\input homework_macros.tex

%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.4  p. 643:   3,  5, 11, 21, 33, 39, 41.
}
\def\Li{\displaystyle\lim_{n\to\infty}}
\def\Sum#1{\displaystyle\sum_{#1}^\infty}
\def\Int{\displaystyle\int}

\SN{\jobname}
\np
$\Sum{n=1}\Frac{n}{n+1}$.
First compute $\Li a_n=\Li\Frac{n}{n+1}=1$.
Hence the series diverges.

\np
$\Sum{n=1}\Frac{3}{\sqrt n}=3\Sum{n=1}\frac1{n^{1/2}}$.
Now $\Sum{n=1}\frac1{n^{1/2}}$ is a $p${--}series with $p=\frac12$
and since $\frac12<1$, it diverges.
Hence the original series diverges.

\np
$\Sum{n=1}\Frac{2^n}{3^n}$ is a geometric series with
$r=\frac23$.
Since $\frac23<1$, this series converges.

\np
$\Sum{n=3}\Frac{1/n}{(\ln n)\sqrt{\ln^2n-1\ }}$.
We want to use the Integral Test with
$f(x)=\Frac{1}{x(\ln x)\sqrt{\ln^2 x-1\ }}$.
First we need to check that $f(x)$ is non{--}negative and
decreasing.
Equivalently, we can check that $x(\ln x)\sqrt{\ln^2 x-1\ }$
is non{--}negative and increasing.
Both $x$ and $\ln x$ are positive and increasing for $x>1$;
it follows that $\ln^2x-1$ is also increasing and hence
$\sqrt{\ln^2x-1\ }$ is also increasing if $\ln x>1$ or $x>e$.
Therefore, for $x\geq3$, the product of
$x(\ln x)\sqrt{\ln^x-1\ }$ is positive and increasing.
Hence the series converges or diverges provided
$\Int^\infty_3\Frac{1}{x(\ln x)\sqrt{\ln^2 x-1\ }}\>dx$ 
converges or diverges.
Rather than do this integral directly, we use the Limit
Comparison Test on the function
$g(x)=\Frac{1}{x\>ln^2x}$.
First compute $\displaystyle\lim_{x\to\infty}\Frac{f(x)}{g(x)}=
\displaystyle\lim_{x\to\infty}
\Frac{\Frac{1}{x(\ln x)\sqrt{\ln^2 x-1\ }}}{\Frac{1}{x\>ln^2x}}=
\displaystyle\lim_{x\to\infty}\Frac{\ln x}{\sqrt{\ln^2x-1\ }}=
\displaystyle\lim_{x\to\infty}\Frac{1}{\sqrt{1-\frac1{\ln^2x}}}=1
$.
The integral $\Int\Frac{dx}{x\ln^2x}$ can be done by the susbtitution
$u=\ln x$, $du=\frac{dx}x$ so
$\Int\Frac{dx}{x\ln^2x}=\Int\Frac{du}{u^2}=-u^{-1}+C=
-\Frac1{\ln x}+C$ so
$\Int^\infty_3\Frac{dx}{x\ln^2x}=
\displaystyle\lim_{t\to\infty}-\frac{1}{\ln t}-(-\frac1{\ln3})=
\frac1{\ln3}$ so the improper integral converges.
It follows from the Limit Comparison Test that
$\Int^\infty_3\Frac{dx}{x(\ln x)\sqrt{\ln^2 x-1\ }}$
converges and therefore the series converges.

The original integral can also be done by the same substitution,
$u=\ln x$: the integral becomes
$\Int\Frac{1}{x(\ln x)\sqrt{\ln^2 x-1\ }}\>dx=
\Int\Frac{1}{u\sqrt{u^2 -1\ }}\>du=\arc{\sec} u+C=
\arc{\sec}(\ln x)+C$.
Hence $\Int^\infty_3\Frac{1}{x(\ln x)\sqrt{\ln^2 x-1\ }}\>dx$ converges
if and only if $\LI x\arc{\sec}(\ln x)$ exits and is finite.
Since $\LI x\ln x=\infty$,
$\LI x\arc{\sec}(\ln x)=\LI t\arc{\sec}(t)=\frac\pi2$.
Hence $\Int^\infty_3\Frac{1}{x(\ln x)\sqrt{\ln^2 x-1\ }}\>dx$ converges,
as does the series.

\np
$a_1=1$, $a_{n+1}=\Frac{1+\ln n}n a_n$:
does $\Sum{n=1}a_n$ converge or diverge?
The Ratio Test appears the way to go since from the
definition,
$\Frac{a_{n+1}}{a_n}=\Frac{1+\ln n}n$.
Hence $\LI n\Frac{a_{n+1}}{a_n}=
\LI n\frac1n+\LI n\frac{\ln n}n$ and
$\LI n\frac 1n=0$ and $\LI n\frac{\ln n}n=0$ so
$\LI n\Frac{a_{n+1}}{a_n}=0$ and the series converges.

\np
$\Sum{n=1}\Frac{(n!)^n}{(n^n)^2}$.
We rewrite the $n$th term of the series so as to get an idea of
how fast it is going to $0$.
First write $\Frac{(n!)^n}{(n^n)^2}=\Bigl(\Frac{n!}{n^2}\Bigr)^n$.
Now apply the Root Test:
$\LI n\root n\of{\Frac{(n!)^n}{(n^n)^2}}=
\LI n\Frac{n!}{n^2}$.
Then note that $\Frac{n!}{n^2}=(n-2)!\cdot \frac{n-1}n=
(n-3)!\cdot \frac{(n-2)(n-1)}n$.
If $n\geq4$, $\frac{(n-2)(n-1)}n>1$ and $(n-3)!\geq n-3$ so
$\Frac{n!}{n^2}>n-3$ and
$\LI n\Frac{n!}{n^2}=\infty$.
In particular, the series diverges.

\np
$\Sum{n=1}\Frac{n^n}{2^{n^2}}$.
As in \#39, let us start by studying $\Frac{n^n}{2^{n^2}}$.
$2^{n^2}=2^{n\cdot n}=\bigl(2^n\bigr)^n$.
Hence $\Frac{n^n}{2^{n^2}}=\Bigl(\Frac{n}{2^n}\Bigr)^n$.
The Root Test again appears the way to go:
$\LI n \root n\of{\Frac{n^n}{2^{n^2}}}=
\LI n\Frac{n}{2^n}=0$ where this last limit can be evaluated by
L'H\^opital's Rule if you don't remember that an exponential function
with a base greater than $1$ increases faster than any polynomial.
Hence the series converges.


\end