\chapter{Infinite Series}
 
 
As we saw in the preceding chapter, a smooth
function can be approximated by its
Taylor series expansion.  And our experience,
based on examples, is that the accuracy of such
approximations improves as the number of terms
increase.  The idea of approximation by 
finite series is a truncated version 
of the more general idea of representation by an infinite 
series.  In this chapter we shall consider the elementary
theory of infinite series in some detail and, in the 
later sections, discuss the representation of functions
by special \index{infinite series} infinite series called power series.
\vspace{.2in}

\section{Infinite Sequences}
 
At the intuitive level a {\em \index{sequence} sequence}
is an ordered succession of real numbers
such as
\[ 1,3,5,7 \hspace{.6in} 1,5,14,30,55 \]
and
\[ 1,2,3,...,n... \hspace{.4in} \mbox{or} \hspace{.4in} 
1^2 , ( 1/2 )^2 , ( 1/3 )^2 ,..., 
( 1/n )^2 ,... \]
The first two sequences stop; they 
are {\em \index{finite sequences} finite sequences}.  The second two
never stop; they are called {\em infinite sequences}.
Only infinite sequences are of interest to us and therefore
we shall use ``sequence'' to mean ``infinite sequence''.
The members of a sequence are called {\em \index{terms} terms} and they
are ordered in the sense that there is a
{\em \index{first term} first term}, a {\em \index{second term} second term}, 
a {\em third term}, and so forth.
It is customary to denote the nth term of a sequence by a 
subscripted symbol such as 
$ a_n , b_n , x_n $, or $ y_n $ and indicate the 
sequence by writing
\vspace{.1in}

\noindent
$ \mbox{(1)} 
\hspace{2in} 
a_1 , a_2 , a_3 ,..., a_n ... ~~ .  $
\vspace{.1in}

\noindent
The sequence (1) can be defined by the compact notation
$ \{ a_n \} $.  Thus $ \{ n^2 \} $ denotes the sequence 
\[ 1^2 , 2^2 , 3^2 ,..., n^2 ,..., \]
and $ \{ \sin \: n \pi / 2 \} $ denotes the sequence
\[ \sin \: \frac{ \pi }{ 2 }
, ~ \sin \: \frac{ 2 \pi }{ 2 }
, ~ \sin \: \frac{ 3 \pi }{ 2 } ,..., \frac{ n \pi }{ 2 } \]
In special situations we may find it convenient not to label
the first term of a sequence with the first counting number.  In
such instances we adopt the following notation:
\[ \{ a_n \} _{ n = 0 }^{ \infty } ~~ 
~~ \mbox{to mean $ a_0 , a_1 , a_2 ,..., a_n ,... $} \]
\[ \{ a_n \} _{ n = 2 } ^{ \infty }
~~ \mbox{to mean $ a_2 , a_3 , a_4 ,..., a_n ,... $} \]
and
\[ \{ a_n \} _{ n = k }^{ \infty } 
\mbox{to mean $ a_k , a_{k+1} , a_{k+2} ,..., a_n ,... $} \]

From the notation it is clear that a sequence is a 
special type of function in which the independent variable
ranges over an infinite block of integers.  This
observation prompts the following definition.
\vspace{.2in}

\noindent 
{\bf Definition}

A {\em \index{sequence} sequence} is a function whose domain
is the set of all integers greater than or equal
to some fixed integer $k$.  The value of
a sequence at $n$ is called the {\em nth term}.
\vspace{.1in}

Sequences are introduced in a variety of ways.
Usually the terms are computed from a formula for 
$ a_n $.  The following table illustrates this
situation for a number of simple examples:
\vspace{.1in}

\begin{tabular}{llclc} 
$ ~~ $ {\sf nth term} & \hspace{.5in} & {\sf first five terms} & \hspace{.5in} & {\sf 100th term} 
\\ [0.1in]
$ a_n = 2n + 1 $ && 3,5,7,9,11 && 201
\\ [0.2in]
$ a_n = 1 + ( - 1 )^n $ && 0,2,0,2,0 && 2 
\\ [0.2in]
$ a_n = {\displaystyle \frac{ ( - 1 )^n n^2 }{ 4n - 1 }} $
&& $ - 1/3 , 4/7 , - 9/11 , 16/15 , - 25/19 $ && 
$ 10,000/399 $ 
\\ [0.2in]
$ a_n = n / ( n + 1 ) $ && 
$ 1/2, \: 2/3, \: 3/4, \: 4/5, \: 5/6 $ && $ 100/101 $ 
\end{tabular}
\vspace{.4in}

\noindent  
In each case the terms are obtained by substituting the values
$ n = 1,2,3,4,5,100 $.  Of course, $ a_n $ need not
be an integer or a simple fraction.  And in many instances,
such as
\[ a_n = n \int_{ 0 }^n \: e^{ t^2 } \: d t ~~
\mbox{or} ~~ a_n = \sin \: ( e^{ \cos \: n } ) , \]
it may not be possible to calculate each term exactly.
Occasionally $ a_n $ is not given by a single formula, as in
the rule
\[ a_n = \left \{ \begin{array}{ll}
n^2 & \mbox{if $n$ is even} 
\\ [0.2in]
\pi & \mbox{if $n$ is odd} 
\end{array} \right. \]
which defines the sequence 
\[ \pi , 4 , \pi , 16 , \pi , 36 ,... \]
And there is the possibility of not being able
to write any kind of formula for $ a_n $.
This is certainly the case for the sequence 
$ \{ a_n \} $ in which $ a_n $ is the nth digit in the 
decimal expansion of the irrational number $ \sqrt{ 2 } $.
Finally, it may be that each term of a sequence 
is defined in terms of its predecessors as, 
for example,
\[ a_{ n + 1 } = a_n + \frac{ 1 }{ a_n } , ~~
a_1 = 1 . \]
The first few terms are
\[ a_1 = 1 , ~ a_2 = 2 , ~
a_3 = 5/2 , ~ a_4 = 20/10 , ~ a_5 = 
941/290 , ~ a_6 = 969,581/272,890 \]
Given in this manner, a sequence is said to be defined
{\em \index{recursively} recursively}.  Newton's formula
$ x_{ n + 1 } = x_n 
- f ( x_n ) / f ^{'} ( x_n ) , ~~ n = 0,1,2 ,..., $ for
finding a root of $ f(x) = 0 $ is a recursively defined
sequence.  For example, using this formula to find the root
of $ x^3 - 4 = 0 $ leads to the recursive sequence 
$ x_{ n + 1 } = \frac{ 2 }{ 3 }
[ x_n + 2 / x_n^2 ] , ~~ n = 0,1,2 ,... $ whose
terms, starting with, say, $ x_0 = 2 $, will approximate 
the cube root of 4.

We are interested in studying sequences $ \{ a_n \} $ with 
the property that the nth term $ a_n $ approaches some
real number as $n$ increases.  If, for example, 
\[ a_n = 1 - \frac{ 1 }{ n } , \]
then $ a_n $ is clearly near 1 for
large $n$,  A  fact we indicate by writing
\[ a_n \rightarrow 1 ~~ \mbox{as $ n \rightarrow \infty $
or $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
a_n = 1 $} \]
or, in words, we say that the sequence 
$ \{ 1 - 1 / n \} $ has limit 1.
More formally, we have the following definition.
\vspace{.1in}

\noindent 
{\bf Definition}

A sequence $ \{ a_n \} $ has the number $L$ as a {\em \index{limit} limit},
written
\vspace{.1in}

\noindent
$ \mbox{(2)} 
\hspace{1.5in}  
{\displaystyle \lim_{ n \rightarrow \infty }} \:  
a_n = L ~~ \mbox{or} ~~ a_n \rightarrow L 
~ \mbox{as} ~ n \rightarrow \infty , $
\vspace{.1in}

\noindent
if, for every positive number $ \epsilon $, there exists
a positive integer $N$ such that 

\noindent
$ \mbox{(3)} 
\hspace{2in} | a_n - L | < \epsilon ~~ \mbox{if} ~~
n \geq N . $
\vspace{.1in}

\noindent
Whenever a sequence has a limit, we say that the sequence 
is {\em \index{convergent sequence} convergent} and that it 
{\em \index{converges to the limit} converges to the limit} $L$.  A sequence that fails
to have a limit is termed {\em \index{divergent} divergent}.

You should think of the above definition
as a technical way of saying that as $n$ increases
the terms $ a_n $ get close and {\em \index{stay close} stay close}
to some number $L$.  One way to obtain a visual image is to
associate the numbers $ a_n $ and $L$ with points on 
a coordinate line.
% as shown in Figure 14.1.  
According to the 
definition of limit, we may choose any positive number $ \epsilon $
we please $ - 1 , ~ 1/2 $, or $ 10^{ - 10 } $,
it doesn't matter -- and be certain that there is some
index $N$ such that each term $ a_n $
with index $ n \geq N $ lies between $ L - \epsilon $ and 
$ L + \epsilon $.  In effect, then, once
%\vspace{2in} 

\noindent 
a segment is drawn with center at $L$, only a finite number 
of the terms are outside that segment.  From this
visualization, that the terms of a sequence with limit 
$L$ must cluster about $L$, it is clear that a sequence 
cannot have two limits.
\vspace{.1in}

\noindent 
{\bf Example 1}.  
Show that $ {\displaystyle \lim_{ n \rightarrow \infty }} \: 
{\displaystyle \frac{ n + 1 }{ n }} = 1 $.
\vspace{.1in}

\noindent 
{\bf Solution}.
Here $ a_n = {\displaystyle \frac{ n + 1 }{ n }} , ~~ L = 1 $ 
and $ a_n - L = a_n - 1 = {\displaystyle \frac{ n + 1 }{n}}
- 1 = \frac{ 1 }{ n } $.

\noindent 
At the intuitive level, it is clear that $ a_n \rightarrow 1 $
as $ n \rightarrow \infty $ because
$ a_n - 1 = \frac{ 1 }{ n } \rightarrow 0 $
as $ n \rightarrow \infty $.  More technically, we must follow
the procedures set forth in the definition of a limit.
Let $ \epsilon $ be a positive number and pose
the following question:  How large must we choose
$n$ in order that
\[ | a_n - 1 | = \frac{ 1 }{ n } < \epsilon \mbox{?} \]
Obviously, $ \frac{ 1 }{ n } < \epsilon $ if
$ \frac{ 1 }{ \epsilon } < n $.  So if we let $N$
be the smallest integer such that $ \frac{ 1 }{ \epsilon } 
< N $, then $ n \geq N $ implies that $ n > 
\frac{ 1 }{ \epsilon } $ and, hence, that
\[ | a_n - 1 | < \epsilon . \]
This analysis amounts to a strategy for finding
$N$ once $ \epsilon $ is chosen.  For example,
if $ \epsilon = 0.5 , ~~ \frac{ 1 }{ \epsilon } = 2 $ and
we choose $ N = 3 $; if $ \epsilon = 0.025 , ~~
\frac{ 1 }{ \epsilon } 
= 40 $ and $ N = 41 $; and if $ \epsilon = 0.00001 , ~~
\frac{ 1 }{ \epsilon } = 100,000 $ and $ N = 100,001 $.  Two
things are obvious:  first, the choice of $N$
depends on the choice of $ \epsilon $ and, second, making
$ \epsilon $ small forces $N$ to be large.  $ \ddagger $
\vspace{.1in}

\noindent
{\bf Example 2}.  
\vspace{.1in} 
Show that the sequence $ a_n $ with 
$ a_n = ( - 1 )^n ( 1 + 1/n ) $ is divergent.  
\vspace{.1in}

\noindent 
{\bf Solution}.
Since $ ( - 1 )^n $ 
is $ - 1 $ if $n$ is odd and $1$ is odd and $1$
if $n$ is even, the terms $ a_n $ oscillate
from slightly larger than $1$ for even $n$ to slightly less than
%\vspace{2in} 

\noindent 
$ - 1 $ for odd $n$.  The terms of this sequence 
clearly can't get close and stay
close to any number.  In fact any two consecutive terms
differ by more than $2$ in absolute value.  % See Figure 14.2.
$ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 3}.  
Show that $ {\displaystyle \lim_{ n \rightarrow \infty }} \: 
{\displaystyle \frac{ 3n^2 - n + 1 }{ 2n^2 + 5n }} 
= 3/2 $.
\vspace{.1in} 

\noindent 
{\bf Solution}.
The nth term of this sequence 
is $ a_n = {\displaystyle \frac{ 3n^2 - n + 1 }{ 2n^2 
+ 5n }} $ and at the intuitive level it is
fairly clear that the limit is $ 3/2 $.  For if
we divide both numerator and denominator by $ n^2 $,
we have
\[ a_n = \frac{ 3 - 1/n + 1/n^2 }{ 2 + 5/n } \]
and since $ 3 - 1/n + 1/n^2 \rightarrow 3 $
and $ 2 + 5/n \rightarrow 2 $ as $ n \rightarrow \infty $,
it follows that
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  a_n =
{\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ 3 - 1/n + 1/n^2 }{ 2 + 5/n } \]
and since $ 3 - 1/n + 1/n^2 \rightarrow 3 $ and $ 2 + 
5/n \rightarrow 2 $ as $ n \rightarrow \infty $, it
follows that
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  a_n = 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  \frac{ 3 -
1/n + 1/n^2 }{ 2 + 5/n } = \frac{ 3 }{ 2 } . \]

To follow a more rigorous approach, we examine the
difference
\[ a_n - \frac{ 3 }{ 2 } = \frac{ 3n^2 - n + 1 }{ 2n^2 + 5n } 
- \frac{ 3 }{ 2 } = \frac{ - 17n + 2 }{ 2 ( 2n^2 
+ 5n ) } \]
and show that the expression on the right tends to $0$
as $ n \rightarrow \infty $.  First, if $ 2 \leq n $,
\[ \begin{array}{rl}
\left | a_n - 
{\displaystyle \frac{ 3 }{ 2 }} \right | & = \left |
{\displaystyle \frac{ - 17n + 2 }{ | 4n^2 + 10 n | }} 
\right | = {\displaystyle \frac{ | - 17n + 2 | }{ | 4n^2
+ 10n | }} \leq {\displaystyle \frac{ | - 17n | + 2 }{ | 4n^2
+ 10n | }} \\ [0.3in]
& = {\displaystyle \frac{ 17n + 2 }{ 4n^2 + 10n }} \leq  
{\displaystyle \frac{ 17n + n }{ 4 n^2 + 0 }} 
= {\displaystyle \frac{ 18n }{ 4n^2 }} =  
{\displaystyle \frac{ 9 }{ 2n }} . \end{array} \] 
Given a positive number $ \epsilon $, we want to be certain than 
$ 9/2n < \epsilon $ for $n$ sufficiently large.  Following
the technical of Example 1, we choose $N$
as the smallest integer such that $ 9/2 \epsilon < N $.
Then, if $ n \geq N $ if follows that
$ 9/2n \leq 9/2N \leq 9/2N < \epsilon $
and hence that
\[ | a_n - \frac{ 3 }{ 2 } | < \epsilon . \]
Again, we have a rule that computes $N$ in terms
of $ \epsilon $.  For example, if $ \epsilon = 0.0017 $,
$ 9/2 \epsilon = 2647.05... $
and $ N = 2648 $. $ \ddagger $

This example illustrates both the intuitive approach and a
strict application of the limit definition.
In practice, we shall not insist on a strict or rigorous approach,
rather preferring to attack limit problems with a blend
of common sense and calculus.

