\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.7  p. 661:  3,  5,  9, 13, 15, 17, 19, 31, 35, 45.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}
\def\Int{\displaystyle\int}

\SN{\jobname}
\np$\Sum{n=1}(-1)^{n+1}\Bigl(\Frac {n}{10}\Bigr)^n$.
To use any of our tests, it is important to understand
how the terms of the  series approach $0$.
As a first step, consider $\Li\Bigl(\Frac {n}{10}\Bigr)^n=\infty$.
This means that the terms are not approaching $0$ at all and
so the series diverges.

\np$\Sum{n=2}(-1)^{n+1}\Frac{1}{\ln n}$.
This time the terms do go to $0$.
Since $\ln x$ is an increasing function, $\Frac1{\ln n}$
is decreasing, so by the Alternating Series Test this series
converges.
The series $\Sum{n=2}\Frac{1}{\ln n}$ diverges since
$\ln n< n$, so $\Frac1{n}<\Frac{1}{\ln n}$ and $\Sum{n=1}\Frac1n$
diverges since it is a $p${--}series with $p=1$.
Hence this is a conditionally convergent series.

\np$\Sum{n=1}(-1)^{n+1}\Frac{\sqrt n+1}{n+1}$.
Check $\Li\Frac{\Frac{\sqrt n+1}{n+1}}{\Frac{1}{\sqrt n}}=
\Li\Frac{n+\sqrt n}{n+1}=\Li\Frac{1+\frac1{\sqrt n}}{1+\frac1n}=1$.
Since $\Sum{n=1}\Frac{1}{\sqrt n}$ is a $p${--}series with $p=\frac12<1$,
it diverges and hence so does 
$\Sum{n=1}\Frac{\sqrt n+1}{n+1}$.
Check that $\Li\Frac{\sqrt n+1}{n+1}=
\Li\Frac{1+\frac1{\sqrt n}}{\sqrt n+\frac1{\sqrt n}}=0$.
To apply the Alternating Series Test, it remains to check
that $\Frac{\sqrt n+1}{n+1}$ is decreasing.
To show this, study the function
$f(x)=\Frac{\sqrt x+1}{x+1}$.
Compute $f^\prime(x)=
\Frac{\dd{\sqrt x+1}x(x+1)-(\sqrt x+1)\dd{x+1}x}{(x+1)^2}=
\Frac{\frac1{2\sqrt x}(x+1)-(\sqrt x+1)}{(x+1)^2}=
\Frac{\frac1{2\sqrt x}-\frac{\sqrt x}2-1}{(x+1)^2}$.
For $x\geq1$, $\sqrt x\geq1$ so $\frac1{2\sqrt x}\leq\frac12$
so $\frac1{2\sqrt x}-\frac{\sqrt x}2-1<0$ and hence
$f^\prime(x)<0$ for all $x\geq1$.
Hence the terms $\Frac{\sqrt n+1}{n+1}$ are decreasing
and the Alternating Series Test applies to show that
$\Sum{n=1}(-1)^{n+1}\Frac{\sqrt n+1}{n+1}$ converges.
Since we have already shown $\Sum{n=1}\Frac{\sqrt n+1}{n+1}$
diverges, $\Sum{n=1}(-1)^{n+1}\Frac{\sqrt n+1}{n+1}$ is a
conditionally convergent series.

\np$\Sum{n=1}(-1)^{n+1}\Frac 1{\sqrt n}$.
To use the Alternating Series Test first check that
$\Li\Frac1{\sqrt n}=0$ and that the terms $\Frac1{\sqrt n}$
are decreasing (since $\sqrt x$ is an increasing function).
These two calculations show that $\Sum{n=1}(-1)^{n+1}\Frac 1{\sqrt n}$
converges.
Since $\Sum{n=1}\Frac 1{\sqrt n}$ is a $p${--}series with $p=\frac12<1$,
this series diverges, so
$\Sum{n=1}(-1)^{n+1}\Frac 1{\sqrt n}$ is a conditionally convergent
series.

