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\title{Math 1b.  Lecture 20 \\
Strategies for Determining Convergence and Divergence of a Series}

\author{T. Judson}

\date{Spring 2006}


\begin{document}

\maketitle


\section{Goals}

\begin{itemize}

\item
To understand and be able to apply the different tests for convergence and divergence of a series.

\end{itemize} 

\section{Strategies}

\begin{enumerate}

\item
If the series is  of the form $\sum 1/n^p$, then it is a $p$-series.  The series converges for $p > 1$ and diverges for $p \leq 1$.

\item
If the series has the form $\sum a r^n$, then it is a geometric series and converges for $|r| < 1$ and diverges for $|r| \geq 1$.

\item
If the series is similar to a $p$-series or a geometric series, consider the Comparison Test.

\item
If $\lim_{n \to \infty} a_n \neq 0$, then the series diverges.

\item
If the series is of the form $\sum (-1)^{n+1} a_n$, consider applying the Alternating Series Test.  You can also test for absolute convergence.

\item
If the series involves products, factorials, or constants raised to the $n$th power, consider the Ratio Test.

\item
If $a_n = f(n)$ and the integral $\int_1^\infty f(x) \, dx$ is easily evaluated, the Integral Test may be useful assuming the hypothesis of the test are satisfied.

\item
Is the series a telescopic series?  If so, convergence or divergence can be determined by computing the limit of the partial sums of the series.

\end{enumerate}




\section{Worksheet Problems}

\begin{enumerate}

\item
$\displaystyle \sum_{n = 1}^\infty \frac{n-1}{n^2 + n}$

\item
$\displaystyle \sum_{n = 1}^\infty \frac{(-3)^{n+1}}{2^{3n}}$

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

\item
$\displaystyle \sum_{k = 1}^\infty \frac{5^k}{3^k + 4^k}$

\item
$\displaystyle \sum_{n = 1}^\infty \frac{3^n n^2}{n!}$

\item
$\displaystyle \sum_{n = 1}^\infty ne^{-n^2}$

\item
$\displaystyle \sum_{n = 1}^\infty \left(\frac{n}{n + 1} \right)^{n^2}$

\item
$\displaystyle \sum_{n = 1}^\infty \tan(1/n)$

\item
$\displaystyle \sum_{n = 1}^\infty \frac{\sqrt{n^2 - 1}}{n^3 - 2n^2 + 5}$

\item
$\displaystyle \sum_{n = 1}^\infty (-1)^n 2^{1/n}$

\item
$\displaystyle \sum_{n = 1}^\infty \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}$

\end{enumerate}





\section*{References}

\begin{itemize}

\item
\S 8.2--8.4 in 
James Stewart.
{\it Single Variable Calculus: Concepts \& Context}, third edition.
Brooks/Cole, Belmont CA, 2005.
ISBN 0--534--41022--7.


\end{itemize}



\section*{Notes}

\subsection*{April 3, 2006}



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