If $f$ is a function whose domain includes the integer
$k$ and all $ x \geq k $, then we can define a sequence 
by the formula 
\[ a_n = f (n) , ~~ n = k , k + 1 , k + 2 ,... ~ . \]
and if it turns out that $f$ has a horizontal asymptote,
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  f(x) 
= L , \]
then the sequence must do the same, that is, we must also have
$ {\displaystyle \lim_{ n \rightarrow \infty }} \:  a_n = L $.
Thus, from the function $ f(x) = ( 1 + 2/x )^x $
comes the sequence 
$ a_n = ( 1 + 2/n )^n $.  This correspondence together
with our previous work on indeterminate forms provides
a useful method for finding limits.
\vspace{.1in}

\noindent 
{\bf Example 4}.  
Find $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
{\displaystyle \frac{ \ln \: n }{ n }} $.
\vspace{.1in} 

\noindent 
{\bf Solution}.
The sequence $ a_n = 
{\displaystyle \frac{ \ln \: n }{ n }} $ is associated
with $ f(x) = {\displaystyle \frac{ \ln \: x }{ x }} $,
a function that assumes the indeterminate 
form $ \infty / \infty $ as $ x 
\rightarrow \infty $.  Hence we may apply
L'Hospital's rule to obtain 
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  \frac{
l^ x }{ x } 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ 1/x }{ 1 } = 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  \frac{ 1 }{x}
= 0 . \]
Therefore,
\[ \hspace{1in} {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ \ln \: n }{ n } = 0 . \hspace{1in} \ddagger  \]
\vspace{.1in} 

As we saw in Example 2, a sequence may diverge
because its terms oscillate, never clustering
around a fixed value.  A sequence also may diverge 
because its terms increase or decrease without bound.
This is certainly the case for the sequences
\[ \begin{array}{l} 1^2 , 2^2 , 3^2 ,..., n^2 ,..., 
\\ [0.1in]
\ln \: 1 , ~ \ln \: 2 , ~ \ln \: 3 ,..., \ln \: n ,... ~ ; 
\end{array} \]
situations which we describe by writing
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  n^2 =
\infty , ~~ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\ln \: n = \infty , ~~ {\displaystyle \lim_{ n \rightarrow \infty }} 
\:  - \sqrt{n} = - \infty \]
In general we write

\noindent
$ \mbox{(4)} \hspace{2in} 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  a_n = \infty $
\vspace{.1in}

\noindent
to mean that given a number $M$ there is an integer $N$ such that 
\[ a_n > M ~~ \mbox{for} ~~ n \geq N . \]
A similar meaning is attached to

\noindent
$ \mbox{(5)} \hspace{2in} 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  a_n = -
\infty . $
\vspace{.1in}

\noindent
Such sequences are divergent; they have no limit.
Nonetheless, we describe (4) by saying that 
$ \{ a_n \} $ diverges to $ \infty $; the sequence in
(5) diverges to $ - \infty $.
It is worth noting that the reciprocal of a sequence 
that diverges to $ \infty $ or $ - \infty $ is a
convergent sequence with limit 0.  Thus,
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  \ln \: n
= \infty ~~ \mbox{but} ~~
{\displaystyle \lim_{ n \rightarrow \infty }} \:
\frac{ 1 }{ \ln \: n } = 0 \]
and
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  n^3 = \infty
~~ \mbox{but} ~~
{\displaystyle \lim_{ n \rightarrow \infty }} \:
\frac{ 1 }{ n^3 } = 0 . \]
The following theorem catalogues some useful limits.
\vspace{.1in}

\noindent 
{\bf Theorem 1}.

\noindent  
(6) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  r^n
= 0 $ if $ | r | < 1 $.
\vspace{.1in} 

\noindent  
(7) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:
{\displaystyle \frac{ 1 }{ n^k }} = 0 ~~ ( k > 0 ) $.
\vspace{.1in}

\noindent  
(8) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
a^{ 1/n } = 1 ~~ ( a > 0 ) $.
\vspace{.1in}

\noindent  
(9) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
n^{ 1/n } = 1 $.
\vspace{.1in}

\noindent  
(10) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left ( 1 + \frac{ x }{ n } \right )^n 
= e^x $ (all $x$) 
\vspace{.1in}

\noindent 
{\bf Proof}.

\noindent  
(6) We want to show that $ | r^n - 0 | 
= | r^n | = | r | ^n $ is small 
for large $n$.  Since $ | r | < 1 , ~
1/ | r | > 1 $ so $ 1 / | r | = 1 + h $ 
for some positive $h$.  An induction argument based on 
\[ \begin{array}{rl}( 1 + h )^2 & \hspace{-.1in} = 1 + 2h + 
h^2 > 1 + 2 h , 
\\ [0.2in] 
(1+h) ^3 & \hspace{-.1in} = (1+h) (1+h) ^2 > 
(1+h) ( 1 + 2h ) 
\\ [0.2in] 
& \hspace{-.1in} = 1 = 3h + 2h^2 > 1 + 3h , 
\\ [0.2in] 
(1+h) ^4 & \hspace{-.1in} = (1+h) (1+h) ^3 
> (1+h) ( 1 + 3h ) = 
1 + 4h + 3 h^2 > 1 + 4h , \end{array} \]
shows that $ (1+h) ^n > 1 + n h $ for 
every positive integer $n$.  Hence
\[ | r |^n = \frac{ 1 }{ (1+h) ^n } 
< \frac{ 1 }{ 1 + n h } 
< \frac{ 1 }{ n h } . \]
But $h$ is fixed and $ 1/n \rightarrow 0 $ as $ n \rightarrow
\infty $.  Therefore
$ 1 / n h \rightarrow 0 $ and, in turn, 
$ | r | ^n \rightarrow 0 $ as $ n \rightarrow \infty $.
\vspace{.1in}

\noindent  
(7) By the definition of exponents, $ n^k = 
e^{ k \ln \: n } $.  Now, if $ k > 0 $,
$ k \: \ln \: n \rightarrow \infty $ as $ n \rightarrow \infty
$ and $ e^x \rightarrow \infty $ as $ x \rightarrow \infty $.
Hence $ n^k \rightarrow \infty $ as $ n \rightarrow \infty $
and, therefore, its
reciprocal $ 1 / n^k $ must tend to 0.
\vspace{.1in}

\noindent  
(8) Let $ b_n = a^{ 1/n } , ~~
a > 0 $.  Then $ \ln \: b_n 
= \frac{ 1 }{ n } \: \ln \: a \rightarrow 0 $ 
as $ n \rightarrow \infty $.  Hence 
\[ b_n = e^{ \ln \: b_n } \rightarrow 
e^0 = 1 ~~ \mbox{as} ~~ n \rightarrow \infty . \]
Here we have used the fact that the function $ e^x $
is continuous at $ x = 0 $.
\vspace{.1in}

\noindent  
(9) Let $ a_n = n^{ 1/n } $.  Then $ \ln \: 
a_n = \frac{ 1 }{ n } \: \ln \: n \rightarrow 0 $ as 
$ n \rightarrow \infty $ by Example 4.
Hence 
\[ a_n = e^{ \ln \: a_n } 
\rightarrow e^0 = 1 ~~ \mbox{as} ~~ n \rightarrow f .  \]

\noindent  
(10) Fix $x$ and consider $ f (t) = 
\left ( 1 + \frac{ x }{ t } \right )^t $
as $ t \rightarrow \infty $.  Since
$ y = \ln \: f(t) = t \: \ln \: 
\left ( 1 + \frac{ x }{ t } \right ) $ has the indeterminate
form $ \infty \cdot 0 $ as 
$ t \rightarrow \infty $, we may apply 
L'Hospital's rule to see that
\[ {\displaystyle \lim_{ t \rightarrow \infty }} \:  
y = {\displaystyle \lim_{ t \rightarrow \infty }} \:  
\frac{ \ln \: \left ( 1 + \frac{ x }{ t } \right ) }{ \frac{ 1 }{t} }
= {\displaystyle \lim_{ t \rightarrow \infty }} \:  
\frac{ \frac{ 1 }{ 1 + \frac{ x }{ t } } \cdot
\frac{ - x }{ t^2 }}{ \frac{ - 1 }{ t^2 } }
= {\displaystyle \lim_{ t \rightarrow \infty }} \:  
\frac{ x }{ 1 + \frac{ x }{ t } } = x \]
and hence that $ f(t) = e^y \rightarrow 
e^x $ as $ t \rightarrow \infty $.
The result now follows in view of the 
correspondence $ \left ( 1 + \frac{ x }{ n } \right )^n
= f (n) $.  $ \ddagger $

Sometimes it is possible to determine the 
limit of a sequence by comparing its terms with the terms
of sequences having known limits.
\vspace{.1in}

\noindent 
{\bf Theorem 2}.  {\em \index{Pinching Principle for Sequences} Pinching Principle for Sequences}

If the sequences $ \{ a_n \} $ and $ \{ b_n \} $
have a common limit $L$ and 
$ \{ b_n \} $
is a sequence such that

\noindent
(11) $ \hspace{2in} \hspace{2in} 
a_n \leq b_n \leq c_n , $
\vspace{.1in}

\noindent
then $ \{ b_n \} $ also has limit $L$.
\vspace{.1in}

\noindent 
{\bf Proof}.  Choose a positive number $ \epsilon $ and draw an 
open interval about $L$ running from $ L - \epsilon $
to $ L + \epsilon $.  Since both $ \{ a_n \} $
and $ \{ c_n \} $ converge to $L$, it
%\vspace{2in}
is possible to find an $N$ such that $ a_n $ and $ c_n $
belong to $ ( L - \epsilon , ~ L + \epsilon ) $
for $ n \geq N $.  But $ b_n $ is {\em between}
$ a_n $ and $ c_n $.  Therefore,
% as we see from Figure 14.3, 
\[ | b_n - L | < \epsilon ~~ \mbox{for} ~~ n \geq N . \]
This proves that $ {\displaystyle \lim_{ n \rightarrow \infty
}} \:  b_n = L $.  $ \ddagger $
\vspace{.1in}

This result is especially useful in the case 
$ a_n = 0 $ for all $n$.  The assertion then amounts to saying
that

\noindent
$ \mbox{(12)} \hspace{2in}
\mbox{if $ 0 \leq b_n \leq c_n $} $ 
\vspace{.1in}

\noindent
and $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  c_n
= 0 $, then $ {\displaystyle \lim_{ n \rightarrow \infty }} \:
b_n = 0 .$
\vspace{.1in}


Computations with limits is sometimes aided by the
application of some almost self-evident rules.
We omit the formal proofs.
\vspace{.1in}

\noindent 
{\bf Theorem 3}.  {\em If $ \{ a_n \} $ 
has limit $A$ and $ \{ b_n \} $ has limit $B$, then}

\noindent 
(13) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
( a_n + b_n ) = A + B $
\vspace{.1in}

\noindent 
(14) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
a_n b_n = A B $,
\vspace{.1in}

\noindent 
(15) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
k a_n = k A ~ $ ($ k = $ constant),  
\vspace{.1in}

\noindent
and 

\noindent 
(16) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
{\displaystyle \frac{ a_n }{ b_n }} = 
\frac{ A }{ B } $ if $ B \neq 
0 $, $ b_n \neq 0 $.
\vspace{.1in}


A straightforward application of (13), 
(15), and (16) is given by
\[ \begin{array}{rl}
{\displaystyle \lim_{ n \rightarrow \infty }} \:
{\displaystyle \frac{ 3n^2 - n + 1 }{ 2n^2 + 5 }} 
& \hspace{-.1in} = {\displaystyle \lim_{ n \rightarrow \infty }} \:
{\displaystyle \frac{ 3 - 1/n +
1/n^2 }{ 2 + 5/n^2 }} = 
{\displaystyle \frac{ {\displaystyle \lim_{ n \rightarrow
\infty }} \:  ( 3 - 1/n + 1 / n^2 }{ {\displaystyle \lim_{ n
\rightarrow \infty }} \:  
( 2 + 5 / n^2 }} \\ [0.3in]
& \hspace{-.1in} = {\displaystyle \frac{ {\displaystyle \lim_{ n
\rightarrow \infty }} \:  
3 - {\displaystyle \lim_{ n \rightarrow \infty }} \:  ( 1 / n
) + {\displaystyle \lim_{ n \rightarrow \infty }} \:  ( 1 / n^2
) }{ {\displaystyle \lim_{ n \rightarrow \infty }} \:  2 + 5
\: {\displaystyle \lim_{ n \rightarrow \infty }} \:  1/n^2 }}
=  {\displaystyle \frac{ 3 - 0 + 0 }{ 2 + 
5.0 }} = 3/2 . \end{array} \]
\vspace{.1in}

