip

To verify convergence of $\sum_{n=1}^{\infty }\frac{n^{2}}{2^{n}}$ using the
integral test, define $f$ by $f(x)=\frac{x^{2}}{2^{x}}$ and note the
following.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate, Evaluate Numerically}
\end{quote}

\begin{quotation}
$\dint_{1}^{\infty }\dfrac{x^{2}}{2^{x}}dx=\allowbreak \dfrac{1}{2}\dfrac{%
2+2\ln 2+\ln ^{2}2}{\ln ^{3}2}=\allowbreak 5.805497209$
\end{quotation}

\strut

Thus, this integral is finite. (Although for $f(x)=\frac{x^{2}}{2^{x}}$, it
is true that $f(1)<f(2)<f(3)$, you can verify that $f$ is decreasing for $%
x>3.$ In fact, 
\[
f^{\prime }(x)=\allowbreak 2\frac{x}{2^{x}}-\frac{x^{2}}{2^{x}}\ln 2\medskip 
\]
is positive only on the interval $0<x<\frac{2}{\ln 2}=\allowbreak
2.885390082,$ so $f$ is decreasing on $3<x<\infty $. Since convergence of a
series depends on the tail end of the series only, it is sufficient that the
sequence of terms be eventually decreasing.)

\subsubsection{Maclaurin Series\label{Maclaurin series}\label{Power series}}

\begin{quotation}
The \emph{Maclaurin series } 
\index{Series!Maclaurin}of a function $f$ is the series 
\[
\sum_{n=0}^{\infty }%
\frac{f^{(n)}(0)}{n!}x^{n} 
\]
where $f^{(n)}(0)$ indicates the $n$th derivative of $f$ evaluated at 0.
\end{quotation}

\strut

To expand a function $f(x)$ in a Maclaurin series (power series about $x=0$%
), choose 
\index{Power series} \textsf{Power Series}, specify desired \textsf{Number
of Terms}, and then specify \textsf{Expand in Powers of} $x.$ With $f(x)=%
\frac{\sin x}{x}$ and 10 terms, the result is as follows.

\begin{quote}
$\blacktriangleright $ \textsf{Power Series}
\end{quote}

\begin{quotation}
$\dfrac{\sin x}{x}=\allowbreak 1-\frac{1}{6}x^{2}+\frac{1}{120}x^{4}-\frac{1%
}{5040}x^{6}+\frac{1}{362880}x^{8}+O\left( x^{9}\right) $
\end{quotation}

\strut

The $O\left( x^{9}\right) $ term indicates that all the remaining terms in
the series contain at least $x^{9}$ as a factor. (In fact, the truncation
error is of order $x^{10}$ in this case.) The odd powers of $x$ have
coefficients of 0.

\textsf{Plot 2D} \label{Plotting series}provides an excellent visual
comparison between a function and an approximating polynomial.

\begin{quote}
$\blacktriangleright $ \textsf{Plot 2D + Rectangular}
\end{quote}

\begin{quotation}
$\dfrac{\sin x}{x},1-\frac{1}{6}x^{2}+\frac{1}{120}x^{4}$
\end{quotation}

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45;theta 45;plottype 4;numpoints 49;axesstyle "normal";xis
\TEXUX{x};var1name \TEXUX{$x$};}}

To determine which graph corresponds to which equation, evaluate one of the
expressions where the graphs show some separation. For example, $\frac{\sin 4%
}{4}=\allowbreak -.1892006238$, and hence the graph of $\frac{\sin x}{x}$ is
the one that is negative at $x=4$.

