\documentclass[fleqn, 12pt]{article}
%\pagestyle{empty}
\setlength{\topmargin}{-.5in}
\setlength{\oddsidemargin}{-.3in}
\setlength{\textheight}{9in}
\setlength{\textwidth}{7.2in}


\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{enumerate}

\newcommand{\adj}{\mathrm{Adj}}
\newcommand{\mbf}{\mathbf}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ra}{\rightarrow}
\newcommand{\proj}{\mathrm{proj}}
\newcommand{\dotprod}{\bullet}

\newcommand{\sol}{{\textbf{Solution: }}}
\newcommand{\inch}{\vspace{1in}}

\newcommand{\lecture}{{\Large\textbf{Lecture Problems}}}

\begin{document}

\begin{center}
{\Large Warm-Up Problems and Lecture Problems

April 9, 2003}
\end{center}

The warm up problems are for fun and to illustrate some ideas about adding up a series.

\begin{enumerate}
\item (This problem is supposed to be fun!) 
	Consider the geometric series below:
	\[
	3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots
	\]
	\begin{enumerate}
	\item  What is the first term $a$?
	\item  What is the common ratio $r$?
	\item  Does the series converge or diverge?  If it converges, find the sum.
	\end{enumerate}
	Here's another way to find this sum.
	Let $S$ be the sum: $S= \sum_{n=0}^\infty	3(\frac{1}{10})^n$.
	Then $10S = 10\sum_{n=0}^\infty 3(\frac{1}{10})^n$.
	Of course $10S-S = 9S$, which we compute below:
	\begin{eqnarray*}
	10S & = & 30 + 3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots \\
	S & = & 3 + \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots \\
	10S - S = 9S & = & 30
	\end{eqnarray*}
	Therefore $S = \frac{30}{9} = \frac{10}{3}$.

\item  Lets try this again, but this time with the series $S = \sum_{n=0}^\infty 3^n$.
	Let $S$ denote the sum and we multipy by $3$ and subtract:
	\begin{eqnarray*}
	3S & = & 3 + 9 + 27 + \cdots \\
	S & = & 1 + 3 + 9 + 27 + \cdots \\
	3S-S = 2S & = & -1
	\end{eqnarray*}
	Therefore $S=-\frac{1}{2}$.
	\\
	Here's the question:  Is there anything wrong with what we just did?  Why or why not?

\item  Here are a few similar problems for you to try 
	(of course they all really diverge, but we're having fun).
	\begin{enumerate}
	\item  $S=\sum_{n=0}^\infty 3(2)^n$
\bigskip

	\item  $S=\sum_{n=0}^\infty 7(-5)^n$
\bigskip
	\end{enumerate}

\newpage
\lecture

\item  Determine which of the following terms describe the sequences:
	\begin{quote}
	\textbf{convergent}, \textbf{divergent}, \textbf{conditionally convergent}, 
	\textbf{absolutely convergent}
	\end{quote}
	\begin{enumerate}
	\item  $\sum_{n=1}^\infty \frac{n}{2n+1}$
\inch

	\item  $\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{2n+1}$
\inch

	\item  $\sum_{n=1}^\infty (-1)^{n}\frac{n}{2n+1}$
\inch

	\item  $\sum_{n=1}^\infty \frac{1}{n^2}$
\inch

	\item  $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n^2}$
\inch

	\item  $\sum_{n=1}^\infty \frac{\ln n}{n}$
\inch

	\item  $\sum_{n=1}^\infty (-1)^{n+1} \frac{\ln n}{n}$

	\end{enumerate}

\end{enumerate}


\end{document}