

\begin{center} 
{\large \bf PROBLEM SET 20.6}
\end{center}
\vspace{.2in}

\noindent
{\bf In Problems 1--12 test the series for absolute 
and for conditional convergence.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $ $ 1 - 1/3 + 1/9 - 1/27 + \cdots $
& \hspace{.2in} & 
$ ~ \: ${\bf 2.} $ ~~ $ 
$ 1 - 1 / \sqrt[2]{2} + 1 / \sqrt[3]{3} - 1 / \sqrt[4]{4}  
+ \cdots $
\\ [0.2in]
$ ~ \: ${\bf 3.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
{\displaystyle \frac{ \cos \: n^2 }{ n^3 }} $
& &
$ ~ \: ${\bf 4.}
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
( - 1 )^n {\displaystyle \frac{ ( \ln \: n )^2 }{ n^3 }} $
\\ [0.2in]

$ ~ \: ${\bf 5.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
{\displaystyle \frac{ 100 - n }{ n^3 }} $  
&& 
$ ~ \: ${\bf 6.} $ ~~ $
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
{\displaystyle \frac{ 10^n }{ n ! }} $
\\ [0.2in]
$ ~ \: ${\bf 7.} $ ~~ $
$ {\displaystyle \sum_{n=1}^{ \infty }} \:
{\displaystyle \frac{ (n+2) ! }{ n^2 n ! 2^n }} $
&& 
$ ~ \: ${\bf 8.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: ( - 1 )^n 
{\displaystyle \frac{ 1 \cdot 3 \cdot 9 
\cdots ( 3n ) }{ ( 2 n ) ! }} $
\\ [0.2in]
$ ~ \: ${\bf 9.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: ( - 1 )^n 
{\displaystyle \frac{ ( \ln \: 2n )^n }{ ( 2n )^n }} $
&& {\bf 10.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: ( - 1 )^n 
{\displaystyle \frac{ 2^{ 3n + 1 }}{ 3^{ 2n + 3 } }} $ 
\\ [0.2in]
{\bf 11.} $ ~~ $ 
$ {\displaystyle \sum_{n=2}^{ \infty }} \: ( - 1 )^n 
{\displaystyle \frac{ \ln \: n^n }{ n^2 }} $
&& {\bf 12.} $ ~~ $ 
$ {\displaystyle \sum_{n=2}^{ \infty }} \: ( - 1 )^n 
{\displaystyle \frac{ \ln \: n }{ ( 1.02 )^n }} $
\end{tabular}
\vspace{.2in}

\noindent  
$ ~ \: ${\bf 13.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
For which values of $a$ does the series 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
\left ( a^{ - 1 } - {\displaystyle \frac{ 1 }{ n }}
\right )^n $ converge absolutely?
\end{minipage}
\vspace{.2in}

\noindent 
{\bf 14.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Find the values of $x$ for which the series 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
( x + 1 )^n ( x + 2 )^n $ is absolutely convergent.
\end{minipage}
\vspace{.3in}

\begin{center}
{\bf CALCULATOR PROBLEMS}
\end{center}
\vskip -2ex

\noindent 
{\bf In Problems 15 and 16, let
$ \boldmath \mbox{$S$} $
denote the sum of the series 
and estimate the error in the approximation 
$ \boldmath \mbox{$S \cong s_N $} $
for the indicated value of
$ \boldmath \mbox{$N$} $.}
\vspace{.2in}

\noindent 
\begin{tabular}{lll} 
{\bf 15.} $ ~~~ $
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
{\displaystyle \frac{ \sin \: n^2 }{ n^3 }} ; ~~ N = 15 $
& \hspace{1in} & 
{\bf 16.} $ ~~~ $
$ {\displaystyle \sum_{n=1}^{ \infty }} \: ( - 1 )^n 
{\displaystyle \frac{ 1 }{ ( 3n + 7 )^5 }} ; ~~ N = 100 $
\end{tabular}
\vspace{.2in}