The following are additional examples of Maclaurin series expansions.

\begin{quote}
$\blacktriangleright $ \textsf{Power Series}
\end{quote}

\begin{quotation}
$\ln (1-x)=\allowbreak -x-\frac{1}{2}x^{2}-\frac{1}{3}x^{3}-\frac{1}{4}x^{4}-%
\frac{1}{5}x^{5}+O\left( x^{6}\right) $

$\tan ^{-1}x=\allowbreak x-\frac{1}{3}x^{3}+\frac{1}{5}x^{5}-\frac{1}{7}%
x^{7}+\frac{1}{9}x^{9}+O\left( x^{10}\right) $

$\sin ^{2}x=\allowbreak x^{2}-\frac{1}{3}x^{4}+\frac{2}{45}x^{6}-\frac{1}{315%
}x^{8}+O\left( x^{10}\right) $

$e^{x}=\allowbreak 1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{24}x^{4}+%
\frac{1}{120}x^{5}+\frac{1}{720}x^{6}+O\left( x^{7}\right) $

$\sin x=\allowbreak x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}-\frac{1}{5040}%
x^{7}+\frac{1}{362880}x^{9}+O\left( x^{10}\right) $

$e^{x}\sin x=\allowbreak x+x^{2}+\frac{1}{3}x^{3}-\frac{1}{30}x^{5}-\frac{1}{%
90}x^{6}+O\left( x^{7}\right) $
\end{quotation}

\strut

Remember that output can be copied and pasted (with ordinary word-processing
tools) to create input for further calculations. In particular, select and
delete the $+O\left( x^{n}\right) $ expression to convert the series into a
polynomial. It is reassuring to note that, if the first few terms of the
Maclaurin series for $e^{x}$ are multiplied by the first few terms of the
Maclaurin series for $\sin x,$ then the result is the same as the first few
terms of the Maclaurin series for $e^{x}\sin x$.

\begin{quote}
$\blacktriangleright $ \textsf{Expand, Polynomials + Sort}
\end{quote}

\begin{quotation}
$\left( 1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{24}x^{4}+\frac{1}{120}%
x^{5}\right) \left( \allowbreak x-\frac{1}{6}x^{3}+\frac{1}{120}x^{5}\right) 
$

$=\allowbreak x+\frac{1}{3}x^{3}-\frac{1}{30}x^{5}+x^{2}-\frac{1}{90}x^{6}-%
\frac{1}{360}x^{7}+\frac{1}{2880}x^{9}+\allowbreak \frac{1}{14400}x^{10}$

$=\allowbreak \frac{1}{14400}x^{10}+\frac{1}{2880}x^{9}-\frac{1}{360}x^{7}-%
\frac{1}{90}x^{6}-\frac{1}{30}x^{5}+\allowbreak \frac{1}{3}x^{3}+x^{2}+x$
\end{quotation}

\strut

\subsubsection{Taylor Series\label{Taylor series}}

\begin{quotation}
The Maclaurin series is a special case of the more general Taylor series.
The \emph{Taylor series} of $f$ expanded 
\index{Taylor series} 
\index{Series!Taylor}about $x=a$ is given by 
\[
\sum_{n=0}^{\infty }%
\frac{f^{(n)}(a)}{n!}(x-a)^{n} 
\]
and hence is expanded in powers of $x-a$.
\end{quotation}

\strut

For Taylor series, enter the number of terms and the point of expansion in
the \textsf{Series} dialog box. To find the Taylor series of $\ln x$
expanded about $x=1$, choose \textsf{Powers Series}. In the dialog box,
select the desired number of terms and expand about the point $x-1$.

\begin{quote}
$\blacktriangleright $ \textsf{Power Series}
\end{quote}

\begin{quotation}
$\ln x=\allowbreak \left( x-1\right) -\frac{1}{2}\left( x-1\right) ^{2}+%
\frac{1}{3}\left( x-1\right) ^{3}-\frac{1}{4}\left( x-1\right) ^{4}+O\left(
\left( x-1\right) ^{5}\right) $
\end{quotation}

\strut

A comparison between $\ln x$ and the polynomial $\allowbreak \left(
x-1\right) -\frac{1}{2}\left( x-1\right) ^{2}+\frac{1}{3}\left( x-1\right)
^{3}-\frac{1}{4}\left( x-1\right) ^{4}$ is illustrated graphically in the
following figure.

