\documentclass[10pt,addpoints]{exam}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usepackage{pgfplots}
\pagestyle{head}
\usepackage{tikz}
\usepackage{hyperref}
\usepackage{cancel}
%\headrule
%\header{MATH 11}{Exam 1}{Fall 2016}
\begin{document}
\begin{coverpages}
\begin{center}\huge MATH 3315 - EXAM 4 - FALL 2017\\
SOLUTION
\end{center}
\begin{flushleft}\vspace{0.5in}
Friday 17 November 2017\\
Instructor: Tom Cuchta
\end{flushleft}
\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}

\vspace{-10pt}

\vspace*{10pt}
\end{coverpages}

\begin{questions} 
\question[13] Find the exact value of the series.
\begin{parts}
\part[6] $\displaystyle\sum_{k=0}^{\infty} \left( \dfrac{2}{3} \right)^k$ \\
\textit{Solution}: This is a geometric series with $r=\dfrac{2}{3}$. So \href{http://www.wolframalpha.com/input/?i=sum+from+k%3D0+to+k%3Dinfinity+of+(2%2F3)%5Ek}{compute}
$$\displaystyle\sum_{k=0}^{\infty} \left( \dfrac{2}{3} \right)^k = \dfrac{1}{1-\frac{2}{3}} = \dfrac{1}{\frac{1}{3}} = 3.$$
\part[7] $\displaystyle\sum_{k=2}^{\infty} \left( \dfrac{1}{k} - \dfrac{1}{k-1}  \right)$ \\
\textit{Solution}: Consider the infinite sum as a limit of a partial sum:
$$\begin{array}{ll} 
\displaystyle\sum_{k=2}^{\infty} \dfrac{(-1)^k x^k}{k} &= \displaystyle\lim_{N \rightarrow \infty} \displaystyle\sum_{k=2}^N \dfrac{(-1)^k x^k}{k} \\
&= \displaystyle\lim_{N \rightarrow \infty} \left( \cancel{\dfrac{1}{2}} - \dfrac{1}{1} \right) + \left( \cancel{\dfrac{1}{3}} - \cancel{\dfrac{1}{2}} \right) + \left( \cancel{\dfrac{1}{4}} - \cancel{\dfrac{1}{3}} \right) + \ldots + \left( \dfrac{1}{N} - \cancel{\dfrac{1}{N-1}} \right) \\
&= \displaystyle\lim_{N \rightarrow \infty} -1 + \dfrac{1}{N} \\
\end{array}$$
\end{parts}

\question[21] Find the interval of convergence and radius of convergence of the following power series.
\begin{parts}
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k x^k}{k}$ \\
\textit{Solution}: Using the ratio test, compute
$$\begin{array}{ll}
\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{(-1)^{k+1}x^{k+1}}{k+1} \dfrac{k}{(-1)^k x^k} \right| &= \displaystyle\lim_{k \rightarrow \infty} \left| x \dfrac{k}{k+1} \right| \\
&= |x|.
\end{array}$$
The ratio test tells us that the series converges provided that $|x|<1$, i.e. $-1<x<1$ and that the series diverges provided that $|x|>1$, i.e. $x<-1$ and $x>1$. We must check the endpoints $x=1$ and $x=-1$ manually. At $x=1$, the series becomes $\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k}{k}$, which converges by the alternating series test. At $x=-1$, the series becomes $\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k (-1)^k}{k} = \displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k}$, which diverges (it is the harmonic series). Therefore the interval of convergence is $(-1,1]$ and the radius of convergence is $1$.

\part[7] $\displaystyle\sum_{k=1}^{\infty} k! x^k$ \\
\textit{Solution}: By the ratio test,
$$\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{(k+1)!x^{k+1}}{k! x^k} \right| = \displaystyle\lim_{k \rightarrow \infty} \left| (k+1)x \right| = \left\{ \begin{array}{ll}
0, & \quad x=0 \\
\infty, & \quad x \neq 0.
\end{array} \right.$$
Therefore the interval of convergence of the series is the set $[0,0]=\{0\}$ and the radius of convergence is $0$.

