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\begin{center}\huge MATH 3315 - EXAM 4 SPRING 2017\\
SOLUTION
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday 28 April 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}


\end{coverpages}

\begin{questions}

\question Do the following sequences have a limit (as $n \rightarrow \infty$)? If yes, what is the limit? If no, why not? For full credit, justify your claim by a calculation or explanation and cite any relevant theorems.
\begin{parts}
\part $a_n=1+\dfrac{1}{n}$ \\
\textit{Solution:} Since $\displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{n}=0$, we have
$$\displaystyle\lim_{n \rightarrow \infty} 1 + \dfrac{1}{n} = 1+0 = 1.$$
\part $a_n = \dfrac{(-1)^n n^2}{n^2+17}$ \\
\textit{Solution:} Since $\displaystyle\lim_{n \rightarrow \infty} \dfrac{n^2}{n^2+17}=1$, we see that $\displaystyle\lim_{n \rightarrow \infty} \dfrac{(-1)^n n^2}{n^2+17}$ does not exist because it oscillates near $1$ and $-1$.
\part $a_n=\dfrac{\cos(n)}{n}$ \\
\textit{Solution:} Since $-1 \leq \cos(x) \leq 1$ for all $x$, we have in particular
$$-1 \leq \cos(n) \leq 1,$$
and so, dividing by $n$, yields
$$-\dfrac{1}{n} \leq \dfrac{\cos(n)}{n} \leq \dfrac{1}{n}.$$
Since 
$$\displaystyle\lim_{n \rightarrow \infty} = 0 = \displaystyle\lim_{n \rightarrow \infty} - \dfrac{1}{n},$$
we conclude by the squeeze theorem that $\displaystyle\lim_{n \rightarrow \infty} \dfrac{\cos(n)}{n}=0$. 
\end{parts}
\question  Find the exact value:
\begin{parts}
\part $\displaystyle\sum_{k=0}^{\infty} \left( \dfrac{3}{4} \right)^k$ \\
\textit{Solution:} Since $-1 < \dfrac{3}{4} < 1$, the geometric series formula $\displaystyle\sum_{k=0}^{\infty} r^k = \dfrac{1}{1-r}$ (valid for $-1 < r < 1$) yields 
$$\displaystyle\sum_{k=0}^{\infty} \left( \dfrac{3}{4} \right)^k = \dfrac{1}{1-\frac{3}{4}} = \dfrac{1}{\frac{1}{4}}=4.$$
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{(k+1)(k+2)}$ \\
\textit{Solution:} We will do partial fractions:
$$\dfrac{1}{(k+1)(k+2)} = \dfrac{A}{k+1} + \dfrac{B}{k+2}.$$
Multiply by the common denominator to get
$$1 = A(k+2)+B(k+1),$$
or equivalently,
$$1 = (A+B)k + (2A+B).$$
Equating coefficients yields the following system of equations:
$$\left\{ \begin{array}{ll}
A+B=0 & (i) \\
2A+B=1 & (ii)
\end{array} \right.$$
From $(i)$, $A=-B$. From $(ii)$, $-2B+B=1$ yields $-B=1$, or $B=-1$. Hence $A=1$. This shows
$$\dfrac{1}{(k+1)(k+2)}=\dfrac{1}{k+1} - \dfrac{1}{k+2}.$$
Therefore the partial sum of the series yields \\
$$\begin{array}{ll}
\displaystyle\sum_{k=1}^n \dfrac{1}{(k+1)(k+2)} &= \displaystyle\sum_{k=1}^n \dfrac{1}{k+1} - \dfrac{1}{k+2} \\
&= \left( \dfrac{1}{2} - \cancel{\dfrac{1}{3}} \right) + \left( \cancel{\dfrac{1}{3}} - \cancel{\dfrac{1}{4}} \right) + \left( \cancel{\dfrac{1}{4}} - \cancel{\dfrac{1}{5}} \right) + \ldots + \left(\cancel{\dfrac{1}{n}} - \cancel{\dfrac{1}{n+1}} \right) + \left( \cancel{\dfrac{1}{n+1}} - \dfrac{1}{n+2} \right) \\
&= \dfrac{1}{2} - \dfrac{1}{n+2}.
\end{array}$$
Therefore
$$\begin{array}{ll}
\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{(k+1)(k+2))} &= \displaystyle\lim_{n \rightarrow \infty} \displaystyle\sum_{k=1}^n \dfrac{1}{k+1} - \dfrac{1}{k+2} \\
&= \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{2} - \dfrac{1}{n+2} \\
&= \dfrac{1}{2}-0 \\
&=\dfrac{1}{2}.
\end{array}$$
\end{parts}

