\section{Absolute Convergence}
\vspace{.1in}

As we have seen, the alternating series 
\vspace{.1in}

\noindent 
(1) \hspace{1.5in} 
$ 1 - {\displaystyle \frac{1}{2}  + \frac{1}{3}  - \frac{1}{4}}  
+ \cdots = 1 + \left ( {\displaystyle \frac{1}{2}} \right )  
+ \left ( {\displaystyle \frac{1}{3} } \right )  
+ \left ( -  {\displaystyle \frac{1}{4}} \right )  + \cdots $
\vspace{.1in}

\noindent 
is convergent, but the harmonic series
\[ | 1 | + \left | - \frac{1}{2} \right |
+ \left | \frac{ 1 }{ 3 } \right |
+ \left | - \frac{ 1 }{ 4 } \right | 
+ \cdots = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots \] 
obtained from it by taking the absolute value of each 
term, is not.  Thus tampering with its signs may destroy
the convergence of a series -- but not necessarily as 
we see by examining the series 
\vspace{.1in}

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

\noindent 
This is a convergent geometric series 
$ {\displaystyle \sum_{ n = 0 }^{ \infty } } \: x^n $
with $ x = - 1/2 $ and the series of absolute values 
\[ | 1 | + \left | - \frac{1}{2} \right |
+ \left | \frac{ 1 }{ 2^2 } \right |
+ \left | - \frac{ 1 }{ 2^3 } \right | 
+ \cdots = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots \] 
is also a convergent geometric series.  The point 
is that some convergent series having both positive and negative 
terms persist in their convergence after each term 
is replaced by its absolute value -- others do not.  The underlying
concept is called {\em absolute convergence}.
\vspace{.1in}

\noindent 
{\bf Definition}.  Absolute Convergence

The series 
\[ \sum_{ n = 1 }^{ \infty } \: a_n = a_1 
+ a_2 + a_3 + \cdots + a_n + \cdots \]
is said to be {\em absolutely convergent} if the series
\vspace{.1in}

\noindent 
(3) \hspace{1.5in} 
$ {\displaystyle \sum_{ n = 1 }^{ \infty }} \:
| a_n | = | a_1 | 
+ | a_2 | + | a_3 | + \cdots + 
| a_n | + \cdots $
\vspace{.1in}

\noindent 
is convergent.

According to his definition and the preceding discussion,
the series (2) is absolutely convergent; the series (1)
is not.  Of course, if $ a_n \geq 0 $ for all $n$, then
\[ \sum_{n=1}^{ \infty } | a_n | 
= | a_1 | + | a_2 | + | a_3 | 
+ \cdots = a_1 + a_2 + a_3 + \cdots . \]
Hence a convergent series of positive terms is always
absolutely convergent.
\vspace{.1in}

\noindent 
{\bf Example 1}.  The series
\vspace{.1in}

\noindent 
(a) \hspace{2in}
$ 1 - {\displaystyle \frac{ 1 }{ 3^2 }
+ \frac{ 1 }{ 5^2 } - \frac{ 1 }{ 7^2 }}
+ \cdots + 
{\displaystyle \frac{ ( - 1)^{ n + 1 } }{ ( 2n
- 1 )^2 }} + \cdots $
\vspace{.1in}

\noindent 
is absolutely convergent because 
\[ 1 + \frac{ 1 }{ 3^2 }
+ \frac{ 1 }{ 5^2 } + \frac{ 1 }{ 7^2 }
+ \cdots + 
{\displaystyle \frac{ 1 }{ ( 2n - 1 )^2 }} + \cdots \]
is convergent.  For instance, apply the integral 
test to $ f(x) = {\displaystyle \frac{ 1 }{ ( 2x - 1 )^2 }} $.
But the series
\vspace{.1in}

\noindent 
(b) \hspace{2in} 
$ 1 - {\displaystyle \frac{ 1 }{ 3 } 
+ \frac{ 1 }{ 5 } - \frac{ 1 }{ 7 } + \cdots + 
\frac{ ( - 1 )^{ n + 1 }}{ ( 2n - 1 ) }} + \cdots $
\vspace{.1in}

\noindent  
is not absolute convergent because 
\[ 1 +  {\displaystyle \frac{ 1 }{ 3 } 
+ \frac{ 1 }{ 5 } + \frac{ 1 }{ 7 } + \cdots + 
\frac{ 1 }{ ( 2n - 1 ) }} + \cdots \]
is a divergent series, as may be seen by applying the integral 
test to $ f(x) = {\displaystyle \frac{ 1 }{ 2x - 1 }} $.
\vspace{.1in}

It is traditional to give a name to a  series which converges but does 
not converge absolutely.
\noindent 
{\bf Definition}  The series $ \sum a_n $ is called {\em conditionally
convergent} if it is convergent but $ \sum | a_n | $ diverges.

This definition simply divides all convergent series 
into two groups -- those that converge absolutely 
and those that do not.  The series 
\[ \ln \: 2 - \ln \: 3/2 + 
\ln \: 4/3 - \ln \: 5/4 + \cdots \]
and
\[ 1 - 1 / \sqrt{ 2 } - 1 / \sqrt{ 3 } + 
1 / \sqrt{ 4 } + 1 / \sqrt{ 5 } -
1 / \sqrt{ 6 } - 1 / \sqrt{ 7 } + \cdots \]
are conditionally convergent:  each series converges
but the corresponding series, with all connecting
signs positive, diverge.  In such series you should think
of convergence as being the result of delicately balanced internal
cancellation, a balance that is shattered by making each connecting
sign positive.  On the other hand, in an absolutely convergent series like
\[ 1 - \frac{ 1 }{ 2 ! }
+ \frac{ 1 }{ 3 ! }
- \frac{ 1 }{ 4 ! } + \cdots , \]
you may change the signs any way you please
and the resulting series remains convergent.

We close this section with an important test for absolute convergence.
\vspace{.1in}

\noindent 
{\bf Theorem 15}.  The Ratio Test

In the series $ \sum a_n $ suppose that
$ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | {\displaystyle \frac{ a_{ n + 1 }}{ a_n }} \right | = L $.  Then

(a) if $ 0 \leq L < 1 $, the series converges absolutely;

(b) if $ L > 1 $, the series diverges; and

(c) if $ L = 1 $, the test gives no information concerning convergence.
\vspace{.1in}