\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k x^{k}}{(k+1)!}$ \\
\textit{Solution}: By the ratio test,
$$\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{x^{k+1}}{(k+2)!} \dfrac{(k+1)!}{x^k} \right| = \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{x}{k+2} \right| = 0,$$
and so the interval of convergence of the series is $(-\infty,\infty)=\mathbb{R}$ and the radius of convergence is $\infty$. 
\end{parts}
\newpage 
\question[10] Find a power series for $f(x)=\dfrac{1}{2-x}$ centered at $c=3$. \\
\textit{Solution}: Think about a simple geometric series centered at $c=3$:
$$\displaystyle\sum_{k=0}^{\infty} (x-3)^k = \dfrac{1}{1-(x-3)} = \dfrac{1}{1-x+3}.$$
So we want to rewrite $\dfrac{1}{2-x}$ to have a denominator with ``$-x+3$" in it. Therefore write
$$\dfrac{1}{2-x} = \dfrac{1}{2-x+3-3} = \dfrac{1}{-1-(x-3)}= \dfrac{-1}{1-(3-x)}=-\displaystyle\sum_{k=0}^{\infty} (3-x)^k.$$

\question[56] Converge or diverge? Why? Explain.

\begin{parts}
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k k}{k^2-5}$ \\
\textit{Solution}: Converges by the alternating series test:
$$\displaystyle\lim_{k \rightarrow \infty} \dfrac{k}{k^2-5} \stackrel{L.H.}{=} \displaystyle\lim_{k \rightarrow \infty} \dfrac{1}{2k} = 0.$$
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{k^4-3k^2+1}{2k^4-k+17}$ \\
\textit{Solution}: Diverges by the test for divergence:
$$\displaystyle\lim_{k \rightarrow \infty} \dfrac{k^4-3k^2+1}{2k^4-k+17} = \dfrac{1}{2} \neq 0.$$
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^{e}}$ \\
\textit{Solution}: Converges because it is a $p$-series with $p=e\approx 2.71\ldots > 1$.
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{5^k}{(2k)!}$ \\
\textit{Solution}: Converges by the ratio test: 
$$\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{5^{k+1}}{(2(k+1))!} \dfrac{(2k)!}{5^k} \right| =\displaystyle\lim_{k \rightarrow \infty} \dfrac{5}{(2k+2)(2k+1)} = 0<1.$$
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{k^k}{(3k+1)^k}$ \\
\textit{Solution}: Converges by the root test:
$$\displaystyle\lim_{k \rightarrow \infty} \sqrt[k]{ \left| \dfrac{k^k}{(3k+1)^k} \right| } = \displaystyle\lim_{k \rightarrow \infty} \dfrac{k}{3k+1} = \dfrac{1}{3} < 1.$$
\part[7] $\displaystyle\sum_{k=1}^{\infty} \left( \dfrac{10}{3} \right)^k$ \\
\textit{Solution}: Diverges -- this is a geometric series with $r=\dfrac{10}{3} > 1$. 
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{k}{e^{k^2}}$ \\
\textit{Solution}: Converges by the integral test: 
$$\displaystyle\int_1^{\infty} \dfrac{x}{e^{x^2}} \mathrm{d}x \stackrel{u=x^2}{=} \dfrac{1}{2} \displaystyle\int_1^{\infty} e^{-u} \mathrm{d}u=\dfrac{1}{2}[-0+e^{-1}]<1.$$
\part[7] $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{\sqrt{k^2-1}}$ \\
\textit{Solution}: Diverges by the limit comparison test. Let $b_k=\dfrac{1}{k}$. Then compute
$$\displaystyle\lim_{k \rightarrow \infty} \dfrac{\frac{1}{k}}{\frac{1}{\sqrt{k^2-1}}} = \displaystyle\lim_{k \rightarrow \infty} \dfrac{\sqrt{k^2-1}}{k} = \displaystyle\lim_{k \rightarrow \infty} \sqrt{ 1-\dfrac{1}{k^2} } = \sqrt{1}=1.$$
The result follows because $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k}$ diverges as it is the harmonic series.
\end{parts}
\end{questions}
\end{document}