\begin{center} 
{\large \bf PROBLEM SET 20.7}
\end{center}
\vspace{.2in}

\noindent
{\bf For the power series in 
Problems 1--6, find the interval of convergence;
in Problems 7 and 8, find only the radius of convergence.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $ 
$ {\displaystyle \sum_{n=0}^{ \infty }} \: 
3^n x^n $
& \hspace{.2in} & 
$ ~ \: ${\bf 2.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
{\displaystyle \frac{ x^n }{ n }} $
\\ [0.2in]
$ ~ \: ${\bf 3.} $ ~~ $ 
$ {\displaystyle \sum_{n=0}^{ \infty }} \: 
{\displaystyle \frac{ x^{ 2n + 1 }}{ 2n + 1 }} $
&& $ ~ \: ${\bf 4.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
e^{ - n } x^n $
\\ [0.2in]
$ ~ \: ${\bf 5.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
{\displaystyle \frac{ x^n }{ n \: \ln \: (n+1) }} $
&& 
$ ~ \: ${\bf 6.} $ ~~ $ 
$ {\displaystyle \sum_{n=0}^{ \infty }} \: 
( - 1 )^n {\displaystyle \frac{ n^3 }{ 5^n }}
( x - 1 )^n $
\\ [0.2in]
$ ~ \: ${\bf 7.} $ ~~ $ 
$ {\displaystyle \sum_{n=0}^{ \infty }} \: 
( - 1 )^n 
{\displaystyle \frac{ 1 \cdot 3 \cdot 6 \cdots 
(3n ) }{ (n+1) ^3 }} x^n $
&& 
$ ~ \: ${\bf 8.} $ ~~ $ 
$ {\displaystyle \sum_{n=0}^{ \infty }} \: 
{\displaystyle \frac{ 2 \cdot 5 \cdot 8 \cdots 
( 3n - 4 ) }{ n ! }} x^n $
\end{tabular}
\vspace{.2in}

\noindent 
{\bf In Problems 9--11, determine the values
of $ \boldmath \mbox{$ x $}$ for which
the series converges.}
\vspace{.2in}

\noindent 
\begin{tabular}{lll} 
$ ~ \: ${\bf 9.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
{\displaystyle \frac{ 1 }{ x^n }} $
& \hspace{1.5in} & {\bf 10.} $ ~~ $ 
$ {\displaystyle \sum_{n=0}^{ \infty }} \: 
{\displaystyle \frac{ 2^n }{ ( x + 1 )^n }} $
\\ [0.2in]
{\bf 11.} $ ~~ $ 
$ {\displaystyle \sum_{n=1}^{ \infty }} \: 
( - 1 )^n \left ( {\displaystyle \frac{ 2x - 1 }{ x + 
2 }} \right )^2 $
&& \end{tabular} 
\vspace{.2in}

\noindent  
$ ~ \: ${\bf 12.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Construct a power series whose interval 
of convergence is $ ( - 2 , 2 ) $.
\end{minipage}
\vspace{.2in}



\begin{center} 
{\large \bf PROBLEM SET 20.8}
\end{center}
\vspace{.2in}

\noindent
{\bf With one of the series (4), (10), or (11), as a starting 
point derive a power series expansion for the functions
in  Problems 1--5.  State the interval of convergence.}
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $ 
$ {\displaystyle \frac{ 1 }{ ( 1 + x )^2 }} $
& \hspace{.2in} & 
$ ~ \: ${\bf 2.} $ ~~ $ 
$ {\displaystyle \frac{ 1 }{ ( 1 - x )^5 }} $
\\ [0.2in]
$ ~ \: ${\bf 3.} $ ~~ $
$ {\displaystyle \frac{ 3x^2 }{ ( 2 - 5x )^2 }} $
&& 
$ ~ \: ${\bf 4.} $ ~~ $ 
$ {\displaystyle \int_{ 0 }^x } \:
\ln \: ( 1 - t ) \: d t $
\\ [0.2in]
$ ~ \: ${\bf 5.} $ ~~ $
$ ( 2 - x ) - 
( 1 - x ) \: \ln \: ( 1 - x ) $
&& \end{tabular}
\vspace{.2in}

\noindent 
{\bf In Problems 6--8, verify the power series expansion.}
\vspace{.2in}