\noindent 
{\bf Proof}.  We begin with (a).  The assertion is that
the rations 
$ | {\displaystyle \frac{ a_{ n + 1 }}{ a_n }} | $
tend to some number $L$ in the interval $ 0 \leq L < 1 $.  Since
the number $ r = {\displaystyle \frac{ L + 1 }{ 2 }} $ satisfies 
the inequality $ L < r < 1 $, it is possible to
choose $N$ large enough so that
\[ \left | {\displaystyle \frac{ a_{ n + 1 }}{ a_n }} \right | 
< r ~~ \mbox{if $ n \geq N $.} \]
In this inequality we give $n$, successively, the values
$ N , ~ N + 1 , ~ N + 2 ,... $ to obtain 
interlocking inequalities which, in tandem, imply that
\[ \begin{array}{l}
| a_{ N + 1 } |
< | a_N | r , 
\\ [0.2in]
| a_{ N + 2 } |
< | a_{ N + 1 } | r 
< | a_N | r^2 , 
\\ [0.2in]
| a_{ N + 3 } |
< | a_{ N + 2 } |
< | a_N | r^3 , 
\\ [0.2in]
~~~~~ \vdots \end{array} \]
At the $k$th level,
\[ | a_{ N + k } | < | a_N | r^k ~~~ ( k \geq 1 ) . \]
These inequalities are the key because, recalling
that $ r < 1 $, they show that the series
\[ | a_{ N + 1 } | + | a_{ N + 2 } |
+ | a_{ N + 3 } | + \cdots \] 
is dominated by the convergent geometric series
\[ | a_N | r + | a_N | r^2 + | a_N | r^3
+ \cdots = | a_N | \: \sum_{ n = 1 }^{ \infty } \: r^n . \]
Hence $ {\displaystyle \sum_{n=1}^{ \infty } } \: | a_{ N + k } | $
converges by the Comparison Test.  Now add on the first $N$
terms, and we see that
\[ \sum_{n=1}^{ \infty } \: | a_n | = \{ | a_1 | 
+ | a_2 | + \cdots + | a_N | \}
+ \sum_{n=1}^{ \infty } \: | a_{ N + k } | \]
converges.

Turning to (b) we need only observe that if $ L > 1 $
or $ L = \infty $ then the ratios ultimately become
and remain greater than 1 -- meaning that, for some $N$,
\[ \left | \frac{ a_{ n + 1 }}{ a_n } \right |
\geq 1 ~~ \mbox{if $ n \geq N$.} \]
Again, by giving $n$ the values $ N , ~ N + 1 , ~ 
N + 2 ,..., $ there results a string of inequalities
\[ | a_N | < 
| a_{ N + 1 } | < 
| a_{ N + 2 } | < \cdots \]
from which it is clear that the positive numbers
$ | a_n | $ increase for $ n \geq N $ and,
in particular, $ {\displaystyle \lim_{ n \rightarrow \infty }} \:
a_n \neq 0 $.
Hence the series $ \sum a_n $ must diverge.

The proof of part (c) is left as a problem.  $ \ddagger $
\vspace{.1in}

In a series of positive terms, the Ratio Test is applied without
absolute values, in the form
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  \frac{ a_{ n
+ 1 }}{ a_n } = L . \]
Then $ \sum a_n $ converges if $ L < 1 $, diverges
if $ L > 1 $, and the test is inconclusive if $ L = 1 $.
\vspace{.1in}

\noindent 
{\bf Example 2}.  Use the Ratio Test to determine the convergence
or divergence of the following series.
\vspace{.1in}

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

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

(c) $ {\displaystyle \sum_{n=1}^{ \infty } } \: 
{\displaystyle \frac{ 1 }{ 3n + 1 } = \frac{ 1 }{ 4 } + \frac{ 1 }{ 7 } +
\frac{ 1 }{ 10 }} + \cdots $
\vspace{.1in}

\noindent 
{\bf Solution}.  (a) Here $ a_n = {\displaystyle \frac{ n }{ 2^n }}
, a_{ n + 1 } = {\displaystyle \frac{ n + 1 }{ 2^{ n + 1 } }} $
and
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ a_{ n + 1 }}{ a_n } 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  \frac{ n + 
1 }{ 2^{ n + 1 }} \frac{ 2^n }{ n } 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  \frac{ 1 }{2}
\frac{ n + 1 }{ n } = \frac{ 1 }{ 2 } . \]
Because $ {\displaystyle \frac{1}{2} } < 1 $,
the series converges.  $ \ddagger $
\vspace{.1in}

\noindent  
(b) We have 
\[ \frac{ a_{ n + 1 }}{ a_n }
= \frac{ ( n + 1 )^{ n + 1 }}{ (n+1) ! } 
\frac{ n ! }{ n^n } 
= \frac{ (n+1) (n+1) ^n }{ (n+1) n ! } \] 
and
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ a_{ n + 1 }}{ a_n } = {\displaystyle \lim_{ n \rightarrow \infty }} \:
\left ( 1 + \frac{ 1 }{ n } \right )^n = e . \]
Because $ e > 1 $, the series diverges.  $ \ddagger $
\vspace{.1in}

\noindent 
(c) In this series the Ratio Test is useless inasmuch as
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ a_{ n + 1 }}{ a_n }
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ 3n + 1 }{ 3 (n+1) + 1 } = 1 . \]
That the series is actually divergent may
be seen by applying the Integral Test with
$ f(x) = {\displaystyle \frac{ 1 }{ 3x + 1 }} $.  $ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 6}.  Use the Ratio Test to show that series 
\[ \sum_{n=1}^{ \infty } \: ( - 1 )^{ n + 1 }  
\frac{ 1 \cdot 3 \cdot 5 \cdots ( 2n - 1 ) }{ ( 2n ) ! }
= \frac{1}{2} - \frac{ 1 \cdot 3  }{ 4 ! }
+ \frac{ 1 \cdot 3 \cdot 5 }{ 6 ! } - 
\frac{ 1 \cdot 3 \cdot 5 \cdot 7 }{ 8 ! } + \cdots \]
is absolutely convergent.
\vspace{.1in}

\noindent 
{\bf Solution}.  In this more complicated example
\[ a_n = ( - 1 )^{ n + 1 }
\frac{ 1 \cdot 3 \cdot 5 \cdots ( 2n - 1 ) }{ ( 2 n ) ! } , 
a_{ n + 1 } = ( - 1 )^{ n + 2 } 
\frac{ 1 \cdot 3 \cdot 5 \cdots ( 2n - 1 ) ( 2n + 1 ) }{ ( 2 n + 2 ) ! } \]
and
\[ \begin{array}{rl}
{\displaystyle \frac{ a_{ n + 1 }}{ a_n }}
& \hspace{-.1in} = - 
{\displaystyle \frac{ 1 \cdot 3 \cdot 5 \cdots ( 2n - 1 ) 
( 2n + 1 ) }{ ( 2n + 2 ) ! } \: 
\frac{ ( 2n ) ! }{ 1 \cdot 3 \cdot 5 \cdots ( 2n - 1 ) }}
\\ [0.3in]
& \hspace{-.1in} = {\displaystyle \frac{ 2n + 1 }{ ( 2n + 1 )
( 2n + 2 ) }} . ~~ \mbox{Note that $ [ ( 2n + 1 ) ! 
= ( 2n + 2 ) ( 2n + 1 ) ( 2n ) ! ] $} 
\end{array} \]
Hence
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | \frac{ a_{ n + 1 }}{ a_n } \right | 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ 1 }{ 2 n + 2 } = 0 \]
and the series converges absolutely.  $ \ddagger $
\vspace{.1in}

At the elementary level, the Ratio Test is 
our best convergence test, especially for 
