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\begin{center}{\huge MATH 2502 - EXAM 3 SPRING 2019 \\
\vspace*{0.2in}
SOLUTION
}
\end{center}

\begin{flushleft}%\vspace{0.5in}
Wednesday, 17 April \\
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}

\begin{center}\underline{Formulas}\end{center}
(\textbf{TFD}) Test for divergence: if $\displaystyle\lim_{n \rightarrow \infty} a_n \neq 0$, then $\displaystyle\sum a_n$ diverges \\
(\textbf{IT}) Integral test: if $a_n \geq 0$ and $a_n=f(n)$ with $f$ continuous and decreasing, then $\displaystyle\int^{\infty} f(x) \mathrm{d}x$ and $\displaystyle\sum^{\infty} a_n$ either both converge or both diverge  \\
(\textbf{CT}) Comparison test: 
\begin{enumerate}[(i)] 
\item if $0 \leq a_n \leq b_n$ and $\displaystyle\sum_{n=1}^{\infty} b_n$ converges, then $\displaystyle\sum a_n$ converges 
\item if $a_n \geq b_n \geq 0$ and $\displaystyle\sum b_n$ diverges, then $\displaystyle\sum a_n$ diverges 
\end{enumerate}
\noindent (\textbf{LCT}) Limit comparison test: suppose $a_n, b_n \geq 0$
\begin{enumerate}[(i)]
\item if $\displaystyle\lim_{n \rightarrow \infty} \dfrac{a_n}{b_n}=L\neq 0$, then $\displaystyle\sum a_n$ and $\displaystyle\sum b_n$ either both converge or both diverge 
\item if $\displaystyle\lim_{n \rightarrow \infty} \dfrac{a_n}{b_n} = 0$ and $\displaystyle\sum b_n$ converges, then $\displaystyle\sum a_n$ converges 
\item if $\displaystyle\lim_{n \rightarrow \infty} \dfrac{a_n}{b_n}=\infty$ and $\displaystyle\sum b_n$ diverges, then $\displaystyle\sum a_n$ diverges  
\end{enumerate}
(\textbf{AST}) Alternating series test: an alternating series of form $\displaystyle\sum (-1)^n b_n$ converges if $0 \leq b_{n+1} \leq b_n$ and $\displaystyle\lim_{n \rightarrow \infty} b_n=0$ \\
(\textbf{RaT}) Ratio test: let $L=\displaystyle\lim_{n \rightarrow \infty} \left| \dfrac{a_{n+1}}{a_n} \right|$
\begin{enumerate}[(i)]
\item If $0 \leq L <1$, then $\displaystyle\sum a_n$ converges absolutely
\item If $L > 1$, then $\displaystyle\sum a_n$ diverges
\item If $L=1$, then the test is inconclusive.
\end{enumerate}
(\textbf{RoT}) Root test: let $L=\sqrt[n]{|a_n|}$.
\begin{enumerate}[(i)]
\item If $0 \leq L < 1$, then $\displaystyle\sum a_n$ converges absolutely
\item If $L>1$ or $L=\infty$, then $\displaystyle\sum a_n$ diverges.
\item If $L=1$, then the test is inconclusive.
\end{enumerate}
(\textbf{Remainder estimate}) If $a_n \geq 0$ and $a_n=f(n)$, then $R_N \leq \displaystyle\int_N^{\infty} f(x) \mathrm{d}x$
\end{coverpages}

\begin{questions}
\question[18] Find the exact value of the series. \\
\begin{parts}
\part[9] $\displaystyle\sum_{n=0}^{\infty} \left( -\dfrac{2}{5} \right)^n$ \\
\textit{Solution}: This is a geometric series with $r=-\dfrac{2}{5}$. Therefore it converges and we have
\[ \displaystyle\sum_{n=0}^{\infty} \left( - \dfrac{2}{5} \right)^n = \dfrac{1}{1-(-\frac{2}{5})} = \dfrac{1}{\frac{7}{5}} = \dfrac{5}{7}. \]
\part[9] $\displaystyle\sum_{n=2}^{\infty} \left( \cos\left( \dfrac{1}{n} \right) - \cos \left( \dfrac{1}{n+1} \right) \right)$ \\
\textit{Solution}: This is a telescoping series. Consider a partial sum: \
\[ \hspace{-75pt} \begin{array}{ll}
S_k &= \left( \cos\left( \dfrac{1}{2} \right) - \cancel{\cos \left( \dfrac{1}{3} \right)} \right) + \left( \cancel{\cos\left( \dfrac{1}{3} \right)} - \bcancel{\cos \left( \dfrac{1}{4} \right)} \right) + \ldots + \left( \cancel{\cos\left( \dfrac{1}{k-1} \right)} - \bcancel{\cos \left( \dfrac{1}{k} \right)} \right) + \left( \bcancel{\cos\left( \dfrac{1}{k} \right)} - \cos \left( \dfrac{1}{k+1} \right) \right) \\
&= \cos \left( \dfrac{1}{2} \right) - \cos\left( \dfrac{1}{k+1} \right) 
\end{array}\]
Therefore we may compute
\[ \begin{array}{ll}
\displaystyle\sum_{n=2}^{\infty} \left( \cos\left( \dfrac{1}{n} \right) - \cos \left( \dfrac{1}{n+1} \right) \right) &= \displaystyle\lim_{k \rightarrow \infty} S_k \\
&= \displaystyle\lim_{k \rightarrow \infty} \cos \left( \dfrac{1}{2} \right) - \cos \left(\dfrac{1}{k+1} \right) \\
&= \cos \left( \dfrac{1}{2} \right) - \cos(0) \\
&= \cos \left( \dfrac{1}{2} \right) - 1.
\end{array}\]
\end{parts}