\begin{quote}
$\blacktriangleright $ \textsf{Plot 2D + Rectangular}
\end{quote}

\begin{quotation}
$\ln x,\left( x-1\right) -\frac{1}{2}\left( x-1\right) ^{2}+\frac{1}{3}%
\left( x-1\right) ^{3}-\frac{1}{4}\left( x-1\right) ^{4}$
\end{quotation}

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Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
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45;plottype 4;numpoints 49;axesstyle "normal";xis \TEXUX{x};var1name
\TEXUX{$x$};}}

You can produce the following power series expansions with \textsl{%
Scientific Notebook}.

\begin{quote}
$\blacktriangleright $ \textsf{Power Series}
\end{quote}

\begin{quotation}
$\frac{1}{x}=\allowbreak \frac{1}{2}-\frac{1}{4}\left( x-2\right) +\frac{1}{8%
}\left( x-2\right) ^{2}-\frac{1}{16}\left( x-2\right) ^{3}+\frac{1}{32}%
\left( x-2\right) ^{4}+O\left( \left( x-2\right) ^{5}\right) $

$\sin x=\allowbreak -\left( x-\pi \right) +\frac{1}{6}\left( x-\pi \right)
^{3}+O\left( \left( x-\pi \right) ^{5}\right) $

$\sqrt{x}=\allowbreak 1+\frac{1}{2}\left( x-1\right) -\frac{1}{8}\left(
x-1\right) ^{2}+\frac{1}{16}\left( x-1\right) ^{3}-\frac{5}{128}\left(
x-1\right) ^{4}+O\left( \left( x-1\right) ^{5}\right) $

$\csc x=\allowbreak 1+\frac{1}{2}\left( x-\frac{1}{2}\pi \right) ^{2}+\frac{5%
}{24}\left( x-\frac{1}{2}\pi \right) ^{4}+O\left( \left( x-\frac{1}{2}\pi
\right) ^{5}\right) $

$2\sin ^{2}x=\allowbreak 2\left( x-\pi \right) ^{2}-\frac{2}{3}\left( x-\pi
\right) ^{4}+O\left( \left( x-\pi \right) ^{6}\right) $

$1-\cos 2x=\allowbreak 2\left( x-\pi \right) ^{2}-\frac{2}{3}\left( x-\pi
\right) ^{4}+O\left( \left( x-\pi \right) ^{5}\right) $
\end{quotation}

\strut

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Limits}{}{}{DM7-2.tex#Limit}

\item  \hyperref{Differentiation}{}{}{DM7-3.tex#Derivative}

\item  \hyperref{Indefinite Integration}{}{}{DM7-4.tex#Indefinite integral}

\item  \hyperref{Definite Integrals}{}{}{DM7-6.tex#Definite integral}

\item  \hyperref{Plotting Curves and Surfaces}{}{}{DM6-1.tex#Plotting}

\item  \hyperref{2D Plots of Functions and Expressions}{}{}{DM6-4.tex#Plot
expression}

\item  \hyperref{3D Plots of Functions and Expressions}{}{}{DM6-6.tex#3D
Plots}

\item  \hyperref{Exercises and Solutions}{}{}{DM7-9.tex#Exercises}
\end{itemize}

\rule{0.67in}{0.01in}

\hyperref{\FRAME{itbpF}{38.875pt}{24.5pt}{6pt}{}{}{index.wmf}{\special%
{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid-file "F";width 38.875pt;height 24.5pt;depth
6pt;original-width 36.125pt;original-height 22.5625pt;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'index.wmf';file-properties
"XNPEU";}}}{}{}{DMIndex.tex#Math Index}\quad {\small Index entries: 
\hyperref{Plotting 2D}{}{}{DMIndex.tex#Plotting 2D}, \hyperref{Plotting 3D}{%
}{}{DMIndex.tex#Plotting 3D}, \hyperref{Power series}{}{}{DMIndex.tex#Power
series}, \hyperref{Sequences}{}{}{DMIndex.tex#Sequences}, \hyperref{Series}{%
}{}{DMIndex.tex#Series}, \hyperref{Summation notation}{}{}{%
DMIndex.tex#Summation notation}}

\end{document}
