%
\def\whichcourse{Math 126}
\def\whichexam{Exam III}
\def\examdate{April 24, 2001}

\font\sm=cmr7
\long\def\newp #1.{
\par\noindent\vtop\bgroup\noindent{\bf #1.}
}
\long\def\problembreak{\par\egroup\vfill\eject\vtop\bgroup}

\def\endp{\par\noindent\vrule width\hsize depth 0pt height .1pt\egroup\par\vfill}
\def\frac#1#2{{{#1}\over{#2}}}
\let\ds=\displaystyle

\long\def\boxit#1#2#3{\vbox{\hrule\hbox{\vrule\kern#2
\vbox{\hsize=#3\kern#2#1\kern#2}\kern#2\vrule}\hrule}}

\magnification1200
\centerline{\whichcourse}
\centerline{\whichexam}
\centerline{\examdate}
\bigskip
\newp 9.
$$\sum_{n=1}^{\infty} \frac{(n!)^n}{(n^{2n})}$$
converge or diverge? Show your reasoning and state clearly any theorems
or tests you are using.
\bigskip
There are many solutions. Easiest is probably the root test:
$\ds\sum a_n$ with $\ds a_n=\frac{(n!)^n}{n^{2n}}$.
\par $\ds\root n\of{a_n}=\frac{n!}{n^2}\to\infty$ \quad as $n\to\infty$.
\par As this limit $\infty$ is $>1$, the root test tells you that the series $\ds\sum a_n$
diverges.

To see why $\ds\lim_{n\to\infty}\frac{n!}{n^2}=\infty$, write it as
$$\ds\lim_{n\to\infty}\frac{n(n-1)\cdot(n-2)!}{n^2}=\lim_{n\to\infty}\frac{n-1}{n}\cdot
\lim_{n\to\infty}(n-2)!=1\cdot\infty$$
It was also OK to have memorized that $\ds\lim_{n\to\infty}\frac{n!}{n^k}=\infty$ for every $k$.

Do NOT use l'H\^opital's rule on $\ds\lim_{n\to\infty}\frac{n!}{n^2}$ since you can not
differentiate $n!$.

Do NOT use the ratio test on $\ds\lim_{n\to\infty}\frac{n!}{n^2}$.
This test would not tell you anything about the {\sl sequence\/} $\ds\biggl\{\frac{n!}{n^2}\biggr\}$.
Rather, it tells you about the {\sl series\/}
$\ds\sum\frac{n!}{n^2}$ (with which you have no business here).

\noindent{\bf Second solution:}
Some people looked at the {\sl series\/} $\ds\sum\frac{n!}{n^2}$ and determined that it diverged
using the ratio test:
$\ds\frac{\frac{(n+1)!}{(n+1)^2}}{\frac{n!}{n^2}}=\frac{n^2}{n+1}\to\infty$ as $n\to\infty$.

By direct comparison, 
$\ds\frac{(n!)^n}{n^{2n}}>\frac{n!}{n^2}$ whenever $\ds\frac{n!}{n^2}>1$ which happens for $n>3$.

\noindent{\bf Third solution:}
Once you have seen $\ds\lim_{n\to\infty}\ds\frac{n!}{n^{2}}=\infty$ and its immediate
consequence $\ds\lim_{n\to\infty}\ds\frac{(n!)^n}{n^{2n}}=\infty$ you can use the fact
that $\ds\lim_{n\to\infty} a_n\neq 0$ to determine that the series diverges.

\problembreak
\noindent{\bf Fourth solution:}
The messiest way to proceed is via the ratio test, but it can be done reasonably well.
$$\eqalign{\frac{a_{n+1}}{a_{n}}=&\frac{\bigl((n+1)!\bigr)^{n+1}n^{2n}}{(n+1)^{2(n+1)}(n!)^n}=
\frac{(n+1)^{n+1}(n!)^{n+1}n^{2n}}{(n+1)^{2n+2}(n!)^n}=
\frac{n^{2n}}{(n+1)^{n+1}}\cdot n!\cr
=&\biggl(\frac{n^2}{n+1}\biggr)^n\cdot\frac{n!}{n+1}=
\biggl(\frac{n^2}{n+1}\biggr)^n\cdot\frac{n}{n+1}\cdot (n-1)!\cr
}$$
The term $\ds\frac{n}{n+1}\to 1$ as $n\to\infty$ and both
$\ds\biggl(\frac{n^2}{n+1}\biggr)^n\to\infty$ and $(n-1)!\to\infty$ as $n\to\infty$.
Hence 
$\frac{a_{n+1}}{a_{n}}\to\infty>1$ and therefore the series diverges.
\endp\vfill\null
\newp 10.
Does the integral

$$\int_{0}^{\infty}\frac{dx}{\sqrt x (x+1)} $$
converge or diverge?