\question For which values of $x$ does the following sum converge?
$$\displaystyle\sum_{k=0}^{\infty} (5x)^k$$
\textit{Solution:} This is a geometric series with $r=5x$. It only converges for
$$-1 < (r=5x) < 1,$$
i.e.
$$-\dfrac{1}{5} < x < \dfrac{1}{5}.$$
\question Converge or diverge? Why? Explain, with appropriate calculations. (For full credit, be sure to state which theorem is used.)
\begin{parts}
\part $\displaystyle\sum_{k=0}^{\infty} \dfrac{1}{\sqrt{k^2+5}}$ \\
\textit{Solution:} It diverges. Let $a_k=\dfrac{1}{\sqrt{k^2+5}}$ and let $b_k=\dfrac{1}{k}$. Compute
$$\begin{array}{ll}
\displaystyle\lim_{k \rightarrow \infty} \dfrac{a_k}{b_k} &= \displaystyle\lim_{k \rightarrow \infty} \dfrac{\frac{1}{\sqrt{k^2+5}}}{\frac{1}{k}} \\
&=\displaystyle\lim_{k \rightarrow \infty} \dfrac{k}{\sqrt{k^2+5}} \\
&= 1.
\end{array}$$
Therefore by the limit comparison test, both $\displaystyle\sum_{k=0}^{\infty} \dfrac{1}{\sqrt{k^2+5}}$ and $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k}$ either both converge or both diverge. Since $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k}$ diverges (it is the harmonic series), we must conclude that $\displaystyle\sum_{k=0}^{\infty} \dfrac{1}{\sqrt{k^2+5}}$ diverges.
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k k^2}{k^3+17}$ \\
\textit{Solution:} Since $a_k =\dfrac{k^2}{k^3+17}$ is decreasing and $\displaystyle\lim_{k \rightarrow \infty} a_k=0$, the alternating series test shows that $\displaystyle\sum_{k=1}^{\infty} \dfrac{(-1)^k k^2}{k^3+17}$ converges.
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{2^k}{k^k}$ \\
\textit{Solution:} Let $a_k = \dfrac{2^k}{k^k}$. Then compute
$$\displaystyle\lim_{k \rightarrow \infty} \sqrt[k]{ \dfrac{2^k}{k^k} } = \displaystyle\lim_{k \rightarrow \infty} \dfrac{2}{k} = 0 < 1,$$
therefore by the root test, $\displaystyle\sum_{k=1}^{\infty} \dfrac{2^k}{k^k}$ converges.
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{k^2+5k+2}{k^2+2k+17}$  \\
\textit{Solution:} Let $a_k = \dfrac{k^2+5k+2}{k^2+2k+17}$. Since $\displaystyle\lim_{k \rightarrow \infty} a_k=1$, we must conclude that the series $\displaystyle\sum_{k=1}^{|infty} a_k = \displaystyle\sum_{k=1}^{\infty} \dfrac{k^2+5k+2}{k^2+2k+17}$ diverges.
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{k}{e^{k^2}}$ \\
\textit{Solution:} Let $a_k=\dfrac{k}{e^{k^2}}$. Let $f(x)=\dfrac{x}{e^{x^2}}$. Then $a_k=f(k)=\dfrac{k}{e^{k^2}}$. Now compute
$$\begin{array}{ll}
\displaystyle\int_1^{\infty} f(x) \mathrm{d}x &= \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_1^b \dfrac{x}{e^{x^2}} \mathrm{d}x \\
&\stackrel{u=x^2}{=}  \dfrac{1}{2} \displaystyle\lim_{b \rightarrow \infty} \displaystyle\int_1^{b^2} e^{-u} \mathrm{d}u \\
&= -\dfrac{1}{2} \displaystyle\lim_{b \rightarrow \infty} e^{-u} \Bigg|_1^{b^2} \\
&=-\dfrac{1}{2} \displaystyle\lim_{b \rightarrow \infty} e^{-b^2}-e^{-1} \\
&=-\dfrac{1}{2} \left( 0 - \dfrac{1}{e} \right) < \infty.
\end{array}$$
Therefore the integral converges. Therefore by the integral test, the sum $\displaystyle\sum_{k=1}^{\infty} \dfrac{k}{e^{k^2}}$ converges.
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{1}{k^{\pi}}$ \\
\textit{Solution:} This series converges because $\pi$ is approximately $3.14 > 1$, and so it converges by the $p$-series test.
\part $\displaystyle\sum_{k=1}^{\infty} \left( \dfrac{5}{4} \right)^k$ \\
\textit{Solution:} This series diverges. It is geometric with ratio $r= \dfrac{5}{4} > 1$. 
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{4^k}{k!}$ \\
\textit{Solution:} Let $a_k=\dfrac{4^k}{k!}$. Compute
$$\displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{a_{k+1}}{a_k} \right| = \displaystyle\lim_{k \rightarrow \infty} \left| \dfrac{4^{k+1}}{(k+1)!} \dfrac{k!}{4^k} \right| = \displaystyle\lim_{k \rightarrow \infty} \dfrac{4}{k+1} = 0 < 1,$$
therefore by the ratio test, the series $\displaystyle\sum_{k=1}^{\infty} \dfrac{4^k}{k!}$ converges.
\part $\displaystyle\sum_{k=1}^{\infty} \dfrac{k^k}{(2k+1)^k}$ \\
\textit{Solution:} Let $a_k = \dfrac{k^k}{(2k+1)^k}$. Compute
$$\displaystyle\lim_{k \rightarrow \infty} \sqrt[k]{\left| \dfrac{k^k}{(2k+1)^k} \right|} = \displaystyle\lim_{k \rightarrow \infty} \dfrac{k}{2k+1} = \dfrac{1}{2}<1,$$
so by the root test, $\displaystyle\sum_{k=1}^{\infty} \dfrac{k^k}{(2k+1)^k}$ converges.
\end{parts}

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