\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.3  p. 638:   3,  7, 11, 17, 23, 29, 35, 43, 45, 57, 59.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}

\SN{\jobname}
\np
$1-\frac12+\frac14-\frac18+\cdots+(-1)^{n-1}\frac1{2^{n-1}}+\cdots$
For this series find the $n$th partial sum and compute the limit
to see if the series converges.
Now $s_n=1-\frac12+\cdots+(-1)^{n-1}\frac1{2^{n-1}}$
since we are supposed to sum $n$ of the terms.
By the usual formula for a geometric series,
$s_n=\Frac{1-(-\frac12)^n}{1-(-\frac12)}=
\frac23\cdot\bigl(1-(-\frac12)^n\bigr)$.
Hence $\Li s_n=\frac23\cdot1$ 
so $1-\frac12+\frac14-\frac18+\cdots+(-1)^{n-1}\frac1{2^{n-1}}+\cdots$
converges and is equal to $\frac23$.

\np
$\Sum{n=0} \Frac{(-1)^n}{4^n}=1-\frac14+\frac1{16}-\frac1{32}+\cdots$.
This is a geometric series with $a=1$ and $r=-\frac14$.
Since $\vert r\vert<1$ this series converges and
$\Sum{n=0} \Frac{(-1)^n}{4^n}=\Frac1{1-\bigl(-\frac14\bigr)}=
\Frac{1}{\frac54}=\frac45$.

\np
$\Sum{n=0}\Bigl(\Frac{5}{2^n}+\Frac{1}{3^n}\Bigr)=
\Sum{n=0}\Frac{5}{2^n} + \Sum{n=0}\Frac{1}{3^n}$.
First $\Sum{n=0}\Frac{5}{2^n}$ is a geometric series with $a=5$ 
and $r=\frac12$.
Since $\vert r\vert<1$, the series converges and
$\Sum{n=0}\Frac{5}{2^n}=\Frac{5}{1-\frac12}=10$.
Next $\Sum{n=0}\Frac{1}{3^n}$ is a geometric series with $a=1$
and $r=\frac13$.
Since $\vert r\vert<1$ this series converges and
$\Sum{n=0}\Frac{1}{3^n}=\Frac{1}{1-\frac13}=\frac32$.
Hence $\Sum{n=0}\Bigl(\Frac{5}{2^n}+\Frac{1}{3^n}\Bigr)$ converges
to $10+\frac32=\frac{23}2$.

\np
$\Sum{n=1}\Frac{40n}{(2n-1)^2(2n+1)^2}$.
First use partial fractions:
$\Frac{40n}{(2n-1)^2(2n+1)^2}=\Frac A{2n-1}+\Frac B{(2n-1)^2}+
\Frac C{2n+1}+\Frac D{(2n+1)^2}$ or
$40n=A(2n-1)(2n+1)^2+B(2n+1)^2+
C(2n-1)^2(2n+1)+D(2n-1)^2$.
Plug in $n=\frac12$;
$20=A\cdot0+B(2)^2+C\cdot0+D\cdot0$ or $B=5$.
Plug in $n=-\frac12$ to see $D=-5$.
This means
$40n=A(2n-1)(2n+1)^2+C(2n-1)^2(2n+1)+5\bigl(
(2n+1)^2-(2n-1)^2\bigr)=
A(2n-1)(2n+1)^2+C(2n-1)^2(2n+1)+40n$,
so $A(2n-1)(2n+1)^2+C(2n-1)^2(2n+1)=0$,
which can only happen if $A=C=0$.
Hence
$\Frac{40n}{(2n-1)^2(2n+1)^2}=\Frac5{(2n-1)^2}-\Frac5{(2n+1)^2}$
so the series is a telescoping one.
Hence $\Sum{n=1}\Frac{40n}{(2n-1)^2(2n+1)^2}=
\Frac{5}{1^2}-\Li\Frac5{(2n+1)^2}=5$.