\noindent  
{\bf Theorem 4}.  If $ {\displaystyle \lim_{ n \rightarrow
\infty }} \:  a_n = L $ and $f$ is a continuous 
function, then 

\noindent
$ \mbox{(17)} \hspace{2in}
{\displaystyle \lim_{ n \rightarrow \infty }} \:  f ( a_n )
= f (L) . $
\vspace{.1in}

\noindent
Thus the composition of a continuous function $f$ and a
convergent sequence $ \{ a_n \} $ is a new sequence 
$ \{ f ( a_n ) \} $ whose limit is given by (17).
\vspace{.1in}


\noindent 
{\bf Example 5}.  
Find $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\cos \: \left ( {\displaystyle \frac{ 2n - 3 }{ 3n 
+ 4 }} \right ) $.
\vspace{.1in} 

\noindent 
{\bf Solution}.
Since $ \cos \: x $ is continuous and 
$ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
{\displaystyle \frac{ 2n - 3 }{ 3n + 4 } = 
\frac{ 2 }{ 3 }} $, it follows
from (17) that
\[ \hspace{.6in} {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\cos \: \left ( \frac{ 2n - 3 }{ 3n + 4 } \right )
= \cos \: 
\left ( {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ 2n - 3 }{ 3n + 4 } \right ) 
= \cos \: \left ( \frac{ 2 }{ 3 } \right ) . 
\hspace{.6in} \ddagger \]



\section{Monotonic and Bounded Sequences}    
\vspace{.1in}

In the preceding section sequences 
were shown to be convergent by actually finding the
limit.  A far more common occurrence is that
a given sequence converges, that is,
it has a limit, but it is quite impossible
to exhibit that limit.  Hence it is important to 
develop methods that guarantee the existence of a limit 
without actually locating it.  For the sequences
of concern to us in this section -- the monotonic 
sequences -- there is a simple convergence test.
\vspace{.1in}

\noindent 
{\bf Definition}.  Monotonic Sequence

\noindent  
A sequence $ \{ a_n \} $ is

{\em \index{increasing} increasing} 
if $ a_1 \leq 
a_2 \leq a_3 \leq \cdots \leq a_n \leq a _{n + 1 }
\leq \cdots $;

{\em \index{strictly increasing} strictly increasing} 
if $ a_1 < a_2 < a_3 < \cdots < \cdots 
< a_n < a_{ n + 1 } < \cdots $;

{\em \index{decreasing} decreasing} 
if $ a_1 \geq a_2 \geq a_3 \geq \cdots \geq a_n \geq a _{n + 1 } $

{\em \index{strictly decreasing} strictly decreasing}
if $ a_1 > a_2 > a_3 > \cdots > \cdots 
> a_n > a_{ n + 1 } > \cdots $.
\vspace{.1in}

A sequence is termed {\em \index{monotonic} monotonic} if any 
one of these conditions hold.
\vspace{.1in}

Technically speaking, the monotonicity
of a sequence cannot be determined except by examining
the formula for the nth term.  Nonetheless, a possible
permanent trend can often be detected by examining the first
few terms.
\vspace{.1in}

\noindent 
{\bf Example 1}.

(a) sequence:  $ 0 , ~ \frac{ 1 }{ 2 } , ~
\frac{ 2 }{ 3 } , ~
\frac{ 3 }{ 4 } , ~
\frac{ 4 }{ 5 } , ... $ type:  strictly increasing

(b) sequence:
$ 1 , ~ 1 , ~ 
\frac{ 3 }{ 2 } , ~
\frac{ 3 }{ 2 } , ~ 2 , ~ 2 , ~
\frac{ 5 }{ 4 } , ~ \frac{ 5 }{ 4 } , ... $ 
type:  increasing

(c) sequence:
$ \frac{ 2^2 }{ 2 ! } , ~ 
\frac{ 2^3 }{ 3 ! } , ~ 
\frac{ 2^4 }{ 4 ! } , ~ 
\frac{ 2^5 }{ 5 ! } , ~ 
\frac{ 2^6 }{ 6 ! } ,... $ 
type:  strictly decreasing

(d) sequence:  $ 0 , ~ 
- 1 , ~ - 1 , ~ - 2 , ~ - 2 , ~ - 3 , ~ - 3 , ~ - 4 ,... $ 
type:  decreasing \hspace{.6in} $ \ddagger $ 
\vspace{.1in}

\noindent 
Each of the sequences of this example are monotonic,
but the sequence 
\[ 1 , ~ 
\frac{ 1 }{ 2 } , ~ 2 , ~
\frac{ 1 }{ 3 } , ~ 3 , ~
\frac{ 1 }{ 4 } ,... \]
is not.
\vspace{.1in}

\noindent 
{\bf Example 2}.
Show that the following sequences are monotonic.

(a) $ a_n = \ln \: n $

(b) $ a_n = {\displaystyle \frac{ n - 1 }{ n + 1 }} $ 

(c) $ a_n = n^{ 1/n } $ with 
domain $ n = 3,4,5,... $

(d) $ a_{ n + 1 }
a_n + 1 / a_n , ~ a_1 = 1 $

(e) $ a_n = {\displaystyle \frac{ 1 }{ n + 1 } 
+ \frac{ 1 }{ n + 2 } + \frac{ 1 }{ n + 3 } + 
\cdots + \frac{ 1 }{ 2 n }} $

(f) $ a_n = {\displaystyle \frac{ n^2 }{ 2^n }} $
with domain $ n = 3,4,5,... $
\vspace{.1in}

\noindent 
{\bf Solutions}.
(a) $ \ln \: n < \ln \: ( n + 1 ) $ because 
$ \ln \: x $ has a positive derivative for all positive $x$.
\vspace{.1in}

\noindent 
(b) $ a_{ n + 1 } - a_n 
= {\displaystyle \frac{ n }{ n + 2 } - 
\frac{ n - 1 }{ n + 1 } = 
\frac{ 2 }{ ( n + 2 ) ( n + 1 ) }} > 0 $
for all $n$ so $ \{ a_n \} $ is strictly increasing.
\vspace{.1in}

\noindent 
(c) Write $ \ln \: a_n = \ln \: n^{ 1/n }
= {\displaystyle \frac{ \ln \: n }{ n }} $.
Let $ f(x) = {\displaystyle \frac{ \ln \: x }{ x }} $
and note that $ f ^{'} (x) 
= {\displaystyle \frac{ 1 - \ln \: x }{ x^2 }} < 0 $
if $ \ln \: x > 1 $ or if $ x > e $.
Hence $ {\displaystyle \frac{ \ln \: x }{ x }} $ is 
decreasing if $ x > e $.  Therefore 
$ \ln \: a_n > \ln \: a_{ n + 1 } $
if $ n \geq 3 $ or, finally,
$ a_n > a_{ n + 1 } $ if $ n \geq 3 $ because 
$ \ln \: x $ increases with $x$.  Consequently the
sequence $ \{ n^{ 1/n } \}_{ n = 3 }^{ \infty } $
is strictly decreasing.
\vspace{.1in}

\noindent 
(d) $ a_1 $ is positive,
and if $ a_n $ is positive, then $ a_{ n + 1 } 
= a_n + 1 / a_n > 0 $.
By induction every term is positive.  Hence 
$ a_{ n + 1 } = a_n 
+ 1 / a_n > a_n $ for all $n$.
The sequence is strictly increasing.
\vspace{.1in}

\noindent 
(e) The sequence is strictly increasing since
\[ a_{ n + 1 } - a_n = \frac{ 1 }{ 2n + 2 } 
+ \frac{ 1 }{ 2n + 1 } - \frac{ 1 }{ n + 1 }
= \frac{ - 1 }{ 2 ( n + 1 ) } + \frac{ 1 }{ 2 n + 1 } 
= \frac{ 1 }{ 2 ( n + 1 ) ( 2 n + 1 ) } > 0 . \]
\vspace{.1in}

\noindent 
(f) The sequence is strictly decreasing because 
\[ \frac{ a_n }{ a_{ n + 1 } } =
\frac{ n^2 }{ 2^n } \cdot 
\frac{ 2^{ n + 1 } }{ ( n + 1 )^2 } =
\frac{ 2 }{ ( 1 + 1 / n )^2 } > 1 \]
if $ n \geq 3 $, that is,
$ ( 1 + 1 / n )^2 < 2 $ if
$ n \geq 3 $. $ \ddagger $
\vspace{.1in}

The question of convergence for monotonic 
sequences is especially easy to answer.  For example,
the strictly increasing sequence $ \{ n^2 \} $ 
diverges to infinity:  its
terms do not lie below some fixed value.  On the other
hand the sequence $ \left \{ \frac{ n - 1 }{ n } \right \} $
is strictly increasing with limit $ 1 $:  its values
do lie below some fixed number, in fact, $ \frac{ n - 1 }{ n }
= 1 - \frac{ 1 }{ n } < 1 $ for all $n$.  The difference
is that one sequence is bounded above, the other is not.
The following definition formalizes this key observation.
\vspace{.1in}

%\newpage

\noindent 
{\bf Definition}.  Bounded Sequence 

A sequence $ \{ a_n \} $ is {\em \index{bounded above} bounded above} if there is a
constant $B$ such that 
\[ a_n \leq B ~~ \mbox{for all $n$;} \]
it is bound below if there is a constant $A$ such that 
\[ A \leq a_n  ~~ \mbox{for all $n$.} \]
A sequence that is both bounded above and 
below is called a {\em \index{bounded sequence} bounded sequence}.
\vspace{.1in}

Not every bounded sequence converges,
as one sees by examining the oscillatory
sequence $ \{ ( - 1 )^n \} $, but as the above
examples suggest, a bounded and monotonic 
sequence must converge.
\vspace{.1in}

\noindent 
{\bf Theorem 5}.  Every increasing sequence 
that is bounded above has a limit;
and every decreasing sequence that is bounded
below has a limit.

This theorem offers no hint as how to find
the limit; it merely asserts that under certain
conditions the limit will exist. 
% Figure 14.4 illustrates the situation for an increasing
% sequence $ \{ a_n \} $ that is bounded above by some 
% constant $B$.  
Once it is shown
% \vspace{2in} 

\noindent 
that $ a_n \leq B $ for all $n$, the theorem
tells us that $ {\displaystyle \lim_{ n \rightarrow \infty }}
\:  a_n = L $.  The bound $B$ need not
be the limit:  all we know
is that $ L \leq B $ and $ a_n \leq L $ for all $n$.
If $ \{ a_n \} $ is strictly increasing, then $ a_n < L $ for
all $n$. 

We shall regard Theorem 5 as being geometrically obvious.
However, it should be noted that its proof rests ultimately
on the Least Upper Bound Axiom, an assumption about
real numbers that undergirds all of calculus.
\vspace{.1in}

%\newpage

\noindent 
LEAST UPPER BOUND AXIOM

\begin{quote}
Every set of real numbers that is bounded above has a
{\em \index{least upper bound} least upper bound}; and every set of real 
numbers that is bounded below has a {\em \index{greatest
lower bound} greatest
lower bound}.
\end{quote}
\vspace{.1in}

For example, the number 2 is an upper bound for
the open interval $ 0 < x < 1 $,
so is $ \frac{ 3 }{ 2 } $; but 1 is at 
least upper bound. 
% As Figure 14.5 suggest, 
The limit $L$
of a bound and increasing sequence is the least number which
is an upper bound for all its terms.
\vspace{.1in}

\noindent 
{\bf Example 3}.
Use Theorem 5 to analyze the convergence 
of divergence of each sequence in Example 2. 
\vspace{.1in}

\noindent 
{\bf Solution}.
(a) The sequence $ \{ \ln \: n \} $ is divergent;
it is increasing but not bounded above.
\vspace{.1in}

\noindent  
(b) The sequence $ \left \{
\frac{ n - 1 }{ n + 1 } \right \} $ is increasing and
bounded above by 1, that is,
$ \frac{ n - 1 }{ n + 1 } \leq 1 $ for 
all $n$.  Hence the sequence has a limit,
by Theorem 5.
\vspace{.1in}

\noindent 
(c) $ \{ n^{ 1/n } \}_{ n = 3 }^{ \infty } $
is a decreasing sequence of positive numbers.
Hence 0 is a lower bound.  Therefore Theorem 5
applies and the sequence has a limit.
\vspace{.1in}

\noindent 
(d) The recursive sequence $ a_{ n + 1 } 
= a_n + 1 / a_n , ~~ a_1 = 1 $ is increasing
and, according to Problem Set 14.1,
$ a_{ n + 1 } > \sqrt{ 2n + 1 } $ for all $n$.
since $ \sqrt{ 2n + 1 } \rightarrow \infty $
as $ n \rightarrow \infty $, $ \{ a_n \} $ is not
bounded above and $ a_n \rightarrow \infty $ as
$ n \rightarrow \infty $.  There is no limit.
\vspace{.1in}

\noindent 
(e) The sequence is increasing, and bounded above because
\[ a_n = \frac{ 1 }{ n + 1 }
+ \frac{ 1 }{ n + 2 } + \frac{ 1 }{ n + 3 } 
+ \cdots + 
\frac{ 1 }{ 2n } < \frac{ n }{ n + 1 } < 1 \]
for all $n$.  By Theorem 5, the sequence has a limit.
\vspace{.1in}

\noindent 
(f) $ \left \{ \frac{ n^2 }{ 2^n } \right \}_{ n
= 3 }^{ \infty } $ is a decreasing sequence of 
positive numbers, hence 0 is a lower bound.
Again Theorem 5 tells us that the sequence 
has a limit.  $ \ddagger $
\vspace{.1in}

