\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \large Math 208} \lfoot{\small \copyright $\;$ Hidegkuti, Powell, 2012} \cfoot{} \chead{\LARGE Series 2 - The Comparison Test} \rfoot{\small Last revised: April 18, 2016} \rhead{ page \thepage } \textwidth 7.6in \textheight 9.6in \setlength{\headheight}{25pt} \setlength{\parindent}{0pt} \begin{document} %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion Theorem: \ If $\dsum a_{n}=A$ \bigskip and $\dsum b_{n}=B$ are convergent series, then \qquad 1) \ Sum Rule \ \ $\dsum \left( a_{n}+b_{n}\right) =\dsum a_{n}+\dsum b_{n}=A+B\medskip $ \qquad 2) \ Difference Rule \ \ \ \ $\dsum \left( a_{n}-b_{n}\right) =\dsum a_{n}-\dsum b_{n}=A-B\medskip $ \qquad 3) \ Constant Multiple Rule: \ $\dsum ka_{n}=k\dsum a_{n}=kA$ \ \ \ \ for any $k\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \medskip \bigskip Consequences: \ \qquad 1) \ Every non-zero constant multiple of a divergent series diverges.$% \medskip $ \qquad 2) \ If $\dsum a_{n}$ converges and $\dsum b_{n}$ diverges, then $% \dsum \left( a_{n}+b_{n}\right) $ and $\dsum \left( a_{n}-b_{n}\right) $ both diverge\bigskip %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion Theorem (The Comparison Test) \ Let $\dsum a_{n}$, $\dsum b_{n}$, and $\dsum c_{n}$ be series with non-negative terms. \ Suppose that for some integer $N$% \begin{equation*} a_{n}\leq b_{n}\leq c_{n}\text{ \ \ for all }n>N \end{equation*} \qquad 1) \ If $\dsum c_{n}$ converges, then $\dsum b_{n}$ also converges.$% \medskip $ \qquad 2) \ If $\dsum a_{n}$ diverges, then $\dsum b_{n}$ also diverges.% %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \medskip \bigskip Proof: \ Suppose that $\dsum a_{n}$, $\dsum b_{n}$, and $\dsum c_{n}$ are series with $a_{n}\leq b_{n}\leq c_{n}$ \ \ for all $n>N$. \ If $c_{n}$ is convergent, then the sequence of all partial sums $\dsum% \limits_{k=1}^{n}c_{k}$ is bounded from above by the sum $C$. \ Then the sequence of partial sums $\dsum\limits_{k=1}^{n}b_{k}$ is also bounded above because (assuming $n>N$)% \begin{equation*} \dsum\limits_{k=1}^{n}b_{k}=\dsum\limits_{k=1}^{N}b_{k}+\dsum% \limits_{k=N+1}^{n}b_{k}\leq \dsum\limits_{k=1}^{N}b_{k}+\dsum\limits_{k=N+1}^{n}c_{k}\leq \dsum\limits_{k=1}^{N}b_{k}+C \end{equation*}% Since $\left\{ b_{n}\right\} $ is non-negative, the sequence $% \dsum\limits_{k=1}^{n}b_{k}$ of partial sums is non-decreasing and so it is also convergent. \ \bigskip Now $D=\dsum\limits_{k=1}^{N}a_{k}$. \ If $\dsum a_{n}$ diverges, then the sequence of partial sums $\dsum\limits_{k=1}^{n}a_{k}$ is not bounded from above. \ \ That means that for any $M\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $, there exists $m\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $ such that (again, assuming $m>N$) \begin{eqnarray*} M+D &\leq &\dsum\limits_{k=1}^{m}a_{k}\leq \dsum\limits_{k=1}^{N}a_{k}+\dsum\limits_{k=N+1}^{n}a_{k}=D+\dsum% \limits_{k=N+1}^{n}a_{k} \\ M+D &\leq &D+\dsum\limits_{k=N+1}^{n}a_{k}~~~~\Longrightarrow ~~~~~M\leq \dsum\limits_{k=N+1}^{n}a_{k} \end{eqnarray*}% Then \begin{equation*} M\leq \dsum\limits_{k=N+1}^{n}a_{k}\leq \dsum\limits_{k=N+1}^{n}b_{k}\leq b_{1}+b_{2}+...