\np$\Sum{n=1}(-1)^{n+1}\Frac{n}{n^3+1}$.
Compare $\Frac{n}{n^3+1}$ to $\Frac{1}{n^2}$:
$\Li\Frac{\Frac{n}{n^3+1}}{\Frac{1}{n^2}}=
\Li\Frac{n^3}{n^3+1}=1$.
Hence $\Sum{n=1}(-1)^{n+1}\Frac{n}{n^3+1}$ absolutely converges
because $\Sum{n=1}\Frac{1}{n^2}$ is a $p${--}series with $p=2>1$
and so converges.

\np$\Sum{n=1}(-1)^n\Frac{1}{n+3}$.
Since $\Sum{n=1}\Frac{1}{n+3}=\Sum{n=4}\Frac{1}{n}$ which is a
$p${--}series with $p=1$ and so diverges.
Since $\Li\Frac1{n+3}=0$ and since $n+3$ is increasing (so
$\Frac1{n+3}$ is decreasing), the Alternating Series Test
shows $\Sum{n=1}(-1)^n\Frac{1}{n+3}$ converges.
Therefore $\Sum{n=1}(-1)^n\Frac{1}{n+3}$ conditionally converges.

\np$\Sum{n=1}(-1)^{n+1}\Frac{3+n}{5+n}$
This  time $\Li\Frac{3+n}{5+n}=\Frac35\neq0$ so the terms of
this series do not go to $0$ so the series diverges.

\np$\Sum{n=1}\Frac{(-1)^{n-1}}{n^2+2n+1}$.
Compute $\Li\Frac{\Frac{1}{n^2+2n+1}}{\Frac{1}{n^2}}=
\Li\Frac{n^2}{n^2+2n+1}=1$.
Since $\Sum{n=1}\Frac{1}{n^2}$ is a $p${--}series with $p=2>1$,
it converges and hence 
$\Sum{n=1}\Frac{(-1)^{n-1}}{n^2+2n+1}$ converges absolutely.
One can also proceed as follows.
Since $n^2+2n+1=(n+1)^2$
$\Sum{n=1}\Frac{(-1)^{n-1}}{n^2+2n+1}=
\Sum{n=1}\Frac{(-1)^{n-1}}{(n+1)^2}=
\Sum{n=2}\Frac{(-1)^{n}}{n^2}$.
Since $\Sum{n=1}\Frac1{n^2}$ is a $p${--}series with $p=2>1$,
it converges, so $\Sum{n=1}\Frac{(-1)^{n-1}}{n^2+2n+1}$
converges absolutely.

\np$\Sum{n=1}\Frac{(-1)^n(n+1)^n}{(2n)^n}$.
Compute $\Li\Frac{\Frac{(n+1)^n}{(2n)^n}}{\Frac1{2^n}}=
\Li\Bigl(\Frac{n+1}{n}\Bigr)^n=
\Li\Bigl(1+\frac1n\Bigr)^n=e$ as we saw back in the L'H\^opital's
Rule section.
The series $\Sum{n=0}\Frac1{2^n}$ is a geometric series with 
$r=\frac12$ and so converges.
Hence $\Sum{n=1}\Frac{(-1)^n(n+1)^n}{(2n)^n}$ converges absolutely.

\np Estimate the difference between $\ln2$ and the first
four terms of the series $\ln2=\Sum{n=1}(-1)^{n+1}\Frac 1{n}$.

The book checks that the Alternating Series Test applies so
$\Sum{n=1}(-1)^{n+1}\Frac 1{n}-\sum^4_{n=1}(-1)^{n+1}\Frac 1{n}=
\Sum{n=5}(-1)^{n+1}\Frac 1{n}$.
The Alternating Series Test says
$0<\Sum{n=5}(-1)^{n+1}\Frac 1{n}<\Frac 15$ so
$$\frac{7}{12}=1-\frac12+\frac13-\frac14<\Sum{n=1}(-1)^{n+1}\Frac 1{n}<
1-\frac12+\frac13-\frac14+\frac15=\frac{47}{60}$$

\end