\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.5  p. 649:   1,  3,  5,  7, 21, 23, 27, 29, 35.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}
\def\Int{\displaystyle\int}

\SN{\jobname}
These are to be done using one of the two comparison tests.
To use these tests you need to know examples.
The two families that you know are
\itemitem{$\bullet$} The geometric series
$\displaystyle\sum^\infty r^n$ converges for $\vert r\vert <1$
and diverges otherwise
\itemitem{$\bullet$} The $p${--}series,
$\displaystyle\sum^\infty \Frac{1}{n^p}$ converges for $p>1$ and
diverges otherwise.

\np$\Sum{n=1}\Frac1{2\sqrt n+\root 3\of n}$.
Compare to $\Sum{n=1}\Frac{1}{\sqrt n}$:
$\Li\Frac{\Frac1{2\sqrt n+\root 3\of n}}{\Frac{1}{\sqrt n}}=
\Li\Frac{\sqrt n}{2\sqrt n+\root 3\of n}=
\Li\Frac{1}{2+\frac{\root 3\of n}{\sqrt n}}=
\Li\Frac{1}{2+\frac{1}{n^{\frac16}}}=\Frac12\neq0$.
Hence the two series diverge or converge together and
$\Sum{n=1}\Frac{1}{\sqrt n}$ is a $p${--}series with $p=\frac12<1$
and so both series diverge.

\np$\Sum{n=1}\Frac{\sin^2 n}{2^n}$.
Compare to $\Sum{n=1}\Frac{1}{2^n}$.
This is a geometric series with $r=\frac12<1$ and so converges.
Since $0\leq \sin^2 n<1$, the first comparison test shows that
$\Sum{n=1}\Frac{\sin^2 n}{2^n}$ comverges.

\np$\Sum{n=1}\Frac{2n}{3n-1}$.
Since $\Li\Frac{2n}{3n-1}=\frac23\neq0$ the series diverges since
the limit of the terms is not $0$.

\np$\Sum{n=1}\Bigl(\Frac{n}{3n+1}\Bigr)^n$.
Compare to $\Sum{n=1}\bigl(\frac{1}{3}\bigr)^n$ which is a geometric
series with $r=\frac13<1$ and hence converges.
Since $0<\Frac{n}{3n+1}=\Frac{1}{3+\frac1n}<\Frac13$, the first comparison 
test shows that $\Sum{n=1}\Bigl(\Frac{n}{3n+1}\Bigr)^n$ converges.

\np$\Sum{n=1}\Frac{1-n}{n2^n}$.
Note $\Sum{n=1}\Frac{1-n}{n2^n}=-\Sum{n=1}\Frac{n-1}{n2^n}$ so
we can determine the behavior of $\Sum{n=1}\Frac{n-1}{n2^n}$
which has non{--}negative terms so our theorems apply.
Compare to $\Sum{n=1}\Frac{1}{2^n}$ which is a geometric series
with $r=\frac12<1$ and hence converges.
Compute $\Li\Frac{\Frac{n-1}{n2^n}}{\Frac{1}{2^n}}=
\Li\Frac{n-1}{n}=1\neq0$ so by the limit comparison test,
$\Sum{n=1}\Frac{n-1}{n2^n}$ converges and hence so does
$\Sum{n=1}\Frac{1-n}{n2^n}$.

\np$\Sum{n=1}\Frac{1}{3^{n-1}+1}$.
Compare to $\Sum{n=1}\Frac{1}{3^{n}}$ which is a geometric series
with $r=\frac13<1$ and hence converges,.
Compute $\Li\Frac{\Frac{1}{3^{n-1}+1}}{\Frac{1}{3^{n}}}=
\Li\Frac{3^n}{3^{n-1}+1}=
\Li\Frac{3}{1+\frac1{3^{n-1}}}=3$ so by the limit comparison test
$\Sum{n=1}\Frac{1}{3^{n-1}+1}$ converges.

\np$\Sum{n=1}\Frac{10n+1}{n(n+1)(n+2)}$.
Compare to $\Sum{n=1}\Frac{n}{n^3}=
\Sum{n=1}\Frac{1}{n^2}$ which is a $p${--}series with $p=2>1$
and hence converges.
Compute $\Li\Frac{\Frac{10n+1}{n(n+1)(n+2)}}{\Frac{1}{n^2}}=
\Li\Frac{10n^3+n^2}{n(n+1)(n+2)}=10\neq0$.
Hence $\Sum{n=1}\Frac{10n+1}{n(n+1)(n+2)}$ converges by the
limit comparison test.

\np$\Sum{n=1}\Frac{\arctan n}{n^{1.1}}$.
To understand the behavior of the terms,
$\Frac{\arctan n}{n^{1.1}}$, we recall the behavior of
$\arctan x$ as $x\to\infty$:
$\displaystyle\lim_{x\to\infty}\arctan x=\Frac\pi2$.
Since $\dd{\arctan x}x=\Frac1{1+x^2}>0$ it follows that
$\arctan n<\Frac\pi2$.
Compare to $\Sum{n=1}\Frac{\frac\pi2}{n^{1.1}}=
\frac\pi2\cdot\Sum{n=1}\Frac{1}{n^{1.1}}$.
Now  $\Sum{n=1}\Frac{1}{n^{1.1}}$ is a $p${--}series with
$p=1.1>1$ and hence convergent.
It follows that $\Sum{n=1}\Frac{\frac\pi2}{n^{1.1}}$ converges and
since $0<\Frac{\arctan n}{n^{1.1}}<\Frac{\frac\pi2}{n^{1.1}}$, it follows
from the first comparison test that 
$\Sum{n=1}\Frac{\arctan n}{n^{1.1}}$ converges.

\np$\Sum{n=1}\Frac{1}{1+2+3+\cdots+n}$.
Again we need to understand the individual terms better.
A useful formula is
$1+2+\cdots+n=\Frac{n(n+1)}{2}$.
Hence
$\Sum{n=1}\Frac{1}{1+2+3+\cdots+n}=
\Sum{n=1}\Frac{2}{n(n+1)}$.
This can be compared to $\Sum{n=1}\Frac{1}{n^2}$ which is
a $p${--}series with $p=2>1$ and hence converges.
Compute $\Li\Frac{\Frac{2}{n(n+1)}}{\Frac{1}{n^2}}=
\Li\Frac{2n^2}{n(n+1)}=2\neq0$.
It follows from the limit comparison that
$\Sum{n=1}\Frac{1}{1+2+3+\cdots+n}$ converges.

\end