\noindent 
\begin{tabular}{l} 
$ ~ \: ${\bf 6.} $ ~~ $ 
$ {\displaystyle \frac{ 1 + x }{ 1 - x }} 
= 1 + 2x + 2x^2 + 2x^3 + \cdots $ if $ | x | < 1 $.
\\ [0.2in]
$ ~ \: ${\bf 7.} $ ~~ $ 
$ {\displaystyle \frac{ x^2 + 6x + 1 }{ ( 1 - x )^3 }}
= 1 + 3^2 x + 5^2 x^2 + 7^2 x^3 + \cdots $
if $ | x | < 1 $.
\\ [0.2in]
$ ~ \: ${\bf 8.} $ ~~ $ 
$ \sin ^{ - 1 } x = 
{\displaystyle \int_{ 0 }^x } \:
{\displaystyle \frac{ d t }{ \sqrt{ 1 - t^2 }} = 
x + \frac{1}{2} \frac{ x^3 }{ 3 } +
\frac{ 1 + 3 }{ 2 \cdot 4 } \frac{ x^5 }{ 5 } 
+ \frac{ 1 + 3 \cdot 5 }{ 2 \cdot 4 \cdot 6 } \frac{ x^7 }{ 7 }}
+ \cdots $ if $ | x | < 1 $.
\end{tabular}
\vspace{.2in}

\begin{center}
{\bf CALCULATOR PROBLEMS}
\end{center}
\vskip -2ex

\noindent 
{\bf In Problems 9--12, use (5), (7), (10),
and (11) to estimate the following with an error less than
0.001.}
\vspace{.2in}

\noindent 
\begin{tabular}{lll} 
$ ~ \: ${\bf 9.} $ ~~ $ 
$ \ln \: 1.2 $
& \hspace{2in} &
{\bf 10.} $ ~~ $ $ e^{ - 0.3 } $
\\ [0.2in]
{\bf 11.} $ ~~ $ 
$ \sqrt[3]{ 1.2 } $
&& {\bf 12.} $ ~~ $ 
$ ( 0.95 )^{ - 1/4 } $
\end{tabular}
\vspace{.2in}

\noindent 
$ ~ \: ${\bf 13.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Because $ \pi / 6 = \sin ^{ - 1 } \: 0.5 $,
the series in Problem 18 can be used to approximate 
$ \pi $.  Construct a table of its first fifteen partial
sums at $ x = 1/2 $.  Compare the resulting approximations to
the calculator value for $ \pi $.
\end{minipage}
\vspace{.2in}

\noindent 
$ ~ \: ${\bf 14.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Repeated applications of the identity
\[ 
\tan ^{ - 1 } \: \frac{ 1 }{ p } 
= \tan ^{ - 1 } \: \frac{ 1 }{ p + q } 
+  \tan ^{ - 1 } \: \frac{ 1 }{ p + r } 
( q r = 1 + p^2 ) \]
of C.L.\ Dodgson (Lewis Carroll) leads to the formula 
\[ \frac{ \pi }{ 4 } = 5 \: \tan ^{ - 1 }
\: \frac{ 1 }{ 7 } + 2 \: 
\tan ^{ - 1 } \: \frac{ 3 }{ 79 } . \]
Use this and the Maclaurin expansion for 
$ \tan ^{ - 1 } x $ to find the first eight digits
in the decimal expansion of $ \pi $.
\end{minipage}
\vspace{.2in}



\begin{center} 
{\large \bf PROBLEM SET 20.9}
\end{center}
\vspace{.2in}

\noindent 
{\bf In Problems 1--4, 
find the Maclaurin series expansion for $ \boldmath \mbox{$ f(x) $}$ 
and use (3) to establish its validity for all
$ \boldmath \mbox{$ x $}$.  
The symbols  $ \boldmath \mbox{$ a $}$ 
and $ \boldmath \mbox{$ b $}$ denote constants.} 
\vspace{.2in}

\noindent  
\begin{tabular}{lll}
$ ~ \: ${\bf 1.} $ ~~ $ 
$ f(x) = e^{ a x } $
& \hspace{.2in} & 
$ ~ \: ${\bf 2.} $ ~~ $ 
$ f(x) = \mbox{sinh} \: (x) $
\\ [0.2in]
$ ~ \: ${\bf 3.} $ ~~ $ 
$ f(x) = {\displaystyle \frac{ x^3 }{ 6 }} - x + \sin \: x $
&& $ ~ \: ${\bf 4.} $ ~~ $ 
$ f(x) = \sin ^2 \: x 
= {\displaystyle \frac{ 1 - \cos \: 2x }{ 2 }} $ 
\end{tabular} 
\vspace{.2in}

\noindent 
{\bf In Problems 5--8  derive a formula for 
the general nth derivative 
$ \boldmath \mbox{$ f^{(n)} $}$ 
and use it to write the Taylor series (4)
for $ \boldmath \mbox{$ f (x) $}$ about 
$ \boldmath \mbox{$ x = a $}$ 
You may omit the verification (3).}
\vspace{.2in}