Whether a sequence $ \{ a_n \} $ has a limit depends on
the behavior of $ a_n $ for large $n$; terms
at the beginning have no effect.  In practice
this means you may ignore any initial block.  Put
technically, if $ \{ a_n \} _{ n = k }^{ \infty } $
has limit $L$, then $ \{ a_n \} _{ n = 1 }^{ \infty } $
will also have limit $L$.  Thus, in reference
to Theorem 5, a sequence that is eventually increasing
(increasing from some term on) and bounded above
has a limit; a sequence that is eventually decreasing
and bound below has a limit.
\vspace{.1in}

\noindent  
{\bf Example 4}.
Use Theorem 5 to establish that the sequence 
$ \left \{ \frac{ n^{ 10 }}{ e^n } \right \}_{ n = 1 }^{
\infty } $
has a limit.
\vspace{.1in}

\noindent 
{\bf Solution}.
The sequence $ a_n = n^{ 10 } / e^n $
increases through the tenth term 
$ a_{ 10 } $ and thereafter decreasing, as may
be seen by differentiating $ f(x) 
= x^{ 10 } / e^x $.  Hence Theorem 5 applies and a 
limit exists.  The limit is 0 
as may be seen by successive applications of L'Hospital's rule.
$ \ddagger $

 
\section{Convergent and Divergent Series, Partial Sums}    
\vspace{.1in}

The most familiar finite series is the {\em finite geometric series} 
\vspace{.1in}

\noindent 
(1) \hspace{2in} $ 1 + x + x^2 \cdots + x^{ n - 1 } $
\vspace{.1in}

\noindent 
whose sum $ s_n $ is given by 
\vspace{.1in}

\noindent 
(2) \hspace{2in} $ s_n 
= {\displaystyle \frac{ 1 - x^n }{ 1 - x } 
= \frac{ 1 }{ 1 - x } - \frac{ x^n }{ 1 - x }} $ if $ x \neq 1 $.
\vspace{.1in}

Of the two terms on the right in (2), one involves
$n$, the other does not; and in the second term,
if $ | x | < 1 , ~~ x^n \rightarrow 0 $
as $ n \rightarrow \infty $, by (6) of Theorem 1.
Hence 
\[ s_n \rightarrow \frac{ 1 }{ 1 - x } ~~
\mbox{as $ n \rightarrow \infty $ if $ | x | < 1 $.} \]
Or, saying the same thing,
\[ 1 + x + x^2 + \cdots + 
x^{ n - 1 } \rightarrow \frac{1}{ 1 - x} ~~ \mbox{as $ n \rightarrow \infty $ 
if $ | x | < 1 $.} \]
We describe this limit by writing an infinite series:
\[ 1 + x + x^2 + \cdots = 
\frac{ 1 }{ 1 - x } ~~
\mbox{if $ | x | < 1 $.} \]
The three dots are important; they serve to indicate
that the established pattern continues without end.
Thus, if $ x = 1/2 $,
\[ + \frac{1}{2} + \frac{ 1 }{ 2^2 } + 
+ \cdots = \frac{ 1 }{ 1 + \frac{1}{2} } = 2 , \]
meaning that 2 is the limit of the partial sums
\[ \begin{array}{rl}
s_1 & \hspace{-.1in} = 1 
\\ [0.2in]
s_2 & \hspace{-.1in} = 1 + \frac{1}{2} 
\\ [0.2in]
s_3 & \hspace{-.1in} = 1 + \frac{1}{2} + \frac{ 1 }{ 2^2 }
\\ & \hspace{-.1in} \vdots
\\ [0.1in]
s_n & \hspace{-.1in} = 1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + 
\frac{ 1 }{ 2^n } \\ & \hspace{-.1in} \vdots
\end{array} \]

This example illustrates the basic concept of an 
infinite series.  
\vspace{.1in}

\noindent
{\bf Definition}.  Any sequence 
$ \{ a_n \} _{ n = 1 }^{ \infty } $ of real
numbers can be used to form an {\em \index{infinite series} infinite series}
\vspace{.1in}

\noindent 
(3) \hspace{2in} $ a_1 + a_2 + a_3 + \cdots , $ 
\vspace{.1in}

\noindent 
which is said to be {\em \index{convergent series} convergent} if its
{\em \index{sequence of partial sums} sequence of partial sums}
\vspace{.1in}

\noindent 
(4) \hspace{2in} $ s_n = a_1 + a_2 + a_3 + \cdots
+ a_n $ 
\vspace{.1in}

\noindent 
converges to a limit $S$.  The limit $S$ is
called the {\em \index{sum of the infinite series} sum} of the infinite series (3) and we write
\vspace{.1in}

\noindent 
(5) \hspace{2in} $ S = a_1 + a_2 + a_3 + \cdots . $
\vspace{.1in}

\noindent 
In case the sequence of partial sums (4) fails to converge,
we say that the infinite series (3) is {\em \index{divergent series} divergent}.

The numbers $ a_n $ are the \index{terms of the
series} {\em terms} of the
series (3); $ a_1 $ is the first term,
$ a_2 $ is the second term, etc.  They can be
computed from the partial sums $ s_n $ by the formula
\[ a_1 = s_1 , ~ a_2 = s_2 - s_1 , ~
a_3 = s_3 - s_2 , \cdots . \]
For clarity, we shall often expand the notation 
in (3) to read
\[ a_1 + a_2 + a_3 + \cdots + \cdots , \]
especially if we wish to remove any doubt as to the
formula for $ a_n $.  And if brevity dictates, we shall adopt
the summation notation
\[ a_1 + a_2 + a_3 + \cdots + 
a_n + \cdots = \sum_{ n = 1 }^{ \infty } \: a_n . \]
In many problems it is convenient to begin 
indexing an infinite series at 0 instead of 1.  We then write
\[ a_0 + a_1 + a_2 + \cdots + 
a_n + \cdots = \sum_{ n = 0 }^{ \infty } \: a_n . \]
In this setting the partial sums are $ s_0 = 
a_0 , ~ s_1 = a_0 + a_1 $, etc.
\vspace{.1in}

\noindent 
{\bf Example 1}.
Find the sum $S$ of the series 
$ 1 + \frac{ 5 }{ 8 } + \left ( \frac{ 5 }{ 8 } \right )^2 +
\left ( \frac{ 5 }{ 8 } \right )^3 + \cdots $.
\vspace{.1in}

\noindent 
{\bf Solution}.
The partial sums are 
\[ \begin{array}{rl} s_1 & \hspace{-.1in} = 1 , 
\\ [0.2in]
s_2 & \hspace{-.1in} = 1 + \frac{ 5 }{ 8 } , \\
\vdots \\ [0.1in]
s_n & \hspace{-.1in} = 1 + \frac{ 5 }{ 8 }
+ \left ( \frac{ 5 }{ 8 } \right )^2 
+ \cdots + \left ( \frac{ 5 }{ 8 } \right )^n
= {\displaystyle \frac{ 1 - \left (
\frac{ 5 }{ 8 } \right )^{ n + 1 }}{ 1 - 
\frac{ 5 }{ 8 } }} \end{array} \]
and, since $ {\displaystyle \lim_{ n \rightarrow \infty }} \:
\left ( \frac{ 5 }{ 8 } \right )^{ n + 1 } = 0 $,
it follows that $ {\displaystyle \lim_{ n \rightarrow \infty
}} \:  s_n = 8/3 $.  Hence the series has sum
$ S = 8/3 $, that is,
\[ \frac{ 8 }{ 3 } = 1 + 
\frac{ 5 }{ 8 } + \left ( \frac{ 5 }{ 8 } \right )^2
+ \left ( \frac{ 5 }{ 8 } \right )^3 + \cdots . ~~~~ \ddagger \]
\vspace{.1in}

\noindent 
{\bf Example 2}.
Show that the geometric series 
$ 1 + x + x^2 + \cdots = 
{\displaystyle \sum_{ n = 0 }^{ \infty }} \:
x^n $ diverges if $ | x | \geq 1 $.
\vspace{.1in}

\noindent 
{\bf Solution}.
At $ x = 1 $ the geometric series reduces to
\[ 1 + 1 + 1 + \cdots , \]
a divergent series because its sequence of partial sums
\[ s_1 = 1 , ~ s_2 = 2 , ~ s_3 = 3 , \cdots , s_n = n \cdots \]
diverge to infinity.  At $ x = - 1 $, the 
geometric series reduces to
\[ 1 - 1 + 1 - 1 + \cdots , \]
and this series diverges because its partial sums
\[ s_1 = 1 , ~ s_2 = 0 , ~ s_3 = 1 , ~
s_4 = 0 , \cdots \]
oscillate between 0 and 1 and therefore fail to have
a limit.

If $ x > 1 $ the geometric series diverges because, by (2),
\[ s_n = \frac{ 1 - x^n }{ 1 - x }
= \frac{ x^n = 1 }{ x - 1 } \rightarrow \infty 
~~ \mbox{as $ n \rightarrow \infty $.} \]
And if $ x < - 1 $, the partial sums oscillate wildly
with $ s_{ 2n } \rightarrow \infty $ and $ s_{ 2n + 1 } 
\rightarrow - \infty $ as $ n \rightarrow \infty $.  $ \ddagger $
\vspace{.1in}

This example and our opening remarks establish the following
important theorem.
\vspace{.1in}

\noindent 
{\bf Theorem 6}.  (Geometric Series).  The geometric series 
\[ 1 + x + x^2 + \cdots = \sum_{ n = 0 }^{ \infty } \:
x^n \]
converges to $ {\displaystyle \frac{ 1 }{ 1 - x }} $
if $ | x | < 1 $; that is,
\vspace{.1in} 

\noindent 
(6) \hspace{2in}   
$ 1 + x + x^2 + \cdots = {\displaystyle \frac{ 1 }{ 1 - x }} $ if $ | x | < 1 $;
\vspace{.1in}

\noindent 
and the geometric series diverges if $ | x | \geq 1 $.

In (6) the number $x$ is called the {\em \index{common ratio} common ratio}.
\vspace{.1in}

\noindent 
{\bf Example 3}.
As a consequence of Theorem 6:
\vspace{.1in}

\noindent 
(a) $ 1 - \frac{ 2 }{ 3 } + \left (
\frac{ 2 }{ 3 } \right )^2
- \left ( \frac{ 2 }{ 3 } \right )^3 + \cdots
= \frac{ 1 }{ 1 - \left ( \frac{ 2 }{ 3 }  \right ) } 
= \frac{ 3 }{ 5 } $ because 
$ \left | - \frac{ 2 }{ 3 } \right | < 1 $;
\vspace{.1in}

\noindent 
(b) $ 1 + e^{ - 1 } + e^{ - 2 } + e^{ - 3 } 
+ \cdots = \frac{ 1 }{ 1 - \frac{ 1 }{ 3 } } = 
\frac{ e }{ e - 1 } $ because $ e^{ - 1 } < 1 $;
\vspace{.1in}

\noindent  
(c) $ 1 + e + e^2 + e^3 + \cdots $ diverges
because $ e > 1 $.  $ \ddagger $
\vspace{.1in}

One of the most illuminating of the many applications
of the geometric series is to the study of decimal expansions of
rational numbers.  The connection arises from the observation
that since $.9$ means $ 9/10 $ and $ .99 $
means $ 9/10 + 9/10^2 $, then the only
logical meaning for the {\em \index{infinite decimal} infinite decimal} $.999 ... $
is the infinite series $ 9/10 + 9/10^2 + 9/10^3 
+ \cdots + 9/10^n + \cdots $.  The sum of
this series is, as always, the limit of its
partial sums, of which the nth is
\[ \begin{array}{rl}
s_n & \hspace{-.1in} = 
\frac{ 9 }{ 10 } + 
\frac{ 9 }{ 10^2 }
+ \frac{ 9 }{ 10^3 }
+ \frac{ 9 }{ 10^n }
\\ [0.2in]
& \hspace{-.1in} = \frac{ 9 }{ 10 }
\left [ 1 + 
\frac{ 1 }{ 10 } +
\frac{ 1 }{ 10^2 }
+ \cdots + \frac{ 1 }{ 10^{n-1} } \right ]
\\ [0.2in]
& \hspace{-.1in} = \frac{ 9 }{ 10 } = 
\frac{ 1 - 1/10^n }{ 1 - 1/10 } = 
1 - 1/10^n . \end{array} \]
Since $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
s_n = 1 $, we conclude that 
\[ .999 ... = \frac{ 9 }{ 10 }
+ \frac{ 9 }{ 10^2 }
+ \frac{ 9 }{ 10^3 } + \cdots = 1 . \]

In general, if $ \{ a_n \} _{ n = 0 }^{ \infty } $
is a sequence of integers taken from 0
through 9, the infinite decimal $ .a_1 , a_2 a_3
\cdots $ denotes the infinite series 
\[ \sum_{ n = 1 }^{ \infty } \:
\frac{ a_n }{ 10^n }
= \frac{ a_1 }{ 10 } + 
\frac{ a_2 }{ 10^2 } + 
+ \frac{ a_3 }{ 10^3 } + 
+ \cdots + \frac{ a_n }{ 10^n } + \cdots \]

And if the $ a_n $ are periodic, in the sense
that a certain pattern is repeated, then the decimal represents 
a rational number which can be found by a procedure similar to 
that just illustrated.
\vspace{.1in}

\noindent 
{\bf Example 4}.
Find the rational number represented by $ .444 ... $.
\vspace{.1in}

