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

Solutions

April 11, 2003}
\end{center}

\begin{enumerate}

\item  Determine if the following series converge or diverge.
	If the series converges, determine if the series converges conditionally or 
	if the series converges absolutely.
	\begin{enumerate}
	\item  $\sum_{n=2}^\infty (-1)^n \ln n$
		\\\sol diverges: $\lim_{n \ra \infty} a_n \neq 0$.
\inch

	\item  $\sum_{n=2}^\infty (-1)^n \frac{1}{\ln n}$
		\\\sol converges by alternating series test.
		But, this series converges conditionally since the series
		$\sum_{n=2}^\infty \frac{1}{\ln n}$ diverges by the comparison test 
		($\frac{1}{n} < \frac{1}{\ln n}$).
\inch

	\item  $\sum_{n=1}^\infty (-1)^{n+1} \frac{n+2}{n^3}$
		\\\sol  converges absolutely by the limit comparison test. 
		Compare $\sum_{n=1}^\infty \frac{n+2}{n^3}$ to $\sum_{n=1}^\infty \frac{n}{n^3}$.
		(Since the series of absolute values converges, we know that the alternating series 
		also converges).
\inch

	\item  $\sum_{n=1}^\infty (-1)^{n+1} \frac{n+2}{n^2}$
		\\\sol  This certainly does not converge absolutely 
		(use a limit comparison test, compare to $\sum_{n=1}^\infty \frac{n}{n^2}$).
		But, it does converge conditionally.
		$\lim_{n \ra \infty} b_n = 0$.  And, if you let $f(x)=\frac{x+2}{x^2}$,
		then $f'(x)=-\frac{(x+4)}{x^3} < 0$ and therefore the terms $\frac{n+2}{n^2}$ are decreasing.

	\end{enumerate}


\newpage
{\Large\textbf{Lecture and homework problems}}

\item  Using the ratio test, 
	determine the radius of convergence and the interval of convergence for the following series:
	\begin{enumerate}
	\item  $\sum_{n=0}^\infty 3^nx^n$
		\\\sol  Letting $a_n=3^nx^n$:
		\[
		\lim_{n \ra \infty} \left|\frac{a_{n+1}}{a_n} \right|
		= \lim_{n \ra \infty} \left|\frac{3^{n+1}x^{n+1}}{3^nx^n} \right|
		= \left|3x \right|
		\]
		Since we want this to be less then one, we have $|3x|<1$ or $|x|<\frac{1}{3}$.
		$R=\frac{1}{3}$.
		To find the interval of convergence, 
		you need to test convergence (by hand) when $x=\frac{1}{3}$ and $x=-\frac{1}{3}$.
		If $x=\frac{1}{3}$ then the series is $\sum_{n=0}^\infty 1$, which diverges.
		If $x=-\frac{1}{3}$ then the series is $\sum_{n=0}^\infty (-1)^n$ which diverges.
		So, the interval of convergence is $(-\frac{1}{3}, \frac{1}{3})$.
\inch

	\item  $\sum_{n=0}^\infty \frac{x^n}{3^n}$
		\\\sol $R=3$ and the interval of convergence is $(-3,3)$.
\inch

	\item  $\sum_{n=1}^\infty \frac{x^n}{n3^n}$
		\\\sol $R=3$ and the interval of convergence is $[-3,3)$.
\inch

	\item  $\sum_{n=0}^\infty \frac{nx^n}{3^n}$
		\\\sol $R=3$ and the interval of convergence is $(-3,3)$.

	\end{enumerate}


\end{enumerate}


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