+b_{N}+\dsum\limits_{k=N+1}^{n}b_{k}=\dsum% \limits_{k=1}^{n}b_{k} \end{equation*}% and so the partial sums $\dsum\limits_{k=1}^{n}b_{k}$ are not bounded from above. \ Thus the series $\dsum b_{k}$ is divergent.\bigskip \bigskip Example 1.) \ \ $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}+2n}$\bigskip Solution: \ if $n\geq 1$, then \begin{eqnarray*} n^{2}+2n &\geq &n^{2} \\ \dfrac{1}{n^{2}+2n} &\leq &\dfrac{1}{n^{2}} \end{eqnarray*}% Since $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}}$ converges, so does $% \dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}+2n}$. \ You may recall that we have seen this series as a telescoping sum. \ To apply the comparison test takes much less work. \ On the other hand, this method does not give us what the sum is, only the fact that it exists.\bigskip Example 2.) \ $\dsum\limits_{n=0}^{\infty }\dfrac{n}{n^{2}+3}$\bigskip Solution: \ If $n$ is a very large number, the $3$ added in the denominator becomes insignificant. \ Then the entire expression is very close to $\dfrac{% 1}{n}$, which is a divergent series. \ When proving divergence using the comparison test, we must find something divergent that is smaller than our expression. \ Often times this can be easily done using a non-zero constant multiplier. \ In this case, $\dfrac{n}{n^{2}+3}$ is slightly less than $% \dfrac{1}{n}.$ \ Consequently, we will not be able to say that $\dfrac{n}{% n^{2}+3}$ is greater than $\dfrac{1}{n}$ but we can easily prove that it is greater than $\dfrac{1}{4}\cdot \dfrac{1}{n}$. \ Since for all $n\geq 1$% \begin{eqnarray*} 1 &\leq &n^{2} \\ 3 &\leq &3n^{2} \\ n^{2}+3 &\leq &n^{2}+3n^{2}=4n^{4} \\ \dfrac{1}{n^{2}+3} &\geq &\dfrac{1}{4n^{2}}\text{ \ \ \ \ \ multiply both sides by }n\geq 1 \\ \dfrac{n}{n^{2}+3} &\geq &\dfrac{n}{4n^{2}}=\dfrac{1}{4n}~~~~\Longrightarrow ~~~\dfrac{n}{n^{2}+3}\geq \dfrac{1}{4n} \end{eqnarray*}% The series $\dsum\limits_{n=1}^{\infty }\dfrac{1}{4n}$ is a constant times $% \dsum\limits_{n=1}^{\infty }\dfrac{1}{n}$, it is divergent. \ Then, by the comparison test, $\dsum\limits_{n=1}^{\infty }\dfrac{n}{n^{2}+3}$ is also divergent.\bigskip Example 3.) \ $\dsum\limits_{n=0}^{\infty }\dfrac{2^{n}}{3^{n}+1}$\bigskip Solution: \ \begin{eqnarray*} 3^{n}+1 &\geq &3^{n} \\ \dfrac{1}{3^{n}+1} &\leq &\dfrac{1}{3^{n}}\text{ \ \ \ \ \ \ multiply by }% 2^{n}>0 \\ \dfrac{2^{n}}{3^{n}+1} &\leq &\dfrac{2^{n}}{3^{n}} \end{eqnarray*}% Since $\dsum\limits_{n=0}^{\infty }\left( \dfrac{2}{3}\right) ^{n}$ is a geometric series with $\left\vert r\right\vert <1$ and it converges. \ Then $% \dsum\limits_{n=0}^{\infty }\dfrac{2^{n}}{3^{n}+1}$ converges by the comparison test.