\question[64] Converge or diverge? Why? Explain using one of the tests on the front page of the exam. Each explanation must be complete for full credit.
\begin{parts}
\part[9] $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n \ln(n)}{n^2}$ \\
\textit{Solution}: Let $b_n = \dfrac{\ln(n)}{n^2}$. Compute 
\[ \begin{array}{ll}
\displaystyle\lim_{n \rightarrow \infty} b_n &= \displaystyle\lim_{n \rightarrow \infty} \dfrac{\ln(n)}{n^2} \Big( = \dfrac{\infty}{\infty} \longrightarrow \text{use L'H\^{o}pital's rule} \Big)\\
&\stackrel{L'H}= \displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{1}{n}}{2n} \\
&= \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{2n^2} \\
&=0.
\end{array} \]
Therefore by the alternating series test, we may conclude that $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n \ln(n)}{n^2}$ \textbf{converges}.
\part[9] $\displaystyle\sum_{n=1}^{\infty} \dfrac{(n^3+2n+1)^n}{(n^5+7n^2)^n}$ \\
\textit{Solution}: Let $a_n = \dfrac{(n^3+2n+1)^n}{(n^5+7n^2)^n}$. Compute
 \[ \begin{array}{ll}
 \displaystyle\lim_{n \rightarrow \infty} \left| \sqrt[n]{a_n} \right| &= \displaystyle\lim_{n \rightarrow \infty} \sqrt[n]{\dfrac{(n^3+2n+1)^n}{(n^5+7n^2)^n}} \\
 &= \displaystyle\lim_{n \rightarrow \infty} \dfrac{n^3+2n+1}{n^5+7n^2} \\
 &= 0,
 \end{array}\]
 the conclusion that the limit is zero follows either from applying L'H\^{o}pital's rule multiple times or by noticing that the polynomial in the denominator is of a higher degree than the polynomial in the numerator. Therefore by the root test, we may conclude that $\displaystyle\sum_{n=1}^{\infty} \dfrac{(n^3+2n+1)^n}{(n^5+7n^2)^n}$ \textbf{converges}. 
\part[9] $\displaystyle\sum_{n=1}^{\infty} n! 2^n$ \\
\textit{Solution}: Let $a_n=n!2^n$ and compute
\[ \begin{array}{ll}
\displaystyle\lim_{n \rightarrow \infty} \left| \dfrac{a_{n+1}}{a_n} \right| &= \displaystyle\lim_{n \rightarrow\infty} \dfrac{(n+1)!2^{n+1}}{n!2^n} \\
&= \displaystyle\lim_{n \rightarrow \infty} 2(n+1) \\
&= \infty.
\end{array}\]
Therefore by the ratio test, we conclude that $\displaystyle\sum_{n=1}^{\infty} n! 2^n$ \textbf{diverges}. 
\part[9] $\displaystyle\sum_{n=1}^{\infty} \dfrac{n^{32}+2n^{17}+9n-1}{3n^{32}-900n^3+2n^2-\pi}$  \\
\textit{Solution}: Compute
\[ \displaystyle\lim_{n \rightarrow \infty} \dfrac{n^{32}+2n^{17}+9n-1}{3n^{32}-900n^3+2n^2-\pi} = \dfrac{1}{3}.\]
Therefore by the test for divergence, $\displaystyle\sum_{n=1}^{\infty} \dfrac{n^{32}+2n^{17}+9n-1}{3n^{32}-900n^3+2n^2-\pi}$ \textbf{diverges}. 
\part[9] $\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{3^n + n}$  \\
\textit{Solution}: Let $a_n=\dfrac{1}{3^n+n}$ and $b_n=\dfrac{1}{3^n}$. Compute
\[ \begin{array}{ll}
\displaystyle\lim_{n \rightarrow \infty} \dfrac{a_n}{b_n} &= \displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{1}{3^n}}{\frac{1}{3^n+n}} \\
&=\displaystyle\lim_{n \rightarrow \infty} \dfrac{3^n+n}{3^n} \\
&= \displaystyle\lim_{n \rightarrow \infty} 1+\dfrac{n}{3^n} \\
&= 1
\end{array} \]
Since $\displaystyle\sum_{n=0}^{\infty} b_n = \displaystyle\sum_{n=0}^{\infty} \dfrac{1}{3^n}$ is a geometric series with $r=\dfrac{1}{3}$, it is a convergence series. Therefore by the limit comparison test, we conclude that $\displaystyle\sum_{n=0}^{\infty} a_n = \displaystyle\sum_{n=0}^{\infty} \dfrac{1}{3^n+n}$ also \textbf{converges}. 
\part[9] $\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n\ln(n)}$ \\
\textit{Solution}: Compute
\[ \begin{array}{ll}
\displaystyle\int_2^{\infty} \dfrac{1}{x\ln(x)} \mathrm{d}x & =\displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_2^{b} \dfrac{1}{x\ln(x)} \mathrm{d}x \\
&\stackrel{u=\ln(x),\mathrm{d}u=\frac{1}{x}\mathrm{d}x}{=} \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_{\ln(2)}^{\ln(b)} \dfrac{1}{u} \mathrm{d}u \\
&= \displaystyle\lim_{b \rightarrow \infty} \ln(u) \Bigg|_{\ln(2)}^{\ln(b)} \\
&= \displaystyle\lim_{n \rightarrow \infty} \ln(\ln(b))-\ln(\ln(2)) \\
&= \infty.
\end{array} \]
Therefore sine the integral $\displaystyle\int_2^{\infty} \dfrac{1}{x\ln(x)} \mathrm{d}x$ diverges, the integral test allows us to conclude that the sum $\displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n\ln(n)}$ also \textbf{diverges}. 
\part[10] $\displaystyle\sum_{n=1}^{\infty} \dfrac{2(n!)^2}{\large((2n)!\large)^2}$ \\
\textit{Solution}: Let $a_n=\dfrac{2(n!)^2}{((2n)!)^2}$. Compute
\[ \begin{array}{ll}
\displaystyle\lim_{n \rightarrow \infty} \left| \dfrac{a_{n+1}}{a_n} \right| &= \displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{2((n+1)!)^2}{((2(n+1))!)^2}}{\frac{2(n!)^2}{((2n)!)^2}} \\
&= \displaystyle\lim_{n \rightarrow \infty} \dfrac{\cancel{2}((n+1)!)^2((2n)!)^2}{\cancel{2}((2n+2)!)^2 (n!)^2 } \\
&= \displaystyle\lim_{n \rightarrow \infty} \dfrac{(n+1)^2 \cancel{(n!)^2}\bcancel{((2n)!)^2}}{(2n+2)(2n+1)\bcancel{((2n)!)^2} \cancel{(n!)^2} } \\ 
&= \displaystyle\lim_{n \rightarrow \infty} \dfrac{n^2+2n+1}{4n^2+6n+2} \\
&= \dfrac{1}{4}.
\end{array}\]
Since $\dfrac{1}{4}<1$, the ratio test allows us to conclude that $\displaystyle\sum_{n=1}^{\infty} \dfrac{2(n!)^2}{\large((2n)!\large)^2}$ \textbf{converges}. 
\end{parts}