\noindent 
{\bf Solution}.
This decimal corresponds to the series 
\[ \sum_{ n = 1 }^{ \infty } \: \frac{ 4 }{ 10^n } = \frac{ 4 }{ 10 } 
+ \frac{ 4 }{ 10^2 } + \frac{ 4 }{ 10^3 } + \cdots \]
which has as its partial sum
\[ \begin{array}{rl}
s_n & \hspace{-.1in} = \frac{ 4 }{ 10 } + \frac{ 4 }{ 10^2 } 
+ \cdots \frac{ 4 }{ 10^n } 
\\ [0.2in]
& \hspace{-.1in} = \frac{ 4 }{ 10 } \left [ 1 + 
\frac{ 1 }{ 10 } + \cdots + \frac{ 1 }{ 10^{ n - 1 }} \right ]
= \frac{ 4 }{ 10 } \frac{ 1 - 1/10^n }{ 1 - 1/10 } 
= \frac{ 4 }{ 9 } ( 1 - 1/10^n ) . \end{array} \]
Since $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  s_n
= 4/9 $, it follows that $ .444 ... = 4/9 $.  $ \ddagger $
\vspace{.1in}


You should have observed by now that the terms of a
convergent series are ultimately very small. 
\vspace{.1in}

\noindent 
{\bf Theorem 7}.  If the series $ {\displaystyle 
\sum_{ k = 1 }^{ \infty }} \: a _k $ tend to 
a finite limit as $ n \rightarrow \infty $.
Therefore, $ a_n = s_{ n + 1 } - s_n \rightarrow S - S = 0 $
as $ n \rightarrow \infty $.  $ \ddagger $

The content of this theorem is of more immediate use when
stated as a
\vspace{.1in}

\noindent 
SUFFICIENT CONDITION FOR DIVERGENCE
\vspace{.1in}

\noindent  
(7) \begin{minipage}[t]{5.9in} 
If the sequence $ \{ a_n \} $ either fails to have a limit or 
has a limit different from 0, then the series $ {\displaystyle 
\sum_{ n = 1 }^{ \infty }} \: a_n $ diverges.
\end{minipage}
\vspace{.1in}

\noindent 
{\bf Example 5}.
In consequence of (7), the following series are divergent.
\vspace{.1in}

\noindent 
(a) \hspace{1in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ n }{ n + 1 } = 
\frac{ 1 }{ 2 } + \frac{ 2 }{ 3 } + \frac{ 3 }{ 4 } 
+ \cdots , ~~ a_n \rightarrow 1 $
as $ n \rightarrow \infty $
\vspace{.1in}

\noindent  
(b) \hspace{1in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ ( - 1 )^n n }{ n + 1 } =  
- \frac{ 1 }{ 2 } + \frac{ 2 }{ 3 } - \frac{ 3 }{ 4 } 
+ \cdots , ~~ a_n \rightarrow 1 $  
as $ n \rightarrow \infty $
\vspace{.1in}

\noindent  
(c) \hspace{1in} 
$ {\displaystyle \sum_{ k = 1 }^{ \infty }} \:
\ln \: k = \ln \: 1 + \ln \: 2 + \ln \: 3 + \cdots , a_n 
\rightarrow \infty $ as $ n \rightarrow \infty $
\vspace{.1in}

\noindent  
(d) \hspace{1in} 
$ {\displaystyle \sum_{ k = 1 }^{ \infty }} \:
{\displaystyle \frac{ k^2 + k + 1 }{ k^2 
- k + 1 }} = \frac{ 3 }{ 1 } + \frac{ 7 }{ 3 }
+ \frac{ 13 }{ 7 } + \cdots , a_n \rightarrow 1 $
as $ n \rightarrow \infty $ 
\vspace{.1in}

\noindent  
(e) \hspace{1in} 
\begin{tabular}{rl}
$ {\displaystyle \sum_{ j = 1 }^{ \infty }} \:
\cos \: ( j \pi ) $ & \hspace{-.1in} $ = 
\cos \: \pi + \cos \: 2 \pi + \cos \: 3 \pi + \cdots $ \\ [0.2in]
& \hspace{-.1in} $  = - 1 + 1 - 1 + \cdots , a_n $
has no limit $ ~~~ \ddagger $ \end{tabular} 
\vspace{.1in}

Even if $ a_n \rightarrow 0 $ as $ n \rightarrow \infty $,
the series $ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
a_n $ may diverge.  There, $ a_n 
= \ln \: ( 1 + 1/n ) \rightarrow 0 $ as 
$ n \rightarrow \infty $.  Yet the series diverges.  
Such examples are easy to construct.  Consider,
for example, the series 
\[ \sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ \sqrt{ n }} = 
1 + \frac{ 1 }{ \sqrt{ 2 }} +
\frac{ 1 }{ \sqrt{ 3 }} + \cdots . \]
The nth term $ a_n = \frac{ 1 }{ \sqrt{ n }} $ certainly
tends to $0$ as $ n \rightarrow \infty $.  Nonetheless,
the series diverges because 
\[ s_n = 1 + \frac{ 1 }{ \sqrt{ 2 }}
+ \frac{ 1 }{ \sqrt{ 3 }} + \cdots + \frac{ 1 }{ \sqrt{ n }}
> \frac{ n }{ \sqrt{ n }} = \sqrt{ n } \rightarrow \infty ~~
\mbox{as $ n \rightarrow \infty $.} \]

The identical phenomenon is present in
the {\em \index{harmonic series} harmonic series}
\vspace{.1in}

\noindent 
(8) \hspace{2in} $ {\displaystyle \sum_{ n = 1 }^{ \infty }}
\: \frac{ 1 }{ n } = 1 + 
\frac{ 1 }{ 2 } + 
\frac{ 1 }{ 3 } + \cdots $
\vspace{.1in}

\noindent 
whose partial sums grow ever so slowly; but they
do grow.  One must add 12,367 terms to obtain 
a partial sum that exceeds 10, and 33,617
terms are needed to exceed 11.  At this latter stage, where
each additional term adds 
less than $ 3 \times 10^{ - 5 } $, it seems far
more likely that the partial sums will level off and
approach a limit.  That they continue to grow may be 
seen by clever grouping:
\[ \begin{array}{rl}
s_2 & \hspace{-.1in} = 1 + \frac{ 1 }{ 2 }
\\ [0.2in]
s_4 & \hspace{-.1in} = \left ( 1 + \frac{ 1 }{ 2 } \right ) + 
\left ( \frac{ 1 }{ 3 } + 
\frac{ 1 }{ 4 } \right ) > 1 + \frac{ 1 }{ 2 } + \frac{ 1 }{ 2 }
\\ [0.2in]
s_8 & \hspace{-.1in} = \left ( 1 + \frac{ 1 }{ 2 } \right ) + \left (
\frac{ 1 }{ 3 } + 
\frac{ 1 }{ 4 } \right ) + \left (
\frac{ 1 }{ 5 } + \frac{ 1 }{ 6 } + \frac{ 1 }{ 7 } + 
\frac{ 1 }{ 8 } \right ) > 1 + \frac{ 1 }{ 2 } + \frac{ 1 }{ 2 }
+ \frac{ 1 }{ 2 } \\ [0.2in]
s_{ 2^m } & \hspace{-.1in} = \left ( 1 + \frac{ 1 }{ 2 } \right )
+ \left ( \frac{ 1 }{ 3 } + \frac{ 1 }{ 4 } \right )
+ \cdots + \left ( \frac{ 1 }{ 2^{ m - 1 } + 1 }
+ \cdots + \frac{ 1 }{ 2^m } \right )
> 1 + \frac{ m }{ 2 } . \end{array} \]
This means that the partial sums $ s_2 , s_4 , 
s_8 , s_{16} , s_{ 32 } , s_{ 64 } ,... $ diverge 
to infinity.  However, $ s_1 < 
s_2 < s_3 < s_4 < \cdots $, so we conclude that
the entire sequence of partial sums diverges to infinity
and the series diverges.  This property of the
harmonic series merits a special statement.
\vspace{.1in}

\noindent 
{\bf Theorem 8}.  The harmonic series diverges.
\vspace{.1in}

The simple rules for adding two convergent series 
and multiplying a series by a constant
are set forth in the following theorem.
The more complicated rules for multiplying and 
dividing infinite series will be touched
on later.
\vspace{.1in}

\noindent 
{\bf Theorem 9}.  Suppose
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
( a_n + b_n ) $ is a convergent series and

\noindent 
(9) \hspace{1in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
( a_n + b_n ) = 
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
a_n + {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
b_n = A + B $.
\vspace{.1in}

\noindent 
Also $ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
c a_n $ is convergent for every constant $c$ and

\noindent  
(10) \hspace{2in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
c a_n = c \: {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
a_n = c A $.
\vspace{.1in}

\noindent 
{\bf Proof}.  By assumption $ s_n = 
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
a _k \rightarrow A $ and $ t_n 
= {\displaystyle \sum_{ k = 1 }^{ \infty }} \:
b _k \rightarrow B $ as $ n \rightarrow \infty $,
and $ s_n + t_n = {\displaystyle \sum_{ k = 1 }^{ \infty }} \:
( a _k + b _k ) $
is the nth partial sum for
$ {\displaystyle \sum_{ k = 1 }^{ \infty }} \:
( a _k + b _k ) $.  The limit of these partial sums is
$ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
( s_n + t_n ) = 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  s_n +
{\displaystyle \lim_{ n \rightarrow \infty }} \:  t_n  
= A + B $, by Theorem 3.  This proves
(9).  For (10), write $ p_n 
= {\displaystyle \sum_{ k = 1 }^{ \infty }} \:
c a _k = c
\: {\displaystyle \sum_{ k = 1 }^{ \infty }} \: a _k 
= c s_n $ so that $ {\displaystyle \lim_{ n \rightarrow
\infty }} \:  c s_n = c \:
{\displaystyle \lim_{ n \rightarrow \infty }} \:  s_n 
= c A $, again by Theorem 3.  This proves (10). 
$ \ddagger $
\vspace{.1in}

Notice that rules (9) and (10) resemble the two basic rules 
of integration 
\[ \int_a^b \: ( f(x) + g(x) \: d x = \int_a^b \:
f(x) \: d x + \int_a^b \: g(x) \: d x \]
and
\[ \int_a^b \: c f(x) \: d x = c \int_a^b \: f(x) \: d x . \]
In practice, they are applied in an identical manner.
For example, consider the series 
\[ \sum_{ n = 1 }^{ \infty } \:
\frac{ 1 }{ n ( n + 1 ) } = 1 ~~ \mbox{and} ~~
{\displaystyle \sum_{ n = 1 }^{ \infty }} \: 
\frac{ 1 }{ n ( n + 2 ) } = \frac{ 3 }{ 4 } . \]
By (9),
\[ \begin{array}{rl}
\frac{ 7 }{ 4 } = 1 + \frac{ 3 }{ 4 } 
& \hspace{-.1in} =  
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 1 }{ n ( n + 1 ) } + 
{\displaystyle \sum_{ n = 1 }^{ \infty }} 
\: \frac{ 1 }{ n ( n + 2 ) } 
\\ [0.2in]
& \hspace{-.1in} = 
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\left [ \frac{ 1 }{ ( n + 1 ) }
+ \frac{ 1 }{ ( n + 2 ) } \right ] 
= {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 2n + 3 }{ n ( n + 1 ) ( n + 2 ) } \end{array} \]
Also,
\[ \begin{array}{rl}
\frac{ 1 }{ 4 } = 1 - \frac{ 3 }{ 4 } 
& \hspace{-.1in} = {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
- \frac{ 1 }{ n ( n + 1 ) } -  
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 1 }{ n ( n + 2 ) } 
\\ [0.2in]
& \hspace{-.1in} = {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\left [ \frac{ 1 }{ n ( n + 1 ) } - 
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 1 }{ n ( n + 2 ) } \right ] 
= {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 1 }{ n ( n + 1 ) ( n + 2 ) } . \end{array} \]
This obvious rule for subtracting two
convergent series results from a 
combination of (9) and (10), using the choice $ c = - 1 $:
\[ \begin{array}{rl}
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
( a_n - b_n ) = {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
( a_n + ( - b_n )) & \hspace{-.1in} = 
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
a_n + {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
( - b_n ) \\ [0.2in]
& \hspace{-.1in} = 
{\displaystyle \sum_{ n = 1 }^{ \infty }} \:
a_n - {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
b_n = A - B . \end{array} \]
\vspace{.1in}

\noindent 
{\bf Example 6}.
Find the sum of the series 
\[ \left ( \frac{ 1 }{ 2 } \right )^3 +  \left ( 
\frac{ 1 }{ 2 } \right )^4 +  \left ( \frac{ 1 }{ 2 } \right )^5 
+ \cdots . \] 
\vspace{.1in}

\noindent 
{\bf Solution}.
This is a geometric series $ a + a x + a x^2 
+ \cdots $ in which $ a = \left ( \frac{ 1 }{ 2 } \right ) ^3 $.
We simply factor $a$ out:
\[ \begin{array}{rl}
\left ( \frac{ 1 }{ 2 } \right ) + \left ( \frac{ 1 }{ 2 } \right )^4 + 
\left ( \frac{ 1 }{ 2 } \right )^5 + \cdots + 
& \hspace{-.1in} = 
\left ( \frac{ 1 }{ 2 } \right )^3 \left [ 1 + \frac{ 1 }{ 2 }  +  
\left ( \frac{ 1 }{ 2 } \right )^2 + \cdots \right ] 
\\ [0.2in]
& \hspace{-.1in} = \left ( \frac{ 1 }{ 2 } \right )^3 
\frac{ 1 }{ 1 - 1/2 } = \frac{ 1 }{ 4 } . ~~~ \ddagger 
\end{array} \]

Rule (10) applies only to convergent series.  But there
is a related idea which is often useful in deciding 
a series diverges.  It is this:
\vspace{.1in}

\noindent 
(11) \hspace{1in}
if $ {\displaystyle \sum_{ n = 1 }^{ \infty }} \: a_n $
diverges and $ c \neq 0 $, then $ {\displaystyle 
\sum_{ n = 1 }^{ \infty }} \: c a_n $ diverges.
\vspace{.1in}

\noindent 
Thus a known divergent series can't be turned into 
a convergent series by multiplying each term 
by a constant, unless that constant is zero.
For each partial sum of the second series is the product
of $c$ and the corresponding partial sum of the first
series; therefore if one diverges so must the other.
\vspace{.1in}

\noindent 
{\bf Example 7}.

