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

Solutions

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  $S=\sum_{n=0}^\infty 3(2)^n$
		\\\sol $2S-S = -3$ so this sum is $-3$.

	\item  $S=\sum_{n=0}^\infty 7(-5)^n$
		\\\sol  $-5S-S=-6S = -7$, so this sum is $\frac{7}{6}$.
	\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  $\sum_{n=1}^\infty \frac{n}{2n+1}$
		\\\sol Divergent!
\bigskip

	\item  $\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{2n+1}$
		\\\sol Divergent!
\bigskip

	\item  $\sum_{n=1}^\infty (-1)^{n}\frac{n}{2n+1}$
		\\\sol Divergent!
\bigskip

	\item  $\sum_{n=1}^\infty \frac{1}{n^2}$
		\\\sol Convergent, absolutely convergent.
\bigskip

	\item  $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n^2}$
		\\\sol Convergent, absolutely convergent.  We already know it is absolutely convergent,
		so therefore this series is convergent.
\bigskip

	\item  $\sum_{n=1}^\infty \frac{\ln n}{n}$
		\\\sol Divergent!  This is a comparison test $\frac{\ln n}{n} \geq \frac{1}{n}$.
\bigskip

	\item  $\sum_{n=1}^\infty (-1)^{n+1} \frac{\ln n}{n}$
		\\\sol Convergent, conditionally convergent.
		All that needs to be checked is that $\lim \frac{\ln n}{n} = 0$ (which is a l'Hopitals rule)
		and that the series is decreasing.  Just look at $f(x) = \frac{\ln x}{x}$ and compute
		$f'(x) = (1-\ln x)/x^2$, which is always negative for $x>1$.  So, the series is decreasing.

	\end{enumerate}

\end{enumerate}


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