series in which $ a_n $ contains $ n ! $; $ ( 2n ) ! $
or some other factorial expression.



\chapter{Power Series}
 

The ideas of the foregoing sections culminate
in the study of infinite series in which 
each term is a function.  Chief among these are 
{\em \index{power series} power series}, series that are important primarily because they
provide a natural and powerful
method of defining and studying functions -- the ordinary
as well as the exotic.

\section{Power Series}
By a {\em power series in} $x$ we mean a series of the form
\vspace{.1in}

\noindent 
(1) \hspace{2in} $ {\displaystyle \sum_{n=0}^{ \infty }} \:  
a_n x^n = a_0 
+ a_1 x + a_2 x^2 + \cdots + 
a_n x^n + \cdots $
\vspace{.1in}

\noindent 
in which the $ a_n $ denote real numbers called
{\em coefficients} and $x$ is considered a {\em variable}.

The first question in discussing a power series is this:
For which values of $x$ does the series converge?
It is the nature of a power series in $x$ to \index{converge absolutely} converge absolutely
in some symmetric open interval of the form $ | x | < r $
and to diverge if $ | x | > r $ or else to converge 
absolutely for $ - \infty < x < \infty $.
%The first possibility is illustrated in Figure 14.8.  
This
open interval of absolute convergence
%\vspace{2in}

\noindent 
is most conveniently found by applying the Ratio Test.
\vspace{.1in}

\noindent 
{\bf Example 1}.  Find the values of $x$ at which
the power series 
\[ {\displaystyle \sum_{n=1}^{ \infty }} \: ( - 1 )^{ n + 1 }
\frac{ x^n }{ n \cdot 3^n } = \frac{ 1 }{ 3 }
x - \frac{ 1 }{ 2 \cdot 3^2 } x^2 
+ \frac{ 1 }{ 3 \cdot 3^3 } x^3 + \cdots \]
converges.
\vspace{.1in}

\noindent 
{\bf Solution}.  In this power series $ a_0 = 0 , ~
a_1 = \frac{ 1 }{ 3 } , ~ a_2 = 
- \frac{ 1 }{ 2 \cdot 3^2 } $, etc.  To apply
the Ratio Test, we put
\[ u_n = ( - 1 )^{ n + 1 } 
\frac{ x^n }{ n \: 3^n } , ~~
u_{n+1} = ( - 1 )^{ n + 2 }
\frac{ x^{ n + 1 }}{ (n+1) 3^{ n + 1 } } \]
and calculate
\[ \frac{ u_{n+1} }{ u_n } = (-1)^{ n + 2 } 
\frac{ x^{ n + 1 }}{ (n+1) 3^{ n + 1 } } 
\frac{ n \: 3^n }{ (-1)^n 3^{n+1} x^n } = 
- \frac{ n }{ 3 (n+1) } x . \]
For fixed $x$,
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | \frac{ u_{n+1} }{ u_n } \right |
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | - \frac{ n }{ 3 (n+1) } x \right | = 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ n }{ 3 (n+1) } \: | x | = 
\frac{ | x | }{ 3 } . \]
According to the Ratio Test, the series converges
absolutely if $ {\displaystyle \frac{ | x | }{ 3 }} < 1 $
and diverges of $ {\displaystyle \frac{ | x | }{ 3 }} > 1 $.  Thus
the series converges absolutely for $x$ in the open interval
\[ | x | < 3 , ~~ - 3 < x < 3 \]
and diverges if
\[ x < - 3 ~~ \mbox{or} ~~ x > 3 . \]

At $ x = 3 $, the power series reduces 
to the conditionally convergent alternating series 
\[ {\displaystyle \sum_{n=1}^{ \infty }} \: (-1)^{n+1} 
\frac{ 3^n }{ n \: 3^n } = 
{\displaystyle \sum_{n=1}^{ \infty }} \: \frac{ (-1)^{n+1} }{ n } = 1 -
\frac{ 1 }{ 2 } + \frac{ 1 }{ 3 } - \cdots \]
and at $ x = - 3 $, it reduces to the divergent series
\[ {\displaystyle \sum_{n=1}^{ \infty }} \: (-1)^{n+1} 
\frac{ ( - 3 )^n }{ n \: 3^n } = 
{\displaystyle \sum_{n=1}^{ \infty }} \: 
\frac{ (-1)^{ 2 n + 1 }}{ n } 
= - 1 - \frac{ 1 }{ 2 } - \frac{ 1 }{ 3 } - \cdots . \]

To sum up:  This power series converges for every $x$
in the interval $ - 3 < x \leq 3 $ and diverge otherwise.
In the open interval $ - 3 < x < 3 $, the convergence
is absolute, at the endpoint $ x = 3 $ the convergence is conditional.
$ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 2}.  Find the values of $x$ at which the power series 
\[ {\displaystyle \sum_{n=0}^{ \infty }} \: 
\frac{ x^n }{ n ! } = 1 
+ x + \frac{ x^2 }{ 2 ! } + 
\frac{ x^3 }{ 3 ! } + \cdots \]
converges.
\vspace{.1in}

\noindent 
{\bf Solution}.  To properly interpret the summation, it is necessary
to remember that $ 0 ! = 1 , ~~ 1 ! = 1 $
and $ x^0 = 1 $.

Applying the Ratio Test with $ u_n = x^n / n ! $ gives
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | \frac{ u_{n+1} }{ u_n } \right |
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | \frac{ x^{n+1} }{ (n+1) ! } 
\frac{ n ! }{ x^n } \right | 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ | x | }{ (n+1) } = 0 , \]
no matter what the values of $x$.  Conclusion:  the series
converges absolutely for all $x$, for $ - \infty 
< x < \infty $.  Later we shall show that this series 
has sum $ e^x $.  $ \ddagger $
\vspace{.1in}

The following theorem provides a precise description of the
possible situations than can occur in analyzing the convergence
of a power series in $x$.  A proof is outlined in the problems.
\vspace{.1in}

\noindent 
{\bf Theorem 18}.  Given a power series in $x$, exactly one of the
following hold:  

\begin{verse}
(a) The series converges only at $ x = 0 $.
\vspace{.1in}

(b) For some positive number $r$, the series 
converges absolutely in the open interval $ | x | < r $
and diverges if $ | x | > r $.
\vspace{.1in}

(c) The series converges absolutely for all $x$.
\end{verse}
\vspace{.1in}

Several remarks are in order.  First, the situation
in (a) is extreme and uninteresting because to be useful a 
power series must converge throughout some interval.
The series
\[ 1 + x + 2 ! x^2 
+ 3 ! x^3 + 4 ! x^4 + \cdots +
n ! x^n + \cdots \]
illustrates the possibility.  Like every power series in $x$,
it converges at $ x = 0 $, but 
if $ x \neq 0 $ the Ratio Test gives
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  \left | 
\frac { (n+1) ! x^{n+1} }{ n ! x^n } \right | 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  (n+1) \:
| x | = \infty \]
and the series diverges.  The problem is caused by the rapid
rate at which the coefficients $ a_n = n ! $ increase.

Second, the endpoint $r$ and $ - r $ of the 
open interval $ - r < x < r $ of absolute 
convergence set forth in part (b) must be 
analyzed separately.  At these two points the series may diverge
or converge, either absolutely or conditionally.

Thus the convergence set may be a single point,
any of the intervals $ ( - r , r ) , ~~
( - r , r ] , ~~ [ - r , r ) , ~~
[ - r , r ] $, or $ ( - \infty , \infty ) $.
In all cases we shall refer to the set of $x$ where a power series 
converges as its {\em \index{interval of convergence} interval of convergence}
and to $r$ as its {\em \index{radius of convergence} radius of convergence}.  The possibilities
corresponding to parts (a), (b), and (c) respectively are that
$r$ is $0$, $r$ is finite, and $r$ is infinite.
\vspace{.1in}

\noindent 
{\bf Example 3}.  Find the interval of convergence and the radius of 
convergence of the power series 
\[ {\displaystyle \sum_{n=1}^{ \infty }} \: 
\frac{ 2^n }{ n^2 } x^n 
= 2 x + x^2 + 
\frac{ 8 }{ 9 } x^3 + \cdots . \]
\vspace{.1in}

\noindent 
{\bf Solution}.  The Ratio Test with
\[ u_n = \frac{ 2^n }{ n^2 } x^n ~~ \mbox{and} ~~
u_{n+1} = \frac{ 2^{n+1} }{ (n+1) ^2 } x^{n+1} \]
gives
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  \left | 
\frac{ u_{n+1} }{ u_n } \right | 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | \frac{ 2^{n+1} }{ (n+1) ^2 } x^{n+1} \cdot 
\frac{ n^2 }{ 2^n x^n } \right | 
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ 2 n^2 }{ (n+1) ^2 } \: | x | 
= 2 \: | x | . \]
Hence the power converges absolutely in the open interval 
$ 2 \: | x | < 1 $ or $ | x | < 1/2 $, that is, for
\[ - 1/2 < x < 1 / 2 ; \]
and diverges if
\[ x < - 1/2 ~~ \mbox{or} ~~ x > 1/2 . \]
Hence the radius of convergence is $ r = 1/2 $.

The series is also absolutely convergent at $ x = 1/2 $,
where it reduces to the convergent $p$-series
\[ {\displaystyle \sum_{n=1}^{ \infty }} \: \frac{ 2^n }{ n^2 }
\left ( \frac{ 1 }{ 2 } \right )^n = 
{\displaystyle \sum_{n=1}^{ \infty }} \: \frac{ 1 }{ n^2 }
= 1 + \frac{1}{2^2} +
\frac{ 1 }{ 2^n3 } + \cdots , \]
and at $ x = - 1/2 $ where it reduces to the absolutely
convergent alternating series
\[ {\displaystyle \sum_{n=1}^{ \infty }} \: 
\frac{ 2^n }{ n^2 } \left ( - \frac{ 1 }{ 2 } \right )^n 
= = {\displaystyle \sum_{n=1}^{ \infty }} \: \frac{ (-1)^n }{ 2^n } 
= - \frac{ 1 }{ 2 } + \frac{1}{2^2} - 
- \frac{1}{3^2} + \cdots . \]

Consequently, the interval of convergence is $ - 1/2 
\leq x \leq 1/2 $ and the convergence is absolute throughout.  $ \ddagger $
\vspace{.1in}


The series
\vspace{.1in}

\noindent 
(2) \hspace{1in} $ {\displaystyle \sum_{n=1}^{ \infty }} \: 
a_n ( x - a )^n = a_0 + a_1 
( x - a ) + a_2 ( x - a )^2 + \cdots + 
a_n ( x-a ) ^n + \cdots $,
\vspace{.1in}

\noindent 
obtained from (1) by replacing $x$ by $ x - a $, is called
a \index{{\em power series in}  $ x - a $} {\em power series in}  $ x - a $.  Its interval of absolute 
convergence is symmetric about $a$.
\vspace{.1in}

\noindent 
{\bf Example 4}.  Find the radius of convergence and the 
interval of convergence of
the power series
\[ {\displaystyle \sum_{n=0}^{ \infty }} \: 
\frac{ 1 \cdot 3 \cdot 5 \cdot 7 \cdots 
( 2 n + 1 ) }{ 2^{ 2 n } n ! } 
( x - 1 )^n = 1 + \frac{ 1 \cdot 3 }{ 2^2 } 
( x - 1 ) + \frac{ 1 \cdot 3 \cdot 5 }{ 2^4 \cdot 2 ! } 
( x - 1 )^2 + \frac{ 1 \cdot 3 \cdot 5 \cdot 7 }{ 2^6 
\cdot 3 ! } ( x - 1 )^3 + \cdots . \]
\vspace{.1in}

\noindent 
{\bf Solution}.  We apply the Ratio Test with
\[ u_n = \frac{ 1 \cdot 3 \cdot 5 \cdots ( 2 n + 1 ) }{ 2^{ 2^n } n ! }
( x - 1 )^n ~~ \mbox{and} ~~
u_{n+1} = \frac{ 1 \cdot 3 \cdot 5 \cdots ( 2 n + 1 ) 
( 2 n + 3 ) }{ 2^{ 2 n + 2 } (n+1) ! } 
( x - 1 )^{n+1} . \]
Note:  The next odd integer after $ 2 n + 1 $ is $ 2 (n+1) + 1 
= 2 n + 3 $ and $ (n+1) ! = (n+1) n ! $.
Thus
\[ \frac{ u_{n+1} }{ u_n } = 
\frac{ 1 \cdot 3 \cdot 5 \cdots ( 2 n + 1 ) ( 2 n + 3 ) }{ 2^{ 2 n + 2 }
(n+1) ! } ( x - 1 )^{n+1} 
\frac{ 2^{ 2 n } n ! }{ 1 \cdot 3 \cdot 
5 \cdots ( 2 n + 1 ) } \: \frac{ 1 }{ ( x - 1 )^n } 
= \frac{ 2 n + 3 }{ 2^2 (n+1) } ( x - 1 ) \]
and
\[ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\left | \frac{ u_{n+1} }{ u_n } \right |
= {\displaystyle \lim_{ n \rightarrow \infty }} \:  
\frac{ 2 n + 3 }{ 4 (n+1) } \: | x - 1 | = \frac{1}{2} 
\: | x - 1 | . \]
Consequently, there is absolute convergence where $ \frac{1}{2} \:
| x - 1 | < 1 $
or $ | x - 1 | < 2 $.  This means that the radius of convergence is $ r = 2 $
and that the interval $ - 1 < x < 3 $, with center $1$ and radius
$2$, is an interval of absolute convergence.

Of course, the series diverges where $ | x - 1 | > 2 $.
It also diverges at the endpoints $ x = 3 $ and $ x = - 1 $
because the terms of the corresponding series 
do not tend to zero as $n$ increases.  For instance, at $ x = 3 $
the original power series reduces to
\[ {\displaystyle \sum_{n=0}^{ \infty }} \: 
\frac{ 1 \cdot 3 \cdot 5 \cdots ( 2 n + 1 ) }{ 2^{ 2 n } n ! }
( 2 )^n \frac{ 1 \cdot 3 \cdot 5 \cdots ( 2 n + 1 ) }{ 2^{ n } n ! } \]
and
\[ {\displaystyle \sum_{n=0}^{ \infty }} \: 
\frac{ 1 \cdot 3 \cdot 5 \cdots ( 2 n + 1 ) }{ 2^{ n } n ! } 
= \frac{ 1 \cdot 3 \cdot 5 \cdots ( 2 n + 1 ) }{ 2 \cdot 4 \cdot 6 \cdots 2 n } 
= \frac{ 3 }{ 2 } \cdot \frac{ 5 }{ 4 } 
\cdot \frac{ 7 }{ 6 } \cdots \frac{ 2 n + 1 }{ 2 n } > 1 \]
for all $n$.  Therefore $ - 1 < x < 3 $ is 
the interval of convergence.  $ \ddagger $ 