\noindent We compare the function $ \frac{1}{\sqrt x(x+1)} $
with
$\frac{1}{x^{\frac{3}{2}}}$ 

\noindent $\ds\lim_{x\rightarrow \infty}\frac{x^{\frac{3}{2}}}{\sqrt x(x+1)}= 
\lim_{x\rightarrow \infty}\frac{x^{\frac{3}{2}}}{
  x^{\frac{3}{2}}+x^{\frac{1}{2}}}=1$

\noindent This implies that the integral above converges if and only if
$\int_{0}^{\infty}\frac{dx}{x^{\frac{3}{2}}}$ converges.
But $\ds\lim_{b\rightarrow \infty}
\int_{0}^{b}\frac{dx}{x^{\frac{3}{2}}}=
\lim_{b\rightarrow \infty}(-2)[x^{-\frac{1}{2}}]^b_1=
\lim_{b\rightarrow \infty}[2-2\frac{1}{\sqrt b}]=2$, so it
converges and so does the original integral.
\endp

\newp 11.
 Find the {\bf interval} of convergence of the series
$\ds\sum_{n=1}^{\infty}\ \frac{x^{2n}}{2^n\>n^2}\ $.

First calculate the radius of convergence.
Use either the $n$th root test or the ratio test:
$$\ds\lim_{n\to\infty}\root{n}\of{\frac{x^{2n}}{2^n\>n^2}}=
\lim_{n\to\infty}\frac{x^{2}}{2\>\root{n}\of{n^2}}=
\lim_{n\to\infty}\frac{x^{2}}{2(\root{n}\of{n})^2}=\frac x2$$
OR
$$\ds\lim_{n\to\infty}\frac{\frac{x^{2(n+1)}}{2^{n+1}\>(n+1)^2}}{\frac{x^{2n}}{2^n\>n^2}}=
\lim_{n\to\infty}\frac{x^2}{2}\cdot\frac{n^2}{(n+1)^2}=\frac{x^2}{2}
$$
Hence $-1<\frac{x^2}{2}<1$ or $-2<x^2<2$.
Several people had issues with going from $x^2$ to $x$.
It is true that $x^2$ can not be negative so $-2<x^2<2$ can be
replaced with $0\leq x^2<2$ {\sl but\/} $x^2<2$ constrains $x$ from
being too negative. In particular, $x^2<2$ implies $-\sqrt2 < x < \sqrt2$.

We now need to check the endpoints: the two series in question are
$$\sum_{n=1}^{\infty}\ \frac{(\pm\sqrt 2)^{2n}}{2^n\>n^2}=
\sum_{n=1}^{\infty}\ \frac{2^{n}}{2^n\>n^2}=\sum_{n=1}^{\infty}\ \frac{1}{n^2}
\ .$$
This latter series is a $p${--}series with $p=2$ and therefore is convergent.
Hence the interval of convergence is $[-\sqrt 2,\sqrt 2]$.

You were not asked, but we know that the convergence is absolute on the entire
interval $[-\sqrt 2,\sqrt 2]$: it converges absolutely on $(-\sqrt 2,\sqrt 2)$ by
our theory and it converges absolutely at the endpoints because that is what we
just showed.
\endp
\newp 12.
\item{(a)} Show that
$$\sum_{n=0}^{\infty} (-1)^n x^{2n} = \frac{1}{1+x^2}$$
provided that $|x|<1$.


\item{(b)} Find
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)(\sqrt{3})^{2n+1}}.$$
\bigskip
For part (a), this is a geometric series with first term $1$
and ratio $-x^2$. Hence, provided $|x|<1$ the series converges
to
$$\frac{1}{1+x^2}.$$

For part (b), we can integrate both sides and use the term-by-term
integration theorem for the series on the left. This gives
$$\sum_{n=0}^{\infty}(-1)^n \frac{x^{2n+1}}{2n+1}=\arctan(x)+C.$$
The constant $C$ must be zero as we see from setting $x=0$.

Now we plug in $x=\frac{1}{\sqrt{3}}<1$ to both sides and find
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(\sqrt{3})^{2n+1}}=
\arctan(\frac{1}{\sqrt{3}})=\frac{\pi}{6}.$$ 

\endp

\end