\pagebreak Example 4.) \ $\dsum\limits_{n=0}^{\infty }\dfrac{2^{n}}{3^{n}-1}$\bigskip Solution: \ For all $n\geq 1$, we have that $3^{n}\geq 2$ \ %TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}% %BeginExpansion \begin{multicols}{2}% %EndExpansion \begin{eqnarray*} 3^{n} &\geq &2\text{ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ \dfrac{3^{n}}{2} &\geq &1\text{ \ \ \ \ \ \ \ \ \ \ \ \ multiply by }-1 \\ -\dfrac{3^{n}}{2} &\leq &-1\text{ \ \ \ \ \ \ \ \ \ \ add }3^{n} \end{eqnarray*}% \bigskip \begin{eqnarray*} 3^{n}-\dfrac{3^{n}}{2} &\leq &3^{n}-1 \\ \dfrac{3^{n}}{2} &\leq &3^{n}-1\text{ \ \ \ \ \ take reciprocal of both sides% } \\ \dfrac{2}{3^{n}} &\geq &\dfrac{1}{3^{n}-1}\text{ \ \ \ multiply by }2^{n} \\ \dfrac{2\cdot 2^{n}}{3^{n}} &\geq &\dfrac{2^{n}}{3^{n}-1} \end{eqnarray*}% %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion Since $\dsum\limits_{n=0}^{\infty }\dfrac{2\cdot 2^{n}}{3^{n}}% =2\dsum\limits_{n=0}^{\infty }\left( \dfrac{2}{3}\right) ^{n}$ \ is convergent, and for all $n\geq 1$, $\dfrac{2^{n}}{3^{n}-1}\leq \dfrac{2\cdot 2^{n}}{3^{n}}\,$, the series $\dsum\limits_{n=0}^{\infty }\dfrac{2^{n}}{% 3^{n}-1}$\ converges by the comparison test\bigskip Example 5.) \ $\dsum\limits_{n=1}^{\infty }\dfrac{\ln n}{n^{3/2}}\medskip $ Solution: \ We will apply the comparison test using the inequality $\ln n\leq n^{1/4}$ that is true for large $n$. \ Let us first prove this statement. \ Consider the function $f\left( x\right) =x^{1/4}-\ln x$. \ We will prove that if $x\geq e^{16}$, then $f\left( x\right) $ is positive for all $x$. Consider first $f\left( e^{16}\right) $.% \begin{equation*} f\left( e^{16}\right) =\left( e^{16}\right) ^{1/4}-\ln \left( e^{16}\right) =e^{4}-16\geq 2.5^{4}-16=23.0625>0 \end{equation*}% Thus $f\left( e^{16}\right) >0$. Now consider $f^{\prime }\left( x\right) =\dfrac{1}{4x^{3/4}}-\dfrac{1}{x}$% \begin{eqnarray*} \dfrac{1}{4x^{3/4}}-\dfrac{1}{x} &>&0 \\ \dfrac{1}{4x^{3/4}} &>&\dfrac{1}{x}\text{ \ \ \ \ \ \ multiply by }4x \\ \sqrt[4]{x} &>&4 \\ x &>&4^{4} \\ x &>&256 \end{eqnarray*}% Note that $e^{16}>256$. \ Consider now the function $f\left( x\right) =\sqrt[% 4]{x}-\ln x$ \ on domain $\left( e^{16},\infty \right) $. \ On this domain, $% f^{\prime }$ is positive and thus $f$ is increasing. \ Since $f\left( e^{16}\right) $ is positive, the function is positive on its entire domain. \ Consequently, if $n>e^{16}$, then \thinspace $f\left( n\right) =n^{1/4}-\ln n>0$ and so $n^{1/4}>\ln n$. \ Thus% \begin{equation*} \dfrac{\ln n}{n^{3/2}}\leq \dfrac{n^{1/4}}{n^{3/2}}=n^{1/4-3/2}=n^{-5/4}=% \dfrac{1}{n^{5/4}} \end{equation*}% Since $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{5/4}}$ converges by the integral test, so does $\dsum\limits_{n=1}^{\infty }\dfrac{\ln n}{n^{3/2}}$ by the comparison test.\pagebreak \begin{center} {\LARGE Practice Problems\bigskip } \end{center} Determine which of the following series converge and which diverge.\bigskip Please note that most of these problems can be solved using several different methods.\bigskip %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}}$ \ \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n\left( n+1\right) }$ $\medskip $\ \item $\dsum\limits_{n=0}^{\infty }\dfrac{2^{n}+1}{3^{n}}\medskip $ \item $\dsum\limits_{n=0}^{\infty }\dfrac{1}{n^{2}+1}\medskip $ \item $\dsum\limits_{n=3}^{\infty }\dfrac{1}{n^{2}-2n}\medskip $ \item $\dsum\limits_{n=0}^{\infty }\dfrac{5^{n}}{4^{n}+1}$ $\medskip $\ \item $\dsum\limits_{n=1}^{\infty }n^{-5/4}\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{\sqrt[n]{n}}$ $\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{8}{n^{2}+1}$ $\medskip $ \item $\dsum\limits_{n=2}^{\infty }\dfrac{\ln n}{n}\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{8\tan ^{-1}n}{1+n^{2}}$ $\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{2}{e^{n}}\medskip $ \item $\dsum\limits_{n=1}^{\infty }\func{sech}n\medskip $ \item $\dsum\limits_{n=2}^{\infty }\dfrac{n+2}{n^{2}-n}\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{3}}\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{\left( n-1\right) !}{\left( n+2\right) !}\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{n}{\left( 1+\dfrac{1}{n}\right) ^{n}}\medskip $ \ \end{enumerate} %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \pagebreak \begin{center} {\LARGE Answers - Practice Problems\bigskip } \end{center} \begin{enumerate} \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}}$ \ converges by the integral test or by grouping of terms \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n\left( n+1\right) }$ \ converges to $1,$ it is a telescoping sum $\ \dfrac{1}{n\left( n+1\right) }=% \dfrac{1}{n}-\dfrac{1}{n+1}$ \ \ \ \ \ \ \ $s_{n}=1-\dfrac{1}{n}$ also by the comparison test $\dfrac{1}{n^{2}+n}\leq \dfrac{1}{n^{2}}$ which is convergent \item $\dsum\limits_{n=0}^{\infty }\dfrac{2^{n}+1}{3^{n}}\medskip $ \ \ \ converges; it is a sum of two geometric series Solution: \ This is a sum of two geometric sequences. \ Not only converges, we can actually figure out the sum.\ $\dsum\limits_{n=0}^{\infty }\dfrac{% 2^{n}+1}{3^{n}}=\dsum\limits_{n=0}^{\infty }\dfrac{2^{n}}{3^{n}}+\dfrac{1}{% 3^{n}}=\dsum\limits_{n=0}^{\infty }\left( \dfrac{2}{3}\right) ^{n}+\dsum\limits_{n=0}^{\infty }\left( \dfrac{1}{3}\right) ^{n}=3+\dfrac{3}{% 2}=\dfrac{9}{2}$ \item $\dsum\limits_{n=0}^{\infty }\dfrac{1}{n^{2}+1}$ converges by the integral test or by the comparison test: $\dfrac{1}{n^{2}+1}\leq \dfrac{1}{% n^{2}}$ and $\dsum\limits_{n=0}^{\infty }\dfrac{1}{n^{2}}$ converges \item $\dsum\limits_{n=3}^{\infty }\dfrac{1}{n^{2}-2n}\medskip $ \ \ converges by the comparison test Solution: \ For all $n\geq 4$, %TCIMACRO{\TeXButton{2col begin}{\begin{multicols}{2}}}% %BeginExpansion \begin{multicols}{2}% %EndExpansion \begin{eqnarray*} n &\geq &4\text{ \ \ \ \ \ \ \ \ \ \ multiply by }n \\ n^{2} &\geq &4n\text{ \ \ \ \ \ \ \ \ divide by }2 \\ \dfrac{n^{2}}{2} &\geq &2n\text{ \ \ \ \ \ \ \ \ subtract }2n \end{eqnarray*}% \begin{eqnarray*} \dfrac{n^{2}}{2}-2n &\geq &0\text{ \ \ \ \ \ \ \ \ \ \ add }\dfrac{n^{2}}{2} \\ n^{2}-2n &\geq &\dfrac{n^{2}}{2}\text{ \ \ \ \ \ \ take reciprocal of both sides} \\ \dfrac{1}{n^{2}-2n} &\leq &\dfrac{2}{n^{2}} \end{eqnarray*}% \ %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion Since $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{2}}$ converges, so does $% \dsum\limits_{n=1}^{\infty }\dfrac{2}{n^{2}}$. \ Since for all $n\geq 4$, $% \dfrac{1}{n^{2}-2n}\leq \dfrac{2}{n^{2}}$, the series $\dsum\limits_{n=1}^{% \infty }\dfrac{1}{n^{2}-2n}$ converges by the comparison test. \item $\dsum\limits_{n=0}^{\infty }\dfrac{5^{n}}{4^{n}+1}$ \ \ \ diverges by the comparison test and the constant multiple rule: $\dfrac{5^{n}}{4^{n}+1}\geq \dfrac{5^{n}}{4^{n}+4^{n}}=\dfrac{5^{n}}{2\cdot 4^{n}}=\dfrac{1}{2}\cdot \left( \dfrac{5}{4}\right) ^{x}$ \item $\dsum\limits_{n=1}^{\infty }n^{-5/4}\medskip $ \ \ \ \ \ \ converges by the integral test \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{\sqrt[n]{n}}$ $\medskip $ \ \ \ \ diverges because it fails the $n$th term test \item $\dsum\limits_{n=1}^{\infty }\dfrac{8}{n^{2}+1}$ $\medskip $ \ \ converges by the comparison or integral test \item $\dsum\limits_{n=2}^{\infty }\dfrac{\ln n}{n}\medskip $ \ \ \ \ diverges by the comparison test If $n\geq 3$, then $\ln n\geq 1$ and so $\dfrac{\ln n}{n}\geq \dfrac{1}{n}$. \ $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n}$ diverges so $% \dsum\limits_{n=2}^{\infty }\dfrac{\ln n}{n}$ diverges as well. \item $\dsum\limits_{n=1}^{\infty }\dfrac{8\tan ^{-1}n}{1+n^{2}}$ \ \ \ \ \ \ converges by the comparison test $\medskip $ For all $n\in %TCIMACRO{\U{2115} }% %BeginExpansion \mathbb{N} %EndExpansion $, \ $\tan ^{-1}n<\dfrac{\pi }{2}$ and so $\dfrac{8\tan ^{-1}n}{1+n^{2}}\leq \dfrac{8\left( \dfrac{\pi }{2}\right) }{1+n^{2}}=4\pi \cdot \dfrac{1}{1+n^{2}% }$ and $\dsum\limits_{n=1}^{\infty }4\pi \cdot \dfrac{1}{1+n^{2}}=4\pi \dsum\limits_{n=1}^{\infty }\dfrac{1}{1+n^{2}}$ is convergent by the integral or comparison test. \item $\dsum\limits_{n=1}^{\infty }\dfrac{2}{e^{n}}\medskip $ \ \ \ \ \ converges because it is a geometric series with $\left\vert r\right\vert <1$ \item $\dsum\limits_{n=1}^{\infty }\func{sech}n\medskip $ \ \ \ \ \ \ \ converges by the comparison test: $\ \func{sech}n=\dfrac{2}{e^{n}+e^{-n}}% \leq \dfrac{2}{e^{n}}$ \item $\dsum\limits_{n=2}^{\infty }\dfrac{n+2}{n^{2}-n}\medskip $ \ \ \ \ \ \ \ diverges by the comparison test diverges by the comparison test: $\dfrac{n+2}{n^{2}-n}\geq \dfrac{1}{n}$ and $\dfrac{1}{n}$ is divergent method 1: \ \ $\dfrac{n+2}{n^{2}-n}\geq \dfrac{n+2}{n^{2}}\geq \dfrac{n}{% n^{2}}=\dfrac{1}{n}$ method 2: \ We want to prove that $\dfrac{n+2}{n^{2}-n}\geq \dfrac{1}{n}$ \ Same as $\dfrac{n+2}{n^{2}-n}-\dfrac{1}{n}\geq 0$ and $\dfrac{n+2}{n^{2}-n}-\dfrac{1}{n}=\dfrac{3}{n\left( n-1\right) }$ is clearly positive for all $n\geq 2$. \item $\dsum\limits_{n=1}^{\infty }\dfrac{1}{n^{3}}\medskip $ \ \ \ \ \ converges by the integral test \item $\dsum\limits_{n=1}^{\infty }\dfrac{\left( n-1\right) !}{\left( n+2\right) !}$ \ \ \ converges by the comparison test\ $\dfrac{\left( n-1\right) !}{\left( n+2\right) !}=\dfrac{\left( n-1\right) !% }{\left( n-1\right) !n\left( n+1\right) \left( n+2\right) }=\dfrac{1}{% n\left( n+1\right) \left( n+2\right) }\leq \dfrac{1}{\left( n\right) \left( n\right) \left( n\right) }=\dfrac{1}{n^{3}}$ \ and $\dsum\limits_{n=1}^{% \infty }\dfrac{1}{n^{3}}$ is convergent$\medskip $ \item $\dsum\limits_{n=1}^{\infty }\dfrac{n}{\left( 1+\dfrac{1}{n}\right) ^{n}}\medskip $ \ \ diverges by the comparison test: \ $\dfrac{n}{\left( 1+% \dfrac{1}{n}\right) ^{n}}>\dfrac{n}{e}$ \ or by the $n$th term test \end{enumerate} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}