\section{Power Series Calculus}
\vspace{.1in}

On its interval of convergence a power series 
$ {\displaystyle \sum_{n=0}^{ \infty }} \: a_n x^n $
defines a function
\vspace{.1in}

\noindent 
(1) \hspace{2in} 
$ f(x) = {\displaystyle \sum_{n=0}^{ \infty }} \: 
a_n x^n = a_0 + a_1 x + a_2 x^2 + \cdots $
\vspace{.1in}

\noindent 
and inside that interval of convergence $f$ can
be differentiated and integrated according to the
following theorem whose proof we omit.
\vspace{.1in}

\noindent 
{\bf Theorem 19}.  Suppose the series (1) has a nonzero
radius of convergence $r$.  Then
\vspace{.1in}

\noindent 
(2) \hspace{1in} $ f ^{'} (x) 
= {\displaystyle \sum_{n=1}^{ \infty }} \: n a_n x^{ n - 1 }
= a_1 + 2 a_2 x^2 + 3 a_3 x^3 + \cdots $ if $ | x | < r $
\vspace{.1in}

\noindent  
and
\vspace{.1in}

\noindent 
(3) \hspace{1in} $ \begin{array}{rl}
{\displaystyle \int_0^x } \: f (t) \: d t 
& \hspace{-.1in} = 
{\displaystyle \int_0^x } \: \left ( {\displaystyle \sum_{n=0}^{ \infty }} 
\: a_n t^n \right ) \: 
d t = {\displaystyle \sum_{n=0}^{ \infty }} \: a_n 
{\displaystyle \int_0^x } \: t^n \: d t 
= {\displaystyle \sum_{n=0}^{ \infty }} \: 
{\displaystyle \frac{ a_n }{ n + 1 }} x^{ n + 1 } \\ [0.3in] 
& \hspace{-.1in} = a_0 x + {\displaystyle \frac{ a_1 }{ 2 } x^2 + 
\frac{ a_2 }{ 3 }} x^3 + \cdots ~~ \mbox{if $ | x | < r $.}
\end{array} $
\vspace{.1in}

\noindent 
Further, the three series (1), (2), and (3) have the same radius of convergence.

The point is that inside its radius of convergence a power series is a 
continuous and differentiable function whose derivatives
and integrals are found by operating on the series 
{\em t\index{erm-by-term} erm-by-term}.  It is important to understand that the 
operations in (2) and (3) can be applied repeatedly.
Thus the derivative of $ f ^{'} (x) $ in (2) is
\[ f ^{''} (x) = {\displaystyle \sum_{n=2}^{ \infty }} \: 
n ( n - 1 ) x^{ (n - 2) } ~~ \mbox{if $ | x | < r $.} \]
In other words, insofar as differentiation and integration
are concerned, we operate on a power series exactly
as if it were a polynomial.  Of course, like
any other convergent series,
power series can be multiplied by a constant and
added together term by term.

In this section we shall apply (2) and (3) to the 
problem of expressing some of the elementary functions
of calculus as power series.  That is, given a 
function $ f(x) $ we will attempt to write it as a 
convergent power series
\[ f(x) = a_0 + a_1 x + a_2 x^2 
+ \cdots + a_n x^n + \cdots \]
in some interval $ | x | < r $.  The problem is to find the
coefficient $ a_n $, and the process is called {\em \index{expanding
$f$ in a power series} expanding
$f$ in a power series} or {\em \index{representing
$ f $ by a power series} representing
$ f $ by a power series}.  Our initial efforts revolve 
around the geometric series
\vspace{.1in}

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

\noindent 
{\bf Example 1}.  Find a power series representation 
for $ {\displaystyle \frac{ 1 }{ 3 - 2 x }} $.
\vspace{.1in}

\noindent 
{\bf Solution}.  First we write the function 
in the form $ {\displaystyle \frac{ 1 }{ 3 }}
\: {\displaystyle \frac{ 1 }{ 1 - \frac{ 2 }{ 3 } x }} $,
then apply the geometric series representation (4) with 
$ {\displaystyle \frac{ 2 }{ 3 }} x $ in place of $x$.  The result is
\[ \begin{array}{rl}
{\displaystyle \frac{ 1 }{ 3 - 2 x } = \frac{ 1 }{ 3 }} \:
{\displaystyle \frac{ 1 }{ 1 - \frac{ 2 }{ 3 } x } } 
& \hspace{-.1in} = {\displaystyle \frac{ 1 }{ 3 }}
\left [ 1 + {\displaystyle \frac{ 2 }{ 3 } x + 
\frac{ 2^2 }{ 3^2 } x^2 
+ \frac{ 2^3 }{ 3^3 } } x^3 + \cdots \right ] 
\\ [0.3in]
& \hspace{-.1in} = {\displaystyle \frac{ 1 }{ 3 }
+ \frac{ 2 }{ 3^2 } x 
+ \frac{ 2^2 }{ 3^3 }} x^3 
+ \cdots \left ( | x | < {\displaystyle \frac{ 3 }{ 2 }} \right )
. ~~~~~~ \ddagger \end{array} \]
\vspace{.1in}

\noindent 
{\bf Example 2}.  Find a power series representation for 
$ {\displaystyle \frac{ 1 }{ ( 1 - x )^2 }} $.
\vspace{.1in}
 
\noindent 
{\bf Solution}.  
Let $ f(x) = {\displaystyle \frac{ 1 }{ 1 - x }} $.
Then, from (4),
\[ f(x) = 1 + x + x^2 + x^3 
+ \cdots + x^n + \cdots ~~ ( | x | < 1 ) . \]
According to (2), the derivative $ f ^{'} (x) = 
{\displaystyle \frac{ 1 }{ ( 1 - x )^2 }} $ is the
term-by-term derivative of the series.  Thus
\[ {\displaystyle \frac{ 1 }{ ( 1 - x )^2 }} 
= 1 + 2 x + 3 x^2 + \cdots + 
n x^{ n - 1 } + \cdots ~~ ( | x | < 1 ) . ~~~ \ddagger \]
\vspace{.1in}

\noindent 
{\bf Example 3}.  Find a power series expansion for
$ \ln \: ( 1 + x ) $.
\vspace{.1in}

\noindent 
{\bf Solution}.  Because $ 1 / ( 1 + x ) $ is the derivative of
$ \ln \: ( 1 + x ) $, we start with the series 
\[ \frac{ 1 }{ 1 + t } = 1 - t + t^2 - t^3 + \cdots + (-1) 
^n t^n + \cdots ~~~ ( | t | < 1 ) \]
and integrate both sides.  First,
\[ \int_0^x \: {\displaystyle \frac{ 1 }{ 1 + t }} \: 
d t = \ln \: ( 1 + t ) ]_{ t = 0 }^x \: = \ln \: ( 1 + x ) . \]
Hence, following (3),
\[ \ln \: ( 1 + x ) = \int_0^x \: \frac{ 1 }{ 1 + t } \: d t 
= \int_0^x \: \left ( \sum_0^{ \infty } \:
(-1) ^n t^n \right ) \: d t = {\displaystyle \sum_{n=0}^{ \infty }}
\: (-1) ^n \int_0^x \: t^n \: d t 
= {\displaystyle \sum_{n=0}^{ \infty }} \: 
(-1) ^n 
\frac{ x^{ n + }}{ n + 1 } , \]
that is,
\vspace{.1in}

\noindent 
(5) \hspace{1in} $ \ln \: ( 1 + x ) 
= x - \frac{ x^2 }{ 2 } + 
\frac{ x^3 }{ 3 } - 
\frac{ x^4 }{ 4 } 
+ \cdots + 
(-1) ^n \frac{ x^{ n + 1 }}{ n + } + \cdots $ 
\vspace{.1in}

\noindent 
if $ | x | < 1 $.  $ \ddagger $
\vspace{.1in}

\noindent 
{\bf Remark 1}.  It is important to understand that the equality
in (5) is good only so long as $ | x | < 1 $.
