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\begin{document}
	
	\mbox{}
	\title{Solutions for Quiz 2}
	
	\maketitle
	
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%%%%%%%%%%%%%%%%%%%%%%%%%%%%
	%\section{Introduction} \label{sec:intro}
	\noindent {\footnotesize Note: Most of the problems were taken from the textbook} \cite{jS15}.
		
	\begin{problem}
		Determine whether the sequence $a_n = \cos \big(\sqrt[n]{\pi^{207 + n}} \big)$ converges or diverges. If it is convergent, find the limit.
	\end{problem}
	
	\noindent {\it Solution:} Since $\cos x$ and $\pi^x$ are both continuous functions we have that
	\[
		\lim_{n \to \infty} a_n = \lim_{n \to \infty} \cos \bigg( \pi^{\frac{207}{n} + 1}\bigg) = \cos \bigg( \pi^{\lim \frac{207}{n} + 1}\bigg)= \cos (\pi) = -1.
	\]
	Hence $\{a_n\}$ is a convergent sequence.
	\hfill \qed
	
	\begin{problem}
		Determine whether the series
		\[
			\sum_{n=1}^{\infty} \sqrt[3]{\frac{n^2}{n^2 + 20n + 9}}
		\]
		is convergent or divergent. If it is convergent, find the sum.
	\end{problem}
	
	\noindent {\it Solution:} As $\sqrt[3]{x}$ is a continuous function, it follows that
	\[
		\lim_{n \to \infty} \sqrt[3]{\frac{n^2}{n^2 + 20n + 9}} = \sqrt[3]{\lim_{n \to \infty}\frac 1{1 + 20/n + 9/n^2}} = \sqrt[3]{1} = 1.
	\]
	Since the general term of the series does not converges to zero, the series diverges.
	\hfill \qed
	
	\begin{problem}
		Find the exact area of the surface of revolution obtained by rotating the curve $x = 1 + 2y^2, \ 1 \le y \le 2$ about the $x$-axis.
	\end{problem}
	
	\noindent {\it Solution:} From $x = 1 + 2y^2$, we find that $f(x) = y = \sqrt{\frac{x-1}{2}}$. Since $1 \le y \le 2$, one has that $3 \le x \le 9$. In addition,
	\[
		1 + [f(x)']^2 = 1 + \bigg(\frac 1{\sqrt{2}} \frac 1{2\sqrt{x-1}} \bigg)^2 = 1 + \frac 1{8(x-1)} = \frac{8x - 7}{8(x-1)}.
	\]
	Therefore the area of the surface of revolution is
	\begin{align*}
		2\pi \int_3^9 \sqrt{\frac{x-1}{2}} \sqrt{\frac{8x-7}{8(x-1)}} \, dx 
		&= \frac {\pi}{2} \int_3^9 (8x - 7)^{1/2} \, dx \\
		&= \frac{\pi}{16} \int_{17}^{65} u^{1/2}\,du = \frac{\pi}{24}\big(\sqrt{65^3} - \sqrt{17^3}\big).
	\end{align*}
	\hfill \qed
	
	\newpage
	
	\begin{problem}
		Use the Integral Test to determine whether the series $\sum_{n=1}^{\infty} n^2\pi^{-n^3}$ is convergent or divergent.
	\end{problem}
	
	\noindent {\it Solution:} The function $f(x) = x^2\pi^{-x^3}$ is continuous and positive for all $x \ge 1$. Moreover,
	\[
		f(x)' = \frac{\pi^{x^3}\big(2x -(3 \ln \pi) x^4\big)}{\pi^{2x^3}} < 0
	\]
	for all $x \ge 1$. Then $f(x)$ is decreasing when $x \ge 1$. Taking $u = x^3$ below, we find that
	\[
		\int_1^\infty f(x) \, dx = \frac 13 \int_1^\infty \bigg(\frac 1{\pi}\bigg)^u \, du = \frac {1}{3 \ln(1/\pi)} (-1/\pi) = \frac 1{3\pi \ln \pi}.
	\]
	Since the above improper integral is finite, the series converges by the Integral Test.
	\hfill \qed
	
	\begin{problem}
		Determine whether the series
		\[
			\sum_{n=1}^{\infty} \bigg( \frac 2{(n+1)(n+3)} + \frac{3}{\pi^{n+1}}\bigg)
		\]
		is convergent or divergent. If it is convergent, find the sum.
	\end{problem}
	
	\noindent {\it Solution:} Notice that
	\begin{align*}
		\sum_{n=1}^{\infty} \bigg( \frac 2{(n+1)(n+3)} + \frac{3}{\pi^{n+1}}\bigg) &= \sum_{n=1}^{\infty} \bigg(\frac 1{n+1} - \frac 1{n+3} \bigg) + \frac{3}{\pi^2} \sum_{n=0}^\infty \bigg(\frac 1{\pi}\bigg)^n \\
		&= \lim_{n \to \infty} \bigg(\frac 12 + \frac 13 - \frac 1{n+2} - \frac 1{n+3} \bigg) + \frac{3/\pi^2}{1 - 1/\pi} \\
		&= \frac 56 + \frac 3{\pi^2 - \pi}.
	\end{align*}
	Hence the series converges to $5/6 + 3/(\pi^2 - \pi)$.
	\hfill \qed
	
	
	\begin{thebibliography}{20}
		
		\bibitem{jS15} J.~Stewart: \emph{Single Variable Calculus} 8th Edition, Cengage Learning, Boston 2015.
		
	\end{thebibliography}
	
\end{document}