(a) \hspace{.6in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 5 }{ n } $ diverges because 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 1 }{ n } $ diverges.
\vspace{.1in}

(b) \hspace{.6in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 3n }{ n + 1 } $ diverges because 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ n }{ n + 1 } $ diverges.  $ \ddagger $
\vspace{.1in}

To this point we have used only series 
starting with $ a_0 $ or $ a_1 $,
but it is also sometimes necessary to use series that
begin with $ a _k $:
\[ \sum_{ n = k }^{ \infty } \: a_n 
= a _k + a _{ k + 1 } + a_{ k + 2 } + \cdots . \]
Indeed, any convergent series can be written as the sum
of an initial block of terms plus another convergent
series.  Thus,
\[ \sum_{ n = 1 }^{ \infty } \: a_n = a_1 + 
\sum_{ n = 2 }^{ \infty } \: a_n , \]
\[ \sum_{ n = 1 }^{ \infty } \: a_n 
= a_1 + a_2 + \sum_{ n = 3 }^{ \infty } \: a_n , \]
and, in general,
\vspace{.1in}

\noindent 
(12) \hspace{1.5in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
a_n = a_1 + a_2 
+ \cdots + a _k + {\displaystyle 
\sum_{ n = k + 1 }^{ \infty }} \:
a_n $
\vspace{.1in}

\noindent 
for each positive integer $k$.  This formula, an easy
consequence of comparing partial sums of the two series,
makes no sense for divergent series.
\vspace{.1in}

\noindent 
{\bf Example 8}.  

(a) \hspace{1in} $ {\displaystyle  
\sum_{ n = 1 }^{ \infty }} \:
x^n = 1 + {\displaystyle 
\sum_{ n = 1 }^{ \infty }} \:
x^n $ if $ | x | < 1 $.
\vspace{.1in}

(b) \hspace{1in} 
$ {\displaystyle  \sum_{ n = 1 }^{ \infty }} \:
\frac{ n }{ n ( n + 1 ) }
= \frac{ 1 }{ 2 } + \frac{ 1 }{ 6 } 
+ {\displaystyle \sum_{ n = 3 }^{ \infty }} \:
\frac{ 1 }{ n ( n + 1 ) } $.
\vspace{.1in}

\noindent 
{\bf Example 9}.  Find the sum of the series 
$ {\displaystyle \sum_{ k = 3 }^{ \infty }} \:
1/2^k $.

By (12),
\[ \sum_{ k = 0 }^{ \infty } \: \frac{ 1 }{ 2^k } = 1 + 
\frac{ 1 }{ 2 } + \frac{ 1 }{ 2^2 } + \sum_{ k = 3 }^{ \infty } \:
\frac{ 1 }{ 2^k } ; \]
also
\[ \sum_{ k = 0 }^{ \infty } \: \frac{ 1 }{ 2^k } = 
\frac{ 1 }{ 1 - 1/2 } = 2 . \]
Therefore,
\[ \begin{array}{rl}
{\displaystyle \sum_{ k = 3 }^{ \infty }} \: \frac{ 1 }{ 2^n } 
& \hspace{-.1in} = 
{\displaystyle \sum_{ k = 0 }^{ \infty }} \:
\frac{ 1 }{ 2^k } - 1 \frac{ 1 }{ 2 } - 
\frac{ 1 }{ 2^2 } 
\\ [0.2in] 
& \hspace{-.1in} = 2 - \frac{ 7 }{ 4 } = 
\frac{ 1 }{ 4 } . \end{array} \]
$ \ddagger $
\vspace{.1in}

\noindent 
Whether a series converges or diverges has nothing
to do with where the indexing starts.
This simple observation saves 
considerable fussing in practice.  Specifically, if
the series $ {\displaystyle \sum_{ n = k }^{ \infty }} \:
a_n $ converges for some starting index $k$,
it will converge for any other starting index; and if
$ {\displaystyle \sum_{ n = k }^{ \infty }} \: 
a_n $ diverges for some starting index,
it will diverge for any other starting index.  To illustrate, since 
\[ \sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ 2^n } 
= 1 + \frac{ 1 }{ 2 } + \frac{ 1 }{ 2^2 } + \cdots \]
is a convergent series, so is the series 
\[ \sum_{ n = 100 }^{ \infty } \:
\frac{ 1 }{ 2^n } = \frac{ 1 }{ 2^{ 100 }} 
+ \frac{ 1 }{ 2^{ 101 }} + \frac{ 1 }{ 2^{ 102 }} + \cdots ; \]
and since the harmonic series
\[ \sum_{ n = 1 }^{ \infty } \:
\frac{ 1 }{ n } = 1 + 
\frac{ 1 }{ 2 } + \frac{ 1 }{ 3 } + \cdots \]
diverges, the series 
\[ \sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ n + 5 }
= \frac{ 1 }{ 6 } + \frac{ 1 }{ 7 } + \frac{ 1 }{ 8 } + \cdots \]
is also divergent.


\section{Series of Positive Terms}
\vspace{.1in}

More often than not the question of convergence
for an infinite series cannot be settled with the
ease suggested by the preceding section, the difficulty
being that it is not usually possible to find a simple
formula for the nth partial sums $ s_n = 
a_1 + a_2 + a_3 + \cdots + 
a_n $ of the series $ \sum a_n $.  Consequently,
the technique of establishing convergence by directly
finding the limit of the sequence $ s_n $ must be
abandoned in favor of various indirect approaches.
The most basic of these methods of {\em testing}
a series for convergence apply to series in which
all the terms are positive.  For brevity we shall
refer to a series $ \sum a_n $ in which $ a_n > 0 $ as a 
{\em series of positive terms}. 
For such a series we always have
\[ s_n = s_{ n - 1 } + a_n > s_{ n - 1 } \]
so that the partial sums form a strictly increasing
sequence.  If the partial sums are also bounded, that is,
if there is a constant $M$ such that $ s_n \leq M $ for all $n$,
then the series must converge because, by Theorem 5,
every increasing and bounded sequence has a limit.
But if the partial sums are not bounded, then the 
series diverges.  These remarks establish a basic principle 
about series of positive terms.
\vspace{.1in}

\noindent 
{\bf Theorem 10}.  A series $ \sum a_n $ of positive 
terms converges if its partial sums are bounded,
and diverges if they are not.

In many series the nth term $ a_n $ is the value 
of a simple function $f$ at $n$,
\[ a_n = f (n) \]
and it becomes fruitful to analyze the partial sums of
\[ a_1 + a_2 + a_3 + \cdots = f (1) + f (2) + f (3) + \cdots \]
by comparing areas.  The comparison is particularly simple if 
the associated function $f$ is continuous, positive, and decreasing.
\vspace{.1in}

\noindent 
{\bf Theorem 11}.  The Integral Test

Let $f$ be continuous, positive, and decreasing
on the interval $ x \geq 1 $.  Then the series 
\vspace{.1in}

\noindent 
(1) \hspace{2in} $ f (1) + f(2) + f (3) + \cdots $
\vspace{.1in}

\noindent 
and the improper integral
\vspace{.1in}

\noindent 
(2) \hspace{2in} $ {\displaystyle \int_{ 1 }^{ \infty }} \:
f(x) \: dx $
\vspace{.1in}

\noindent 
either both converge or both diverge.
\vspace{.1in}

\noindent 
{\bf Proof}. 
% The comparisons are suggested by Figures 14.5 and 14.6. 
 Since $f$ is decreasing on each
of the unit intervals $ n \leq x \leq n + 1 $, the area
under the graph from $n$ to $ n + 1 $ is less 
than that of a rectangle of area $ f(n) $
(Figure 14.6) and greater than that of 
a rectangle of area $ f ( n + 1 ) $ (Figure 14.5).  In particular,
% \vspace{4in}

\[ \begin{array}{l}
f(2) < {\displaystyle \int_{ 1 }^{ 2 }} \: f(x) \: d x < f (1) ,
\\ [0.2in]
f(3) < {\displaystyle \int_{ 2 }^{ 3 }} \: f(x) \: d x < f (2) , 
\\ [0.2in]
f(4) < {\displaystyle \int_{ 3 }^{ 4 }} \: f(x) \: d x < f (3) ,
\\ [0.2in]
f(n + 1) < {\displaystyle \int_{ n }^{ n + 1 }} \: f(x) \: 
d x < f (n) , \end{array} \]
Addition yields
\[ f (2) + f (3) 
+ \cdots + f (n+1) < \int_{ 1 }^{ n + 1 } \:
f(x) \: d x < f (1) + f (2) + \cdots + f (n) . \]
In terms of the partial sums $ s_n = {\displaystyle 
\sum_{ k = 1 }^n } \: f (k) $, these
two inequalities state that
\vspace{.1in}

\noindent 
(3) \hspace{1in} $ s_{ n + 1 } - f (1) < 
{\displaystyle \int_{ 1 }^{ n + 1 }} \: f(x) \: d x ~~
\mbox{and} ~~ 
{\displaystyle \int_{ 1 }^{ n + 1 }} \: f(x) \: d x < s_n $
\vspace{.1in}

\noindent 
These comparisons form the basis of the integral test.

The second inequality shows that if the improper
integral (2) diverges, that is, if $ {\displaystyle 
\int_{ 1 }^{ n + 1 }} \: f(x) \: d x \rightarrow 
\infty $ as $ n \rightarrow \infty $, then $ s_n 
\rightarrow \infty $ as $ n \rightarrow \infty $ and the
series diverges.  But if the improper integral converges,
then $ {\displaystyle \int_{ 1 } ^{ n + 1 }} \:
f(x) \: d x $ increases to a limit $L$ as $ n \rightarrow
\infty $ and 
\[ s_{n+1} < f (1) + 
\int_{ 1 }^{n+1} \: f(x) \: d x 
< f (1) + L \]
for all $n$.  Hence $ \{ s_n \} $, being increasing
and bounded, has a limit by Theorem 10. $ \ddagger $
\vspace{.1in}

As a first application of the integral test, consider
the harmonic series $ \sum 1/n $, a known divergent
series.  Clearly, the test function is $ f(x) 
= 1/x $ because $ f (n) = 1/n $.  And since $ 1/x $
is continuous, positive and decreasing on the
range $ x > 0 $, we conclude from 
\[  \int_{ 1 } ^{ \infty } \:
\frac{ 1 }{ x } \: d x = 
{\displaystyle \lim_{ b \rightarrow \infty }} \:  
\int_{ 1 }^b \frac{1}{x}\: d x  =
{\displaystyle \lim_{ b \rightarrow \infty }} \:  
\ln \: x  ]_1^b = 
{\displaystyle \lim_{ b \rightarrow \infty }} \:  
\ln \: b = \infty \]
that the harmonic series diverges.  Moreover, the second
inequality in (3) provides a rough estimate
of the rate at which its partial sums grow:
\[ \ln \: (n+1) = \int_{ 1 }^{n+1} \:
\frac{ 1 }{ x } \: d x < 1 + 
\frac{1}{2} + \frac{1}{3} + \cdots + 
\frac{ 1 }{ n } = s_n . \]

It is interesting to compare the estimates
given by this inequality with some computed values
of $ s_n $ for selected $n$:
\[ \begin{array}{rr}
\mbox{Estimates} $ ~ $ & \mbox{Computed Values} 
\\ [0.1in]
s_{ 83 } > 4.4308 & s_{ 83 } = 5.002068... 
\\ [0.1in]
s_{ 227 } > 5.4293 & s_{ 227 } = 6.004366...  
\\ [0.1in]
s_{ 616 } > 6.4248 & s_{ 616 } = 7.001274...
\\ [0.1in]
s_{ 1,674 } > 7.4235 & s_{ 1,674 } = 8.000485...
\\ [0.1in]
s_{ 4,550 } > 8.4231 & s_{ 4,550 } = 9.000208... 
\\ [0.1in]
s_{ 12,367 } > 9.4228 & s_{ 12,367 } = 10.000043... 
\end{array} \]

The harmonic series is a special case of the more 
general \index{$p$-{\em series}} $p$-{\em series}
\vspace{.1in}

\noindent 
(4) \hspace{2in} $ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
\frac{ 1 }{ n^p } = 
1 + \frac{1}{2^p} + \frac{1}{3^p} + \cdots . $  
\vspace{.1in}

\noindent 
{\bf Example 1}.
Show that the $p$-series converges if $ p > 1 $
and diverges if $ p \leq 1 $.
\vspace{.1in}

\noindent 
{\bf Solution}.
Here the test function is $ f(x) = 1/x^p $
because $ f (n) = 1/n^p $.  It is continuous,
positive, and since $ f ^{'} (x) 
= - p/x^{ p + 1 } < 0 $, it is decreasing
on the interval $ x > 0 $.  If $ p = 1 $, the
series (4) is the divergent harmonic series.
If $ p \neq 1 $, the $p$-series converges or
diverges with the improper integral
\[ \int_{ 1 }^{ \infty } \:
\frac{ 1 }{ x^p } \: d x = 
{\displaystyle \lim_{ b \rightarrow \infty }} \:  
\int_{ 1 }{ b } \: x^{ - p } \: d x = 
{\displaystyle \lim_{ b \rightarrow \infty }} \:  
\frac{ x^{ - p + 1 }}{ - p + 1 } \left |_1^b  \right.  = 
{\displaystyle \lim_{ b \rightarrow \infty }} \:  
\left ( \frac{ b^{ - p + 1 }}{ 1 - p } - 
\frac{ 1 }{ 1 - p } \right ) . \]
Whether this limit is finite or infinite depends on the 
exponent $ - p + 1 $.  If $ - p + 1 > 0 $
or $ p < 1 , ~~ b^{ - p + 1 } $ will increase 
without bound as $b$ increases; but if
$ - p + 1 < 0 $, the exponent on $b$ is negative and 
$ b^{ - p + 1 } \rightarrow 0 $
as $ b \rightarrow \infty $.  Hence 
\[ \int_{ 1 }^{ \infty } \: \frac{ 1 }{ x^p } \: d x 
= \frac{ 1 }{ p - 1 } ~~ \mbox{if $ p > 1 $ and} ~~
\int_{ 1}^{ \infty }\frac{1}{x^p} \: d x 
~~ \mbox{diverges if $ p < 1 $.} \]
According to the integral test this means that the
$ p $-series (4)
converges if $ p > 1 $ and diverges if $ 0 < p < 1 $.  $ \ddagger $
\vspace{.1in}

In particular, this means that
\[ \sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ n^2 } ~~ \mbox{and} ~~ 
\sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ n^4 } \]
are convergent series ($ p = 2 $ and $ p = 4 $) while
\[ \sum_{ n = 1 }^{ \infty } \:
\frac{ 1 }{ \sqrt{ n }} ~~ \mbox{and} ~~
\sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ n^{ 1/3 }} \]
are divergent series ($ p = 1/2 $ and $ p = 1/3 $).
the conclusions appear reasonable because,
for instance, $ 1/n^2 $ certainly goes to 0 much 
faster than $ 1 / \sqrt{ n } $.  But, intuitively,
no meaningful distinction is apparent between the series 
\[ \sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ n^{ 1.001 }} 
\mbox{and} ~~ 
\sum_{ n = 1 }^{ \infty } \: \frac{ 1 }{ n^{ .999 }} . \]
The first converges, the second diverges;
yet, for large $n$ their nth terms would seem to be going
to zero at about the same rate.  But there is a difference,
and that difference is enough to force radically
different behavior on the respective partial sums.