This is because the starting point 
was the geometric series which itself converges only where
$ | x | < 1 $.  In contrast, the function 
$ \ln \: ( 1 + x ) $ is defined and has derivatives
of all orders where $ x > - 1 $.
\vspace{.1in}

\noindent 
{\bf Remark 2}.  The series expansion in (5) provides a reasonable
procedure for computing $ \ln \: 
( 1 + x ) $ if $ x $ is near zero.  For example,
being an alternating series we know that 
\[ \begin{array}{rl}
\ln \: {\displaystyle \frac{ 25 }{ 24 }} 
= \ln \: \left ( 1 + {\displaystyle \frac{ 1 }{ 24 }} \right )
& \hspace{-.1in} \approx {\displaystyle \frac{ 1 } 24 
- \frac{1}{2} - \frac{1}{2}} \left ( {\displaystyle \frac{1}{24} }
\right )^2 
+ {\displaystyle \frac{ 1 }{ 3 } } \left ( 
{\displaystyle \frac{ 1 }{ 24 }} \right )^3 
\\ [0.3in]
& \hspace{-.1in} \approx 0.04166667 - 
0.00086806 + 0.00002411 
\\ [0.3in]
& \hspace{-.1in} \approx 0.04082272 \end{array} \]
with an error in absolute value less than
$ {\displaystyle \frac{ 1 }{ 4 }} \left ( {\displaystyle \frac{ 1 }{24}}
\right )^4 < 0.0000008 $.
Hence $ \ln \: {\displaystyle \frac{ 25 }{ 24 }} 
= 0.04082 $, accurate to five decimal digits.
\vspace{.1in}

\noindent 
{\bf Example 5}.  Find a power series representation for $ \ln \: 
{\displaystyle \frac{ 1 + x }{ 1 - x }} $.
\vspace{.1in}

\noindent 
{\bf Solution}.  In representation (5) 
\[ \ln \: ( 1 + x ) = x - 
\frac{ x^2 }{ 2 } 
+ \frac{ x^3 }{ 3 } 
- \frac{ x^4 }{ 4 } + \cdots , \]
we replace $x$ by $ - x $ to obtain 
\[ \ln \: ( 1 - x ) = - x - 
\frac{ x^2 }{ 2 } 
-  \frac{ x^3 }{ 3 } 
- \frac{ x^4 }{ 4 } + \cdots \]
and subtract the second series from the first.
The resulting series representation is
\vspace{.1in}

\noindent 
(6) \hspace{1in} 
$ \ln \: {\displaystyle \frac{ (1+x) }{ (1-x) }} = 2 \left ( x + 
{\displaystyle \frac{ x^3 }{ 3 }} + {\displaystyle \frac{ x^5 }{ 5 }} + 
\cdots + {\displaystyle \frac{ x^{ 2 n - 1 }}{ 2 n - 1 }} + 
\cdots \right ) , $
\vspace{.1in}

\noindent
valid if $ | x | < 1 $.
\vspace{.1in}

\noindent 
{\bf Example 6}.  Find a power series representation 
for $ \tan^{-1} \: x $.
\vspace{.1in}

\noindent 
{\bf Solution}.  Again the geometric series 
is involved because $ 1 / ( 1 + x^2 ) $
is the derivative of $ \tan^{-1} \: x $.  Integration of
both sides of
\[ \frac{ 1 }{ 1 + t^2 }
= 1 - t^2 + t^4 - t^6 + \cdots \]
from $0$ to $x$ gives
\[ \begin{array}{rl}
{\displaystyle \int_0^x } \: {\displaystyle \frac{ 1 }{ 1
+ t^2 }} \: d t & \hspace{-.1in} = 
{\displaystyle \int_0^x } \: ( 1 - t^2 
+ t^4 - t^6 + \cdots ) \: d t \\ [0.3in]
& \hspace{-.1in} = x - {\displaystyle \frac{ x^3 }{ 3 } 
+ \frac{ x^5 }{ 5 } - \frac{ x^7 }{ 7 } } + \cdots . \end{array} \]
Hence
\vspace{.1in}

\noindent 
(7) \hspace{1in} 
$ \tan^{-1} \: x = x 
- {\displaystyle \frac{ x^3 }{ 3 }
+ \frac{ x^5 }{ 5 } - \frac{ x^7 }{ 7 }} + \cdots + (-1) ^n \: 
{\displaystyle \frac{ x^{ 2 n + 1 }}{ 2 n + 1 }} + \cdots $
\vspace{.1in}

\noindent 
if $ | x | < 1 $. $ \ddagger $
\vspace{.1in}

An analysis of the remainder in (7) shows that the
representation is valid at the endpoints $ x = 
\pm 1 $.  Since $ \tan^{-1} \: 1 = \pi / 4 $, this means that
\vspace{.1in}

\noindent 
(8) \hspace{2in} 
$ {\displaystyle \frac{ \pi }{ 4 } = 1 
- \frac{ 1 }{ 3 } + \frac{ 1 }{ 5 } -
\frac{ 1 }{ 7 } + \frac{ 1 }{ 9 } - \frac{ 1 }{ 11 }} + \cdots $,
\vspace{.1in}

\noindent 
but its agonizingly slow convergence makes it a 
poor vehicle for computing $ \pi $.
It is better to observe that $ \tan^{-1} \: ( 1 / \sqrt{ 3 } ) = \pi / 6 $ and
hence that
\vspace{.1in}

\noindent 
(9) \hspace{2in} 
$ {\displaystyle \frac{ \pi }{ 6 }} 
= {\displaystyle \frac{ 1 }{ \sqrt{ 3 }}
- \frac{ 1 }{ 3 }} 
\left ( {\displaystyle \frac{ 1 }{ \sqrt{ 3 } }}
\right ) ^3 + 
{\displaystyle \frac{ 1 }{ 5 }} \left ( 
{\displaystyle \frac{ 1 }{ \sqrt{ 3 } }} \right )^5 - \cdots . $
\vspace{.1in}

\noindent 
Another approach to computing $ \pi $ exploits the 
trigonometric identity
\[ \frac{ \pi }{ 4 } = \tan^{-1} \: \frac{1}{2} 
+ \tan^{-1} \: \frac{ 1 }{ 3 } \]
Using the first six terms in (7), we find that
\[ \tan^{-1} \: \frac{1}{2} \approx 0.45363989 
~~ \mbox{with error $ < 0.0000094 $,} \]
\[ \tan^{-1} \: \frac{1}{3} \approx 0.32175051
~~ \mbox{with error $ < 0.00000005 $} \]
and hence that $ \pi \approx 3.141565 $ with error
$ < 0.00004 $.  If we note further that the seventh
term in (7) is positive if $ x > 0 $, we see that our
estimate is too small and that $ \pi \approx 3.1416 $ accurate
to four decimal places.


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

As we have just seen, special purpose techniques
can be used to find power series representations
for certain of the elementary calculus functions.
For other such functions it is preferable to adopt
a more theoretical approach based on Taylor's formula
  About some point $ x = a $
that formula represents $f$ in the form
\vspace{.1in}

\noindent 
(1) \hspace{1in} $ f(x) = f(x) 
+ f ^{'} (x) ( x-a ) + {\displaystyle \frac{ f ^{''} (a) }{ 2 ! }}
( x-a ) ^2 + \cdots + f^{ (n) } (a) 
{\displaystyle \frac{ ( x-a ) ^n }{ n ! }} + R_n (x) $
\vspace{.1in}

\noindent 
where the remainder $ R_n $ is given by
\vspace{.1in}

\noindent 
(2) \hspace{2in} $ R_n (x) 
= f^{ (n+1) } (c) 
{\displaystyle \frac{ ( x-a ) ^{n+1} }{ (n+1) ! }} $
\vspace{.1in}

\noindent 
for some $c$ between $a$ and $x$.  Recall that we 
have frequently used the polynomial part to approximate
$ f(x) $ and it is natural to examine that approximation 
as $ n \rightarrow \infty $.  Specifically, suppose that
$f$ is a function that has derivatives of all orders