\question[9] Conditionally convergent, absolutely convergent, or neither? Explain. \\
\[\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n}{\sqrt{n}}\]
\textit{Solution}: First we show that this series converges using the alternating series test: let $b_n=\dfrac{1}{\sqrt{n}}$, then
\[ \displaystyle\lim_{n \rightarrow \infty} b_n = \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{\sqrt{n}} = 0,\]
and so by the alternating series test, $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n}{\sqrt{n}}$ converges. So, at worst, it wil be conditionally convergent. \\
\newpage
\underline{\textbf{Is $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n}{\sqrt{n}}$ absolutely convergent?}} \\
To decide this, we have to investigate whether or not $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}}$ converges or not. We will use the integral test to show that it diverges: compute 
\[ \begin{array}{ll}
\displaystyle\int_1^{\infty} \dfrac{1}{\sqrt{x}} \mathrm{d}x &= \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_1^b x^{-\frac{1}{2}}  \mathrm{d}x \\
&= \displaystyle\lim_{b \rightarrow \infty} \dfrac{x^{\frac{1}{2}}}{\frac{1}{2}} \Bigg|_1^b \\
&= \displaystyle\lim_{b \rightarrow \infty} 2\sqrt{b} - 2\sqrt{1} \\
&= \infty.
\end{array}\]
Therefore by the integral test, we say that $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}}$ diverges. Therefore we say that the alternating series $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n}{\sqrt{n}}$ is \textbf{NOT} absolutely convergent. Hence it is conditionally convergent.
\question[9] How many terms of the series $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^2}$ would guarantee the value of the sum with a remainder error of $< 0.00001$? \\
\textit{Solution}: Using the remainder estimate, we see that 
\[ R_N \leq \displaystyle\int_N^{\infty} \dfrac{1}{x^2} \mathrm{d}x = \ldots = \dfrac{1}{N}. \]
Therefore to guarantee a remainder error $<0.00001$, it suffices to choose an $N$ so large that 
\[ R_N \leq \dfrac{1}{N} < 0.00001.\]
Therefore taking reciprocals shows that taking 
\[ N > \dfrac{1}{0.00001} =100000\]
terms of the sum will guarantee the desired accuracy.
\end{questions}

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