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\def\dd#1#2{\frac{d#1}{d#2}}
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8.1  p. 619:   1,  9, 15, 33, 37, 39.
}

\SN{\jobname}
\np
$a_n=\Frac{1-n}{n^2}$ so
$a_1=\Frac{1-1}{1^2}=0$;
$a_2=\Frac{1-2}{2^2}=-\Frac{1}{4}$;
$a_3=\Frac{1-3}{3^2}=-\Frac{2}{9}$;\par
$a_4=\Frac{1-4}{4^2}=-\Frac{3}{16}$.

\np
$a_1=2$, $a_{n+1}=(-1)^{n+1}a_n/2$.
$a_1=2$; 
$a_2=(-1)^{1+1}\Frac{2}{2}=1$;\par
$a_3=(-1)^{2+1}\Frac{1}{2}=-\frac12$;
$a_4=(-1)^{3+1}\Frac{-\frac12}{2}=-\frac14$;
$a_5=(-1)^{4+1}\Frac{-\frac14}{2}=\frac18$;\par
$a_6=(-1)^{5+1}\Frac{\frac18}{2}=\frac1{16}$;
$a_7=(-1)^{6+1}\Frac{\frac1{16}}{2}=-\frac1{32}$;
$a_8=(-1)^{7+1}\Frac{-\frac1{32}}{2}=-\frac1{64}$;\par
$a_9=(-1)^{8+1}\Frac{-\frac1{64}}{2}=\frac1{128}$;
$a_{10}=(-1)^{9+1}\Frac{\frac1{128}}{2}=\frac1{256}$.

\np
The sequence $1$, $-4$, $9$, $-16$, $25$, \dots\ {\sl Squares of the
positive integers with alternating signs}.
\boxit{\vrule height 10pt depth 4pt width 0pt\ 
$a_n=(-1)^{n+1}n^2$.\ $n\geq1$.}
Note $a_n=(-1)^nn^2$ has the same verbal discription but does
not have the correct first value.

\np
$a_n=\Frac{2^n3^n}{n!}$.
We can rewrite this as
$a_n=\Frac{2\cdot3}{1}\cdot\Frac{2\cdot3}{2}\cdots\Frac{2\cdot3}{n}$
so $a_{n+1}=a_n\cdot\Frac{6}{n+1}$.
For $n\geq 6$, $\Frac{6}{n+1}<1$ so $a_{n+1}<a_n$.
In particular, the sequence is bounded from above by 
$a_5=6\cdot3\cdot2\cdot\frac32\cdot\frac65=\frac{36\cdot 9}5=
\frac{314}5=62.8$.
For $n\geq 6$ the sequence is decreasing, 
hence {\sl not\/} non{--}decreasing.
More interestingly, the sequence is bounded below by $0$ and
since it is decreasing, it does have a limit.
Once you know it has a limit, you can compute the limit as follows.
\def\Li{\displaystyle\lim_{n\to\infty}}
Let $\Li a_n=L$.
Then $\Li a_{n+1}=L$ also 
(sequences with the same tails have the same limit) and hence
$L=\Li a_{n+1}=\Li a_n\Frac{6}{n+1}=(\Li a_n)\cdot(\Li\frac 6{n+1})=
L\cdot 0$ so $L=0$.

\np
$a_n=\Frac{2^n-1}{2^n}$.
There are at least two ways to proceed.
One way is to rewrite $a_n=1-\Frac{1}{2^n}$, so
$\Li a_n=\Li 1-\Frac{1}{2^n}=1-\Li \Frac{1}{2^n}=1-0=1$
since you are supposed to know $\Li \Frac{1}{2^n}=0$.
The second way is via L'H\^opital's Rule since 
$\Li 2^n-1=\Li 2^n=\infty$.
Hence $\Li\Frac{2^n-1}{2^n}=
\Li\Frac{(\ln 2)2^n-0}{(\ln2)2^n}=\Li 1=1$.
Hence the sequence converges because we have evaluated the limit.
We have used theorems from section 8.2 so these techniques are
not what was expected at the time but this is how you should now do these
problems.
At the time you could have argued as follows.
The sequence $a_n$ is increasing which we can show in any of several ways.
One way is to write $a_n=1-\frac1{2^n}$ and remark that
$\frac 1{2^n}$ is decreasing.
A second way is to take the derivative of the function
$f(x)=\Frac{2^x-1}{2^x}$;
$f^\prime(x)=
\Frac{\bigl((\ln2)2^x\bigr)\cdot2^x-\bigl(2^x-1\bigr)\cdot 
\bigl((\ln2)2^x\bigr)}{\bigl(2^x\bigr)^2}=
\Frac{(\ln2)2^x}{2^{2x}}=\Frac{\ln2}{2^x}>0$,
so $f(x)$ is an increasing function and hence $a_n$ is an
increasing sequence.
Since $2^n-1<2^n$, $a_n<1$ so the sequence is an increasing, bounded
sequence and hence converges.

\np
$a_n=\bigl(1+(1-)^n\bigr)\Bigl(\Frac{n}{n+1}\Bigr)$.
Start by noticing that for $n$ odd, $a_n=0$ since
$1+(-1)^n=0$.
Hence, if $\Li a_n$ exists, it must be $0$ since the subsequence
$\{a_{2n+1}\}$ converges to $0$. ({\bf1}. on page 618).
Next consider the subsequence $\{a_{2n}\}$.
$a_{2n}=2\cdot\Frac{2n}{2n+1}$ and
$\Li 2\cdot\Frac{2n}{2n+1}$ can be computed via L'H\^opital's Rule 
or by $\Li 2\cdot\Frac{2n}{2n+1}=\Li\Frac{4}{2+\frac1n}=\Frac{4}{2+0}=2$.
Since we have two subsequences which converge to two different
limits, $\Li a_n$ does not exist. ({\bf 2}. on page 618).


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