\noindent 
\begin{tabular}{lll} 
$ ~ \: ${\bf 5.} $ ~~ $
$ f(x) = {\displaystyle \frac{ 1 }{ 1 + x }} ; ~~
a = 0 $
& \hspace{1in} & 
$ ~ \: ${\bf 6.} $ ~~ $ 
$ f(x) = {\displaystyle \frac{ 1 }{ 2x + 3 }} ; ~~
a = - 1 $
\\ [0.2in]
$ ~ \: ${\bf 7.} $ ~~ $ 
$ f(x) = x^5 ; ~~ a = 1 $
&& $ ~ \: ${\bf 8.} $ ~~ $ 
$ f(x) = x^2 ; ~~ a = 100 $
\end{tabular} 
\vspace{.2in}

\noindent 
{\bf Use known series expansions from this and the preceding
section together with
algebraic manipulation and substitution to obtain 
the following Maclaurin expansions.}
\vspace{.2in}

\noindent 
\begin{tabular}{l}
$ ~ \: ${\bf 9.} $ ~~ $ 
$ x^2 e^{ - x } = x^2 - x^3 + 
{\displaystyle \frac{ x^4 }{ 2 ! } - \frac{ x^5 }{ 3 ! }} + \cdots $
\\ [0.2in]
{\bf 10.} $ ~~ $ 
$ ( 1 + x )^{ - 1/3 }
= 1 - {\displaystyle \frac{ 1 }{ 3 } x 
+ \frac{ 1 \cdot 4 }{ 3^2 \cdot 2 } x^2
- \frac{ 1 \cdot 4 \cdot 7 }{ 3^3 \cdot 3 ! }}
x^3 + \cdots $
\\ [0.2in]
{\bf 11.} $ ~~ $ 
$ \sin ^{ - 1 } \: x - x = 
{\displaystyle \frac{1}{2} \frac{ x^3 }{ 3 } 
+ \frac{ 1 \cdot 3 }{ 2 \cdot 4 } \frac{ x^5 }{ 5 } 
+ \frac{ 1 \cdot 3 \cdot 5 }{ 2 \cdot 4 \cdot 6 }
\frac{ x^7 }{ 7 }} + \cdots $
\\ [0.2in]
{\bf 12.} $ ~~ $ 
$ ( 1 - x )^2 ) \: \ln \: ( 1 - x^2 ) 
= - x^2 + {\displaystyle \frac{1}{2 x^4 }
+ \frac{ 1 }{ 6 } x^6 + \frac{ 1 }{ 12 } x^8 +
\frac{ 1 }{ 20 }} x^{ 10 } + \cdots $
\end{tabular} 
\vspace{.2in}


\noindent  
$ ~ \: ${\bf 13.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Integrate both sides of 
\[ {\displaystyle \frac{ 1 }{ 1 + t }} 
= 1 - t + t^2 - t^3 + \cdots (-1) ^{ n - 1 } \:
t^{ n - 1 } + \frac{ t^n }{ 1 + t } \]
from $ 0 $ to $1$ to obtain
\[ \ln \: 2 = 1 - \frac{1}{2} + 
\frac{1}{3} - \frac{1}{4} 
+ \cdots + (-1)^{ n - 1 } \:
\frac{ 1 }{ n } + (-1) ^n 
\int_0^1 \: \frac{ t^n }{ 1 + t } \: d t . \]
Then show that
\[ R_n = (-1) ^n \int_0^1 \: \frac{ t^n }{ 1 + t } \: d t
\rightarrow 0 ~~ \mbox{as} ~~ n \rightarrow \infty , \]
thereby verifying that $ \ln \: 2 = 1 
- 1/2 + 1/3 - 1/4 + \cdots $.
\end{minipage}
\vspace{.2in}

\noindent 
$ ~ \: ${\bf 14.} $ ~~~ $
\begin{minipage}[t]{5.9in} 
Use integration by parts in the formula
\[ \tan^{-1} \: x = \int_0^x \: 
\frac{ 1 }{ 1 + t^2 } \: d t \]
to obtain the series expansion 
\[ \tan^{-1} \: x = \frac{ x }{ 1 + x^2 } 
+ \frac{ 2 }{ 3 } \frac{ x^3 }{ ( 1 + x^2 )^2 } 
+ \frac{ 2 }{ 3 } \cdot \frac{ 4 }{ 5 } 
\frac{ x^5 }{ ( 1 + x^2 )^3 } + \cdots \]
\end{minipage}