(infinitely differentiable) in some interval about $a$
and that
\vspace{.1in}

\noindent 
(3) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \: 
R_n (x) = 0 $
\vspace{.1in}

\noindent 
for every $x$ in that interval.  Denoting the polynomial
part in (1) by $ P_n (x) $, we have 
\[ f(x) = P_n (x) + R_n (x) \]
and
\[ \begin{array}{rl}
f(x) & \hspace{-.1in} = 
P_n (x) + R_n (x) 
\\ [0.2in]
& \hspace{-.1in} = 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  [ 
P_n (x) + R_n (x) ] 
\\ [0.2in]
& \hspace{-.1in} = 
{\displaystyle \lim_{ n \rightarrow \infty }} \:  P_n (x) . 
\end{array} \]
This mean that the polynomials $ P_n (x) $ form
the partial sums of a series whose limit 
is $ f(x) $.
\vspace{.1in}

\noindent 
(4) \hspace{1in} $ f(x) 
= f (a) + f ^{'} (a) ( x-a ) + f ^{''} (a) 
{\displaystyle \frac{ ( x-a ) ^2 }{ 2 ! }}
+ \cdots + f^{ (n) }
(a) {\displaystyle \frac{ ( x-a ) ^n }{ n ! }} 
+ \cdots . $
\vspace{.1in}

\noindent 
Put differently, the series converges throughout some interval
about $a$ and its sum is $ f(x) $.  Note that the 
expansion (4) is a power series in $ x - a $;
it is called the \index{{\em Taylor series for $f$ at} $ x = a $} {\em Taylor series for $f$ at} $ x = a $.
If $ a = 0- $ in (4) the expansion reduces to
\vspace{.1in}

\noindent 
(5) \hspace{1in} $ f(x) = f (0) + f ^{'} (0)
x + f ^{''} (0) 
{\displaystyle \frac{ x^2 }{ 2 ! }} + \cdots + 
f^{ (n) } (0) {\displaystyle \frac{ x^n }{ n ! }} + 
\cdots . $
\vspace{.1in}

\noindent 
This representation of $f$ as a power series in $x$ is
called the {\em \index{Maclaurin series for $f$} Maclaurin series for $f$}.

The crucial point is that the validity of (4) or (5)
depends directly on the limit calculation in (3).
We begin by rederiving the power series expansion 
for $ e^x $.
\vspace{.1in}

\noindent 
{\bf Example 1}.  Find the Maclaurin series for $ e^x $.
\vspace{.1in}

\noindent 
{\bf Solution}.  Since $ f(x) = e^x $ is its own derivative,
$ f^{ (n) } (x) = e^x $ for all $n$ and 
$ f^{ (n) } (0) = e^0 = 1 $.  Consequently, the Maclaurin series 
(5) becomes
\[ e^x = 1 + x + \frac{ x^2 }{ 2 ! }
+ \cdots + \frac{ x^n }{ n ! } + \cdots . \]
To justify the claim that this expansion is valid for all
$x$, we must show that
\[ R_n (x) = 
f ^{ (n+1) } (c)
\frac{ x^{n+1} }{ (n+1) ! } = e^c 
\frac{ x ^{n+1} }{ (n+1) ! } \]
tends to zero as $ n \rightarrow \infty $.  Since $ 0 < 
c < x $, we see that $ e^c 
< e^x $ if 
$ x > 0 $,
and $ e^c < 0 $
if $ x < 0 $.  In either case $ e^c < e^{ | x | } $;
hence
\[ | R_n (x) | \leq e ^{ | x | } \:
\frac{ | x | ^{n+1} }{ (n+1) ! } . \]
At this point, we need to recall that the ratio test
shows the series 
$ \sum \: {\displaystyle \frac{ x^n }{ n ! } } $
to be a convergent series for all $x $.  Hence
\vspace{.1in}

\noindent 
(6) \hspace{2in} $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
{\displaystyle \frac{ x ^{n+1} }{ (n+1) ! }} = 0 $
\vspace{.1in}

\noindent 
because the terms of a convergent series must tend to 
zero.  Therefore,
\[ 0 \leq {\displaystyle \lim_{ n \rightarrow \infty }} \:  
| R_n (x) | \leq {\displaystyle \lim_{ n \rightarrow \infty }} \:
e^{ | x | } \:
\frac{ | x | ^{n+1} }{ (n+1) ! } = 0 \]
or $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
| R_n (x) | = 0 $.

It should be emphasized that there are two 
phases to finding the Taylor or Maclaurin 
series for a function.  The first is formal -- just compute
the derivatives $ f^{ (n) } (a) $ 
or $ f^{ (n) } (0) $ and substitute into the 
appropriate formula; then, to prove the validity of the
representation, you must wrestle
with the problem of showing that $ {\displaystyle \lim_{ n \rightarrow
\infty }} \:  R_n (x) = 0 $.
\vspace{.1in}

\noindent 
{\bf Example 2}.  Find the Maclaurin series for $ \sin \: x $.
\vspace{.1in}

\noindent 
{\bf Solution}.  The even derivatives of $ \sin \: x $ at $ x 
= 0 $ are zero and its odd derivatives
of order $ 2 n + 1 $ are $ (-1) ^n $.  Substituted
in (5), this information gives the Maclaurin series 
\vspace{.1in}

\noindent 
(7) \hspace{1in} $ \sin \: x = x 
- {\displaystyle \frac{ x^3 }{ 3 ! } + 
\frac{ x^5 }{ 5 ! } -  
\frac{ x^7 }{ 7 ! }}
+ \cdots + (-1) ^n {\displaystyle \frac{ x^{ 2 n + 1 }}{ (
2 n + 1 ) ! }} + \cdots ~~
( - \infty < x < \infty ) $
\vspace{.1in}

\noindent 
The validity of this expansion rests on the 
observation that if $ f $ denotes $ \sin \: x $,
then $ | f^{ (n) } (c) | \leq 1 $ for any number 
$c$ and any positive integer $n$ and hence that
\[ | R_n (x) | = | f^{ (n+1) }
(c) | ~ \left | 
\frac{ x ^{n+1} }{ (n+1) ! } 
\right | = | f^{ (n+1) } (c) | ~
\frac{ | x | ^{n+1} }{ (n+1) ! } \leq
\frac{ | x | ^{n+1} }{ (n+1) ! } . \]
Therefore $ {\displaystyle \lim_{ n \rightarrow \infty }} \:  
| R_n (x) | = 0 $, by (6).  $ \ddagger $
\vspace{.1in}

\noindent 
{\bf Example 3}.  Find the Taylor series expansion 
for $ \sin \: x $ about $ a = \pi / 6 $.
\vspace{.1in}

\noindent 
{\bf Solution}.  Let $ f(x) = \sin \: x $.
The derivative pattern is
\[ f(x) = \sin \: x , ~~ f ^{'} (x) = \cos \: x \]
\[ f ^{''} (x) = - \sin \: x , ~~ f ^{ (3) } 
(x) = - \cos \: x \]
and, in general,
\[ f^{ ( 2 n ) } (x) 
= ( - 1 )^n \: \sin \: x , ~~
f^{ ( 2 n + 1 ) } (x) = 
(-1) ^n \: \cos \: x . \]
And at $ a = \pi / 6 $,
\[ f^{ ( 2 n ) } ( \pi / 6 ) = 
(-1) ^n \: \frac{ 1 }{ 2 } , ~~
f^{ ( 2 n + 1 ) } ( \pi / 6 ) 
= (-1) ^n \sqrt{ 3 } / 2 . \]
Accordingly, the Taylor series is
\[ \sin \: x = \frac{1}{2} + 
\frac{ \sqrt{ 3 }}{ 2 } 
( x - \pi / 6 ) 
- \frac{ 1 }{ 2 \cdot 2 ! }
( x - \pi / 6 )^2 
- \frac{ \sqrt{ 3 }}{ 2 \cdot 3 ! }
( x - \pi / 6 )^3 + \cdots . \]