If a series is shown to converge by the integral test,
the sum of the series is {\em never exactly} the value  of the 
integral.  In fact, if $ a_n = f(n) $ and $f$ satisfies 
the conditions of the integral test, then, by (3),
the best we can say is that
\vspace{.1in}

\noindent 
(5) \hspace{2in} $ {\displaystyle \int_{ 1 }^{ \infty }} \: 
f(x) \: d x <  \sum_{ n = 1 }^{ \infty } \:
a_n < a_1 + {\displaystyle \int_{ 1 }^{ \infty }} \: f(x) \: d x $.
\vspace{.1in}

\noindent 
Hence

\[ 1 = {\displaystyle \int_{ 1 }^{ \infty }} \:
\frac{ 1 }{ x^2 } < \sum_{ n = 1 }^{ \infty } \:
\frac{ 1 }{ n^2 } < 1 + {\displaystyle \int_{ 1 }^{ \infty }} \:
\frac{ 1 }{ x^2 } \: d x = 1 + 1 = 2  \]
and
\[ \frac{ 1 }{ p - 1 } = {\displaystyle \int_{ 1 }^{ \infty }} \:
\frac{ 1 }{ x^p } \: d x <  \sum_{ n = 1 }^{ \infty } \:
\frac{ 1 }{ n^p } < 1 + {\displaystyle \int_{ 1 }^{ \infty }} \:
\frac{ 1 }{ x^p } \: d x = 1 + \frac{ 1 }{ p
- 1 } = \frac{ p }{ p - 1 } ~~ \mbox{if $ p > 1 $.} \]

Used in this fashion the integral test provides an upper and 
lower bound for the sum of a convergent series, but 
the estimates it provides aren't very accurate,
as you can tell by examining finite 14.5.  A more useful 
approach consists of calculating the sum of the first few 
terms and using an integral to estimate the tail-end or remainder.
Specifically, we write 
\[ S = \sum_{ n = 1 }^{ \infty } \: a_n =a_1 + a_2 + a_3 + \cdots + 
a_n + \sum_{ k = n + 1 }^{ \infty } \: a _k \]
and call
\vspace{.1in}

\noindent 
(6) \hspace{2in} $ R_n = 
{\displaystyle \sum_{ k = n + 1 }^{ \infty }} \: a _k $
\vspace{.1in}

\noindent 
the \index{{\em remainder after} $n$-terms} {\em remainder after} $n$-terms.
In the setting of the integral test, a comparison of areas shows that 
\vspace{.1in}

\noindent 
(7) \hspace{2in} $ R_n < {\displaystyle \int_{ n }^{ \infty }} \:
f(x) \: d x $.
\vspace{.1in}

\noindent 
{\bf Example 2}.
Estimate the sum of $ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
1/n^2 $ with an error less than or equal to $ .005 $.
\vspace{.1in}

\noindent 
{\bf Solution}.  Write 
\[ S = \sum_{ n = 1 }^{ \infty } \:
\frac{ 1 }{ n^2 } = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + 
\frac{ 1 }{ n^2 } + R_n , \] 
where $ R_n = {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
1/k^2 $.  Using $ f(x) = 1/x^2 $ we find that
\[ R_n < \int_{ n }^{ \infty } \:
\frac{ 1 }{ x^2 } \: d x = \frac{ 1 }{ n } . \]
Now the idea is to choose $ n $ so large
that $ R_n < .005 $ and this certainly occurs
when $ n = 200 $.  Since the partial sums increase 
to $S$ we have
\[ \frac{ 1 }{ 2^2 } + \cdots + 
\frac{ 1 }{ 200^2 } < S 
< 1 + \frac{1}{2^2} 
+ \cdots + \frac{1}{200^2} + .005 \]
and, with the aid of a calculator, find that
\[ 1.639946... < S < 1.644946... \]
Actually it can be shown by advanced methods that
\[ S = \sum_{ n = 1 }^{ \infty } \: 
\frac{ 1 }{ n^2 } = \frac{ \pi ^2 }{ 6 }
= 1.64493406... \]
\vspace{.1in}

\noindent 
{\bf Example 3}.
Use the integral test to show that the series 
$ {\displaystyle \sum_{ n = 2 }^{ \infty }} \: 
\frac{ 1 }{ n \: \ln \: n } $ diverges.
\vspace{.1in}

\noindent 
{\bf Solution}.
The test function $ f(x) = 
{\displaystyle \frac{ 1 }{ x \ln \: x }} $ is 
continuous, positive, and decreasing on the 
range $ x \geq 2 $, and
\[ \begin{array}{rl}
{\displaystyle \int_{ 2 }^{ \infty }} \: 
{\displaystyle \frac{ d x }{ x \: \ln \: x }} 
& \hspace{-.1in} = 
{\displaystyle \lim_{ b \rightarrow \infty }} \:  
{\displaystyle \int_{ 2 }^{ b }} \:
{\displaystyle \frac{ d x }{ x \: \ln \: x }} = 
{\displaystyle \lim_{ b \rightarrow \infty }} \: 
\ln \: ( \ln \: x ) \left |_2^b \right. \\ [0.2in]
& \hspace{-.1in} = 
{\displaystyle \lim_{ b \rightarrow \infty }}
\: [ \ln \: ( \ln \: b ) -
\ln \: ( \ln \: 2 ) ] = \infty . \end{array} \]
Hence the series diverges, even though its terms 
tend to zero faster than the terms of the divergent series 
$ \sum 1/n $.  $ \ddagger $
\vspace{.1in}

In applying the integral test in the preceding example 
we deviated from the integral test recipe as given in 
Theorem 10 in that everything was shifted to the right
by one unit.  A quick review of the idea behind the integral
test should convince you that nothing is amiss.  Indeed,
the integral test applies equally well if the beginning
index in (1) and the lower limit in (2) are changed from
1 to any positive integer $k$.
\vspace{.1in}

\noindent 
COMPARISON TESTS
\vspace{.1in}

The integral test is of no use in showing that the series 
\vspace{.1in}

\noindent 
(8) \hspace{2in} $ 1 + 
{\displaystyle \frac{1}{ 2 ! } 
+ \frac{1}{ 3 ! } + \frac{1}{ 4 ! } 
+ \cdots + \frac{1}{ n ! }} + \cdots $
\vspace{.1in}

\noindent 
converges, even though the terms are positive
and decreasing.  The difficulty, of course, is that 
there is no apparent candidate for $ f(x) $.
Nonetheless, this series can be analyzed by the
idea upon which the integral test is based; namely,
that of comparing the terms of one series with those of
another.  One comparison arises from 
observing that $ 2 ! \geq 2 , ~ 3 ! > 2^2 , ~ 4 !
> 2^3 $ and, in general, if $ k > 1 $
and one knows that if
\[ k ! > 2^{ k - 1 } , ~~ \mbox{then $ ( k + 1 ) ! = 
k ! ( k + 1 ) > 2^{ k - 1 } ( k + 1 ) > 2^k $.} \]
This inductive argument shows that $ n ! \geq 
2^{ n - 1 } $ for each positive integer (in fact, equality
holds only at $ n = 1 $).  Hence $ 1 / n ! \leq 1/2^{ n - 1 } $ and
\[ \begin{array}{rl}
s_n & \hspace{-.1in} = 1 + 
{\displaystyle \frac{ 1 }{ 2 } + \frac{ 1 }{ 2^2 }
+ \cdots + \frac{ 1 }{ 2 ^{n-1} }}
= {\displaystyle \frac{ 1 - 1/2^n }{ 1 - 1/2 }} \\ [0.3in]
& \hspace{-.1in} = 2 - 
{\displaystyle \frac{ 1 }{ 2^{n-1} }} < 2 
\end{array} \]
for each positive integer $n$.  This last inequality 
shows that the partial sums of (8) are bounded
above by 2 and that, according to Theorem 10,
it converges.  Later on we will show that the sum 
in (8) is $ e - 1 = 1.718281828... $ 

The preceding argument shows that the series 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
1/n ! $ converges because its terms are less 
than the corresponding terms of the convergent series 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
1/2^{ n - 1 } $.  In general, given a series 
$ \sum a_n $ of positive terms and inequalities
\[ 0 < a_n \leq b_n , \]
for all $n$, we say that the series $ \sum b_n $ {\em \index{dominates} dominates}
the series $ \sum a_n $.
\vspace{.1in}

\noindent  
{\bf Theorem 12}.  Comparison Test

Let $ \sum a_n $ and $ \sum b_n $ be series of positive terms.
\vspace{.1in}

\noindent 
{\em Convergence Test}.  If $ 0 \leq a_n \leq b_n $ for all $ n $ 
and $ \sum b_n $ converges, then $ \sum a_n $ is also convergent.
\vspace{.1in}

\noindent 
{\em \index{Divergence Test} Divergence Test}.  If $ 0 \leq
b_n \leq a_n $ for all $n$ and $ \sum b_n $ diverges, then
$ \sum a_n $ is also divergent.
\vspace{.1in}

\noindent 
{\bf Proof}.  Let 
$ s_n = {\displaystyle \sum_{ k = 1 }^{ n }} \:
a _k $ and $ t_n = s_n = {\displaystyle \sum_{ k = 1 }^{ n }} \:
b _k $.  For the Convergence Test,
$ s_n \leq t_n $ for all $n$ and the partial sums
$ t_n $ increase to the sum $ T = 
\sum b_n $.  Hence $ s_n \leq t_n \leq T $ and the partial
sums $ \{ s_n \} $ form an increasing and bounded sequence, that is,
\[ s_1 \leq s_2 \leq s_3 
\leq \cdots \leq s_n \leq \cdots \leq T . \]
The convergence of $ \sum a_n $ is guaranteed
by Theorem 8.

In the Divergence Test, the relation between 
$ s_n $ and $ t_n $ is reversed.
Now, $ t_n \leq s_n $ because $ b_n \leq a_n $ for
all $n$ and it is clear that $ s_n $ must grow large if $ t_n
$ does.  But $ t_n $ tends to infinity because 
$ \sum b_n $ is divergent; hence
$ s_n \rightarrow \infty $ as $ n \rightarrow \infty $ and 
$ \sum a_n $ diverges.  $ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 4}.
Use the comparison test to determine whether the series
$ {\displaystyle \sum_{ n = 1 }^{ n }} \:
{\displaystyle \frac{ \sqrt{ n }}{ n^2 + 1 }} $ converges.
\vspace{.1in}

\noindent 
{\bf Solution}.
The series is convergent because 
\[ 0 \leq \frac{ \sqrt{ n }}{ n^2 + 1 } 
= \frac{ 1 }{ n^{ 3/2 } + 1 / \sqrt{ n } } 
< \frac{ 1 }{ n^{ 3/2 }} \]
and $ {\displaystyle \sum_{ n = 1 }^{ n }} \:
1 / n^{ 3/2 } $ is a known convergent series -- the $p$-series 
for $ p = 3/2 $.  $ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 5}.
Apply the comparison test to the
series $ {\displaystyle \sum_{ n = 1 }^{ n }} \: 
{\displaystyle \frac{ n^{ 3/2 }}{ 3 n^2 - n }} $.
\vspace{.1in}

\noindent 
{\bf Solution}.
The key observation is that the nth term 
of this series behaves like $ 1 / 3 \sqrt{ n } $ for
large $n$ and that $ \sum \frac{ 1 }{ 3 \sqrt{ n }} 
= \frac{ 1 }{ 3 } \sum \frac{ 1 }{ \sqrt{ n }} $
is a divergent series.  This suggests we use the comparison test
in its divergence form.  Specifically, the inequality 
\[ \frac{ n^{ 3/2 }}{ 3 n^2 - n } 
= \frac{ 1 }{ 3 \sqrt{ n } - \frac{ 1 }{ n } } >
\frac{ 1 }{ 3 \sqrt{ n }} > 0 \]
shows that the given series is divergent.  $ \ddagger $
\vspace{.1in}