\np
$\Sum{n=0}\Bigl(\Frac{1}{\sqrt2}\Bigr)^n$.
This is a geometric series with $a=1$ and
$r=\Frac1{\sqrt2}$.
Since $\sqrt2>1$, $\Frac1{\sqrt{2}}<1$ so this series converges.
From the formula
$\Sum{n=0}\Bigl(\Frac{1}{\sqrt2}\Bigr)^n=\Frac{1}{1-\frac1{\sqrt2}}=
\Frac{\sqrt{2}}{\sqrt{2}-1}=\Frac{\sqrt{2}(\sqrt{2}+1)}{2-1}=
2+\sqrt{2}$.

\np
$\Sum{n=0}e^{-2n}$ is a geometric series with $a=1$ and
$r=e^{-2}$.
Since $e>1$, $e^2>1$ and $\vert r\vert<1$.
Hence the series converges and
$\Sum{n=0}e^{-2n}=\Frac{1}{1-e^{-2}}=\Frac{e^2}{e^2-1}$.

\np
$\Sum{n=0}\Frac{n!}{1000^n}$.
You should always wonder about $\Li a_n$.
In this case,
$\Frac{n!}{1000^n}=\frac1{1000}\cdot\frac2{1000}\cdots
\frac{n-1}{1000}\cdot\frac n{1000}$.
Hence $a_{n+1}=a_n\cdot\frac{n+1}{1000}$.
Hence the sequence $a_n$ is increasing for $n\geq 1000$.
Hence either $\Li a_n=L$ if the sequence $a_n$ is bounded or
else $\Li a_n=\infty$.
As usual, $\Li a_n=\Li a_{n+1}$
and $\Li a_{n+1}=(\Li a_n)\cdot\Bigl(\Li\frac{n+1}{1000}\Bigr)$.
Since $\Li\frac{n+1}{1000}=\infty$, the only choice for
$\Li a_n$ is $\infty$.
Hence the series diverges.

\np
$\Sum{n=0}3\Bigl(\Frac{x-1}2\Bigr)^n$ is a geometric series with
$a=3$ and $r=\frac{x-1}2$.
The series converges whenever $\vert r\vert<1$ or
$\Big\vert\frac{x-1}2\Big\vert<1$.
Hence $\vert x-1\vert <2$ so $-2<x-1<2$ or
$-1<x<3$.
Hence the series converges for any number $x$, $-1<x<3$.
We weren't asked, but we can sum this series,
$\Sum{n=0}3\Bigl(\Frac{x-1}2\Bigr)^n=\Frac{3}{1-\frac{x-1}2}=
\Frac{6}{2-(x-1)}=\Frac{6}{3-x}$.

\np
$\Sum{n=0}2^nx^n$ is a geometric series with $a=1$ and $r=2x$.
Hence the series converges precisely when $\vert2x\vert<1$ or
$-\frac12<x<\frac12$.
For such $x$, $\Sum{n=0}2^nx^n=\Frac{1}{1-2x}$.

\np
$1.24\overline{123}=1.24+\Sum{n=1}\frac{123}{100\cdot1000^{n}}$.
The series $\Sum{n=1}\frac{123}{100\cdot1000^{n}}$
 is a geometric series
with $a=\frac{123}{100\cdot1000}$ and $r=\frac{1}{1000}$.
Hence $\Sum{n=1}\frac{123}{100\cdot1000^n}=
\Frac{\frac{123}{100\cdot1000}}{1-\frac{1}{1000}}=
\frac{123}{100000-100}=
\frac{123}{99900}$, so
$1.24\overline{123}=\frac{124}{100}+\frac{123}{99900}=
\frac{124*999+123}{99900}=\frac{124000-1}{99900}=
\frac{123999}{99900}=\frac{41333}{33300}$.

\np
$\Sum{n=1}\Frac{1}{(n+1)(n+2)}=\frac{1}{2\cdot3}+
\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots$.
(a)
$\Sum{n=1}\Frac{1}{(n+1)(n+2)}=
\Sum{n=-2}\Frac{1}{(n+4)(n+5)}$;
(b)
$\Sum{n=1}\Frac{1}{(n+1)(n+2)}=
\Sum{n=0}\Frac{1}{(n+2)(n+3)}$;
(c)
$\Sum{n=1}\Frac{1}{(n+1)(n+2)}=
\Sum{n=5}\Frac{1}{(n-3)(n-2)}$.


\end