\documentclass[12pt]{article}
\usepackage{latexsym, amscd, %graphicx, 
amssymb}
\newcommand{\SECT}[1]
		{\vspace{.3cm}
		$\!\!\!\!\!\!\!\!\!\!\!${\bf {#1}}
		\vspace{.2cm}}

\begin{document}


\begin{center}

{\large \bf Math 152, Fall, 2004, Workshop 9\par}
\vspace{.5cm}
{{\bf Honors Section} 
\par}
\end{center}

\begin{enumerate}

\item For each of the following series, determine whether the 
three conditions of the alternating series 
test are satisfied. If the answer is `YES', determine how many 
terms of the series need be added 
up to get an approximation to the exact sum of the series 
within an error tolerance of $10^{-20}$.
$$(i) \; \sum_{n=1}^{\infty}\frac{(-1)^{n}}{(n^{3}+4)^{2}},\;\;\;\;\;\;\;\;\;\;
(ii)\;  \sum_{n=2}^{\infty}\frac{(-1)^{n}n}{n+\ln n}, \;\;\;\;\;\;\;\;\;\;
(iii) \; \sum_{n=1}^{\infty}(-1)^{n}\frac{\sin^{2}n}{n}.$$

\item Consider the two series
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$
and 
$$\sum_{n=1}^{\infty}\frac{1}{n2^{n}}.$$

\begin{enumerate}

\item Explain why each series converges.

\item It happens that the
sum of each of these series equals $\ln 2$. 
(We shall learn why, in another week 
or so.) Granted this fact, suppose that you have to compute 
$\ln 2$ to an accuracy of $10^{-6}$. Which series 
is better suited for this calculation? Justify your answer
by figuring out how many terms of each series you have to add up
to get a partial sum of the desried accuracy, that is, to
have $|R_{N}|<10^{-6}$. (Hint for the second series: 
Estimate the error $|R_{N}$ by comparison with the error
of a certain geometric series.)

\end{enumerate}

\item Consider the alternating series 
$$\sum_{n=1}^{\infty}(-1)^{n}\left(\frac{1}{n}+\frac{(-1)^{n}}{n}\right).$$

\begin{enumerate}

\item Write down explicitly the first $10$ terms of the series. 

\item Verify that $a_{n}\ge 0$ and $\lim_{n\to \infty}a_{n} = 0$. 

\item Does the series converge? Explain. 

\item Is the alternating series test applicable? Explain. 

\end{enumerate}


4. For which positive integers $k$ does 
$$\sum_{n=1}^{\infty}\frac{k^{n}n!}{n^{n}}$$
converge? Hint: 
$$\lim_{n\to \infty} \left(1+\frac{1}{n}\right)=e.$$

\end{enumerate}

\end{document}