\noindent 
WARNING.  In applying the comparison test
$ 0 < a_n \leq b_n $, you can conclude nothing about $ \sum
a_n $ if the series $ \sum b_n $ is divergent.
Thus, $ 0 < 1/n^2 < 1/n $ and $ 0 < 1/2n 
< 1/n $; yet $ \sum 1/n^2 $ converges and $ \sum 1/2n $
diverges.  Similarly, if $ 0 < a_n \leq b_n $ and $ \sum 
b_n $ and $ \sum a_n $ converges, you can conclude
nothing about the series $ \sum b_n $.
\vspace{.1in}


\section{Alternating Series}
\vspace{.1in}

In this section we turn our attention to series 
that have positive as well as negative terms.
Of these, the simplest are those whose terms are
alternating positive and negative.  An {\em \index{alternating series} alternating}
series may be written conveniently in the form
\[ \sum_{ n = 1 }^{ \infty } \:
( - 1)^{ n-1 } a_n = a_1 
- a_2 + a_3 - a_4 + \cdots \]
where each $ a_n > 0 $.  The series 
\[ \frac{ 1 }{ \sqrt{ 3 }} - \frac{ 1 }{ \sqrt{ 5 }}
+ \frac{ 1 }{ \sqrt{ 7 }} - \frac{ 1 }{ \sqrt{ 9 }} + \cdots , \]
\[ 1 - \frac{ 1 \cdot 2 }{ 1 \cdot 3 }+
\frac{ 1 \cdot 2 \cdot 3 }{ 1 \cdot 3 \cdot 5 }-
\frac{ 1 \cdot 2 \cdot 3 \cdot 4 }{ 1 \cdot 3 \cdot 5 \cdot 7 }
+ \cdots , \]
and
\[ 1 - 1 + 1 - 1 + \cdots \]
are examples of alternating series.
Also, for $x$ in the range $ - 1 < x < 0 $,
the geometric series
\[ 1 + x + x^2 + \cdots 
= \frac{ 1 }{ 1 - x } 
~~ ( | x | < 1 ) \]
becomes a convergent alternating series; at 
$ x = - 1/2 $, it reduces to
\[ 1 - \frac{ 1 }{ 2 } + \frac{ 1 }{ 2^2 }
- \frac{ 1 }{ 2^3 } + \cdots = \frac{ 2 }{ 3 } . \]

The remarkable feature of an alternating series is that it
will converge on the {\em sole assumption
that the $ a_n $ decrease to zero}.
\vspace{.1in}

\noindent 
{\bf Theorem 14}.  Alternating Series Test 

If 

\noindent 
(1) \begin{minipage}[t]{5.9in} 
$ a_n \geq a_{ n + 1 } > 0 $ and $ a_n \rightarrow 0 $ 
as $ n \rightarrow \infty $, then the alternating series 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
(-1)^{ n - 1 } a_n $ converges.  Moreover, if $ S = 
{\displaystyle \sum_{ n = 1 }^{ \infty }} \: ( - 1 )^{ n
- 1 } a_n $ and $ s_n $ is the nth partial sum, then 
\end{minipage}
\vspace{.1in}

\noindent 
(2) \hspace{2in} 
$ | S - s_n | < a_{ n + 1 } $
\vspace{.1in}

\noindent 
{\bf Proof}. 
% Figure 14.7 gives 
the idea:
partial sums with an odd index decrease toward $S$
while partial sums with an even index increase toward 
$S$.  Let us first look at the even partial 
%\vspace{2in} 
\noindent 
sums -- those with even indices.  We have
\[ \begin{array}{l}
s_4 = s_2 + ( a_3 - a_4 ) \geq s_2 
~~ \mbox{because} ~~
a_3 \geq a_4 , \\ [0.2in]
s_6 = s_4 + ( a_5 - a_6 ) \geq s_4 
~~ \mbox{because} ~~
a_5 \geq a_6 , \end{array} \]
and, in general,
\[ s_{ 2n + 2 } = s_{ 2n }
+ ( a_{ 2n + 1 } - 
a_{ 2n + 2 } ) \geq s_{ 2n }
~~ \mbox{because} ~~
a_{ 2n + 1 } \geq a_{ 2n + 2 } \]
For the odd partial sums,
\[ \begin{array}{l}
s_3 = s_1 - a_2 + a_3 = 
s_1 - ( a_2 - a_3 ) \leq s_1 
~~ \mbox{because} ~~
a_2 \geq a_3 ,  \\ [0.2in]
a_5 = s_3 - a_4 + a_5 = s_3 - ( a_4 
- a_5 ) \leq s_3 
~~ \mbox{because} ~~
a_4 \geq a_5 , \end{array} \]
and, in general,
\[ s_{2n+1} = s_{ 2n - 1 }
= a_{ 2n } + a_{2n+1} 
= s_{ 2n - 1 } -
( a_{ 2n } - a _{2n+1} ) 
\leq s _{2n-1} 
~~ \mbox{because} ~~
a_{ 2n } \geq a _{2n+1} \]
Further, $ s_2 + a_3 = s_3 , ~~
s_4 + a_4 = s_5 $, and
\vspace{.1in}

\noindent 
(3) \hspace{2in} $ s_{ 2n } + a _{2n+1} = s _{2n+1} $
\vspace{.1in}

\noindent 
for all $n$, so that $ s_{ 2n } < s _{2n+1} $.
Hence the situation is exactly as shown in the figure:
\[ s_2 \leq s_4 \leq s_6 
\leq \cdots \leq s_{ 2n - 2 } 
\leq s_{ 2n } \leq s_{ 2n - 2 } \leq 
s_{ 2n } < s _{2n+1} 
\leq s _{2n-1} \leq \cdots \leq 
s_5 \leq s_3 \leq s_1 . \]
The information we wish to extract from this string
of inequalities is that the even
partial sums $ \{ s_{ 2n } \} $ form an 
increasing sequence bounded above by
$ s_1 $ and that the odd partial sums $ \{ s _{2n+1} \} $ 
form a decreasing sequence bounded below by $ s_2 $.
We now call on Theorem 5; it tells us that both
sequences converge, one to a number $ \underline{S} $, say, the
other to $ \overline{S} $: 
\vspace{.1in}

\noindent 
(4) \hspace{2in} $ 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  
s_{ 2n } = \underline{S} , ~~
{\displaystyle \lim_{ n \rightarrow \infty }} \:  = \overline{S} $.
\vspace{.1in}

To this point, everything we have said depends on the
assumptions $ a_n \geq a_{ n + 1 } > 0 $.  The
next step uses the assumption $ a_n \rightarrow 
0 $ and, in particular, that $ a _{2n+1} \rightarrow 0 $
as $n\rightarrow \infty $.  From the identities (3) 
and the limits in (4) it follows that
\[ \overline{S} =
{\displaystyle \lim_{ n \rightarrow \infty }} \:  
s _{2n+1} = {\displaystyle \lim_{ n \rightarrow \infty }} \:  
( s_{ 2n } + a _{2n+1} ) = \underline{S} + 0 = \underline{S} , \]
that is, $ \overline{S} = \underline{S} $.  This common value, call it $S$,
is clearly the limit of the entire sequence 
$ \{ s_n \} $.  We have proved that the partial sums
have a limit and hence that the alternating series is
convergent.

Proof of inequality (2) rests on the observation 
%from  Figure 14.7 
that $S$, the sum of the series, always lies
between any two consecutive partial sums; precisely
\vspace{.1in}

\noindent 
(5) \hspace{2in} $ s_{ n + 1 } 
< S < s_n $ if $n$ is odd.
\vspace{.1in}

\noindent 
and

\noindent 
(6) \hspace{2in} $ s_n 
< S < s_{ n + 1 } $ if $n$ is even. 
\vspace{.1in}

\noindent 
Since $ | s_{ n + 1 } - s_n | 
= a _{ n+1 } $, we see that the inequality 
$ | S - s_n | < a _{n+1} $
is merely the statement that the distance from a point inside
an interval to one of its endpoints is less than the length
of the interval.  $ \ddagger $
\vspace{.1in}

The first step in applying the Alternating
Series Test is visual:  you have
to check to be the terms actually alternate in sign.
Clearly, the series 
\vspace{.1in}

\noindent 
(7) $ 1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} 
- \frac{1}{6} - \frac{1}{7} - \frac{1}{8} + \frac{1}{9} + \frac{1}{10} 
+ \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} 
+ \frac{1}{15} + \frac{1}{16} - \frac{1}{17} - \cdots $ 
\vspace{.1in}

\noindent
has positive and negative terms;
yet, it is not an alternating series and the 
Alternating Series Test cannot be applied.  Actually, the series 
(7) diverges. Nor does the alternating series 
\[ e - e^{ 1/2 } + e^{ 1/3 } - e^{ 1/4 } + \cdots \]
meet the requirements of the Alternating Series Test,
even though $ a_n = e^{ 1/n }$ is positive and decrease
with $n$.  The difficulty is that $\displaystyle  \lim_{n\rightarrow \infty}
\: a_n \neq 0 $ (see Theorem 1).  Thus the series diverges
by Theorem 7. But in the series 
\vspace{.1in}

\noindent 
(8) \hspace{2in} $ 1 - \frac{ 1 }{ 2 } + \frac{ 1 }{ 3 }-
\frac{ 1 }{ 4 } + \cdots , $ 
\vspace{.1in}

\noindent 
$ a_n = 1/n $ is positive, $ a_n > a _{n+1} $
and $ \displaystyle \lim_{ n \rightarrow \infty }a_n \:  = 0 $.
The series is therefore convergent
because all conditions in (1)
are satisfied; accordingly, we say that the series (8) 
converges by the Alternating Series Test.

Further, according to (2), if $S$ denotes the sum
of the series in (8) and we truncate the series
after the 199th term, then the partial sum 
$ s_{ 199 } $ approximates $S$ with an error less than
$ a_{ 200 } = 1 / 200 = .005 $.
More accurately, by (5), $ s_{ 199 } $ is an overapproximation
and $ s_{ 200 } $ is an underapproximation to $S$.
Aided by a calculator, we find that
\[ .690653... = s_{ 200 } < S < s_{ 199 } = .695653... \]
Naturally, more accurate estimates require more terms;
it takes more than a million terms to estimate $S$
to six place accuracy.  Later we shall show that
\[ S = \ln \: 2 = .693147... \]
\vspace{.1in}

\noindent
{\bf Example 1}.  Show that the series
\[ \sum_{ n = 1 }^{ \infty } \: ( - 1 ) ^{ n - 1 }
\frac{ n }{ n^2 + 1 } = \frac{1}{2} - \frac{ 2 }{ 5 } 
+ \frac{ 3 }{ 10 } - \frac{ 4 }{ 17 } + \cdots \]
is convergent.
\vspace{.1in}

\noindent 
{\bf Solution}.  In this alternating series 
$ a_n = \frac{ n }{ n^2 + 1 } $ and the conditions in
(1) are most easily verified by examining the continuous
function $ f(x) = \frac{ x }{ 1 + x^2 } $.
It is clearly positive in the interval $ x \geq 1 $
and decreasing there since 
\[ f ^{'} (x) = \frac{ 1 - x^2 }{ ( x^2 + 1 )^2 } < 0 \]
if $ 1 < x $.  Because $ a_n = f(n) $, it follows that $ a_n > 
a _{n+1} > 0 $.  And it is clear that 
$ {\displaystyle \lim_{ n \rightarrow \infty }} \:  a_n = 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ n }{ n^2 + 1 } = 0 $.  Therefore 
the series is convergent by the Alternating Series Test.  $ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 2}.  Show that the series 
\[ 1 - \frac{ 1 }{ 2 ! } + \frac{ 1 }{ 3 ! } - \frac{ 1 }{ 4 ! }
+ \cdots \]
is convergent and estimate the error in approximating
its sum $S$ by the first nine terms.
\vspace{.1in}

\noindent 
{\bf Solution}.  Here $ a_n = 1 / n ! $
and $ a_n > a _{n+1} > 0 $ because $ n ! $ increases with $n$.
Also $ a_n \rightarrow 0 $ because $ n ! $
increases without bound.  [Recall that $ n ! \geq 2^{ n - 1 } $.]
Hence the series converges by the Alternating Series Test.

The error in approximating the sum $S$ by
\[ s_9 = 1 - \frac{1}{2 ! } -   \frac{1}{3 ! } -   
\frac{1}{4 ! } +   \frac{1}{5 ! } -   \frac{1}{6 ! } +  
\frac{1}{7 ! } -   \frac{1}{8 ! } +  \frac{1}{9 ! } = 
.63212081... \]
is less than
\[ a_{ 10 } = \frac{ 1 }{ 10 ! } = 
.000000275... ; \]
therefore
\[ S \approx .632121 , \]
rounded to six decimal places.  $ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 3}.  Show that the series 
\[ \sum_{ n = 1 }^{ \infty } \:
( - 1 )^{ n - 1 } \frac{ 2n + 1 }{ n ( n + 1 ) }
= \frac{ 3 }{ 2 } - \frac{ 5 }{ 2.3 } +  \frac{ 7 }{ 3.4 }
-  \frac{ 9 }{ 4.5 } + \cdots \]
is convergent.
\vspace{.1in}

\noindent 
{\bf Solution}.  In this alternating series 
$ a_n = \frac{ 2n + 1 }{ n (n+1) } $.  Clearly,
$ a_n > 0 $ and $ {\displaystyle \lim_{ n \rightarrow \infty
}} \:  a_n = 0 $.  Rather than proceed as in Example 1,
we can establish the inequality $ a_n \geq a _{n+1} $ directly 
by computing the ratio
\[ \frac{ a _{n+1} }{ a_n } = \frac{ 2n + 3 }{ (n+1) (n+2) }
\dot \frac{ n (n+1) }{ 2n + 1 }
= \frac{ 2n^2 + 3n }{ 2n^2 + 5n + 2 } \]
and then noting the denominator is larger than the 
numerator if $ n \geq 1 $.  Hence 
$ a _{n+1} / a_n < 1 $ or $ a_n > a _{n+1} $, and by 
Theorem 14 the series converges.  $ \ddagger $