The proof that this expansion is valid for all $x$ is exactly
as in the previous example because $ | f^{ (n) }
(c) \leq 1 $ for all $x$.
\vspace{.1in}

\noindent 
{\bf Example 4}.  Find the Maclaurin series for $ ( 1 
+ x ) e^x $.
\vspace{.1in}

\noindent 
{\bf Solution}.  There are two ways to attack this problem.
\vspace{.1in}

\noindent 
(A) Put $ f(x) = ( 1 + x ) e^x $ and calculate
\[ \begin{array}{rl} f ^{'} (x) = e^x + 
( 1 + x ) e^x = ( 2 + x ) e^x , 
\\ [0.2in]
f ^{''} (x) = e^x + ( 2 + x ) e^x 
= ( 3 + x ) e^x ,
\\ [0.2in]
f^{ ''' } (x) = e^x + ( 3 + x ) e^x = 
( 4 + x ) e^x . \end{array} \]
The pattern is clear:  for each $n$
\[ f^{ (n-1) } (x) = 
( n + 1 + x ) e^x . \]
Hence $ f^{ (n-1) } (0) = n + 1 $
for $ n \geq 0 $ and the Maclaurin series is
\[ ( 1 + x ) e^x 
= 1 + 2 x + \frac{ 3 x^2 }{ 2 }
+ \frac{ 4 x^3 }{ 3 ! }
+ \frac{ 5 x^4 }{ 4 ! } + \cdots + 
\frac{ (n+1) x^n }{ n ! } + \cdots . \]
This expansion is valid for all $x$ because 
\[ | R_n
(x) | = | ( n + 2 + c ) e^c 
\frac{ x ^{n+1} }{ (n+1) ! } | 
\leq ( n + 2 ) 
\: \frac{ | x | ^{n+1} }{ (n+1) ! }
+ | x | e^{ | x | } \:
\frac{ | x | ^{n+1} }{ (n+1) ! } \]
and both terms on the right tend to zero as $ n \rightarrow \infty $.
\vspace{.1in}

\noindent 
(B) Start with
\[ e^x = 1 + x + \frac{ x^2 }{ 2 ! }
+ \frac{ x^3 }{ 3 ! }
+ \cdots + \frac{ x^n }{ n ! } + \cdots , \]

multiply by $x$ to get
\[ x e^x = x + x^2 
+ \frac{ x^3 }{ 2 ! }
+ \frac{ x^4 }{ 3 ! } + \cdots + 
\frac{ x^n }{ (n+1) ! } + \cdots . \]
Then add:
\[ \begin{array}{rl}
( 1 + x ) e^x & \hspace{-.1in} = 
( 1 + x + {\displaystyle \frac{ x^2 }{ 2 ! }
+ \frac{ x^3 }{ 3 ! } + \cdots + \frac{ x^n }{ n ! }}
+ \cdots ) + 
\left ( x + x^2 + 
{\displaystyle \frac{ x^3 }{ 2 ! } + 
\frac{ x^4 }{ 3 ! }
+ \cdots + 
\frac{ x^n }{ (n+1) ! }} + 
\cdots \right ) 
\\ [0.2in]
& \hspace{-.1in} = 
1 + 2 x + \left ( {\displaystyle \frac{1}{2 ! } }
+ 1 \right ) x^2 
+ \left ( {\displaystyle \frac{ 1 }{ 3 ! } + 
\frac{1}{2} } \right ) x^3 
+ \cdots + \left ( {\displaystyle \frac{ 1 }{ n ! }
+ \frac{ 1 }{ (n+1) ! }} \right )
x^n + \cdots 
\\ [0.2in]
& \hspace{-.1in} = 1 + 2 x + 
3 {\displaystyle \frac{ x^2 }{ 2 ! }} + 
4 {\displaystyle \frac{ x^3 }{ 3 ! }} 
+ \cdots + (n+1) 
\: {\displaystyle \frac{ x^n }{ n ! }} + \cdots . 
\end{array} \]

In the preceding example, it is no accident that the
two methods yield identical results.
\vspace{.1in}

\noindent 
{\bf Theorem 21}.  If a function $f$ can be represented by
a  power series 
\vspace{.1in}

\noindent 
(9) $ \begin{array}{rl}
f(x) & \hspace{-.1in} = 
{\displaystyle \sum_{n=0}^{ \infty }} \: 
b_n ( x-a ) ^n 
\\ [0.3in]
& \hspace{-.1in} = 
b_0 + b_1 ( x-a ) 
+ b_2 ( x-a ) ^2 
+ \cdots + b_n ( x-a ) ^n + \cdots 
\end{array} $
\vspace{.1in}

\noindent 
in some interval about $ x = a $, then the series
is the Taylor series expansion of $f$ 
at $ x = a $ and the coefficients $ b_n $ are given by 
\vspace{.1in}

\noindent 
(10) \hspace{2in} 
$ b_n = {\displaystyle \frac{ f^{ (n-1) } (a) }{ n ! }} 
~~ \mbox{for all $n$.} $
\vspace{.1in}

\noindent 
{\bf Proof}.  Inside its interval of convergence the power series
on the right in (9) represents an infinitely differentiable 
function that can be differentiated term-by-term.
Clearly, $ f (a) = b_0 $ and from
\[ f ^{'} (x) = b_1 + 2 b_2 
( x-a ) + 3 b_3 ( x-a ) ^2 + \cdots +
n b_n ( x-a ) ^{ n - 1 } + \cdots \]
it follows that $ f ^{'} (a) = b_1 $.  Letting $ x = a $
in the second derivative
\[ f ^{''} (x) = 2 b_2 + 2 \cdot 3 b_3 ( x-a ) 
+ \cdots + n ( n - 1 ) b_n 
( x-a ) ^{ n - 2 } + \cdots \]
gives
\[ f ^{''} (a) = 2 b_2 ~~ \mbox{or} ~~
b_2 = \frac{1}{2} f ^{''} (a) . \]
From the third derivative,
\[ f^{ (3) } (x) = 2 \cdot 3 b_3 
+ 2 \cdot 3 \cdot 4 b_4 ( x-a ) 
+ \cdots + n ( n - 1 ) ( n - 2 ) b_n 
( x-a ) ^{ n - 3 } + \cdots , \]
we conclude that
\[ f^{ (3) } (a) = 2 \cdot 3 b_3 ~~
\mbox{or} ~~ b_3 = \frac{ f^{ (n-1) } (a) }{ 3 ! } . \]
In general,
\[ f^{ (n-1) } (x) = n ( n - 1 ) 
( n - 2 ) \cdots (2) b_n + 
(n+1) (n) ( n - 1 ) \cdots 
(2) b_{ n + 1 } ( x-a ) + \cdots \]
and 
\[ f^{ (n) } (a) = n ( n - 1 ) ( n - 2 ) 
\cdots (2) b_n = \frac{ f^{ (n) } (a) }{ n ! } . ~~~~ \ddagger \]
\vspace{.1in}

The point of this theorem is that the Taylor series 
representation for a function is unique, and that no matter
how you arrive at a power series representation 
for a function, the series in hand {\em is} the
Taylor series.  In particular, this means that the series
representations of the
last section are Maclaurin expansions.
\vspace{.1in}

\noindent 
{\bf Example 5}.  Find the Maclaurin series expansion for $ e^{ - x^2 } $.
\vspace{.1in}

\noindent 
{\bf Solution}. In the series
\[ e^x = 1 + x + \frac{ x^2 }{ 2 ! } + 
\cdots + \frac{ x^n }{ n ! } + \cdots \]
replace $x$ by $ - x^2 $ to get
\[ e^{ - x^2 } = 1 - x^2
+ \frac{ x^4 }{ 2 ! } - \frac{ x^6 }{ 3 ! }
+ \cdots + (-1) ^n \: \frac{ x^{ 2 n } }{ n ! } 
+ \cdots \]
The power series converges for all $ x $ and for all $x$ it
converges to $ e^{ - x^2 } $.
Conclusion:  It must be the Maclaurin series for
$ e^{ - x^2 } $.  $ \ddagger $
\vspace{.1in}

Knowledge of coefficients in a Maclaurin expansion often provides 
an efficient procedure for computing derivatives at the
origin.  For example, suppose you want to know the
sixth derivative of $ f(x) = e^{ -x^2 } $ at $ x = 0 $.
Straightforward differentiation is tedious; however, according to (10),
\[ f^{ (6) } (0) = 6 ! b_6 = 6 ! 
\left ( - \frac{ 1 }{ 3 } \right ) = - 120 \]
because $ - {\displaystyle \frac{ 1 }{ 3 ! } } $ is the 
coefficient of $ x^6 $ in the Maclaurin expansion of $ e^{ -x^2 } $.








