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\documentclass{ximera}
\input{../preamble}
\title{Exercises: Series}
%%%%%\author{<NAME>}

\begin{document}
\begin{abstract}
Exercises relating to fundamental properties of series.
\end{abstract}
\maketitle


\begin{exercise}
Compute the sum of the series:
\begin{hint} Split the series into two separate geometric series and evaluate each separately. \end{hint}
\[ \sum_{n=0}^\infty \frac{3^n - 2^n}{4^n}  = \answer{2}. \]
\end{exercise}


\begin{exercise}
Compute the sum of the series:
\begin{hint} One way to proceed is to reindex the series so that it starts at $n = 0$. \end{hint}
\[ \sum_{n=3}^\infty \frac{(-1)^{n+1} 2^n}{3^n} = \answer{ \frac{8}{45} } \]
\end{exercise}

\begin{exercise}
Compute the sum of the series:
\[ \sum_{n=1}^\infty \frac{3^{2n+1}}{2^{3n} \cdot 9^n} = \answer{\frac{3}{7}} \]
\begin{hint} Simplify as much as possible and write in standard form for a geometric series.\end{hint}
\end{exercise}

\begin{exercise}
\begin{itemize}
\item Give an example of a geometric series whose first term is $3$ and which sums to $6$. If no such series exists, enter N/A in both spaces below:
\[ 6 = \sum_{n=0}^\infty \answer{3} \left( \answer{\frac{1}{2}} \right)^n \]
\item Give an example of a geometric series whose first term is $3$ and which sums to $2$. If no such series exists, enter N/A in both spaces below:
\[ 2 = \sum_{n=0}^\infty \answer{3} \left( \answer{- \frac{1}{2}} \right)^n \]
\item Give an example of a geometric series whose first term is $3$ and which sums to $1$. If no such series exists, enter N/A in both spaces below:
\[ 1 = \sum_{n=0}^\infty \answer{N/A} \left( \answer{N/A} \right)^n \]
\end{itemize}
\end{exercise}

\begin{exercise}
Find the correct value of the constant in the blank below which makes the sum of the series equal to zero:
\[ 0 = \sum_{n=1}^\infty \frac{(-1)^{n-1}\answer{8} - 2^n}{3^n} \]
\begin{hint} One reasonable strategy is to separate the series into two geometric series and reindex to apply the standard formula. \end{hint}
\end{exercise}

\begin{exercise}
Compute the sum of the infinite series below:
\[ \sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)} = \answer{\frac{1}{4}}. \]
\begin{hint}
There is a closely-related telescoping series:
\[ \frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)} = \frac{2}{n(n+1)(n+2)} \]
\end{hint}
\end{exercise}

\begin{exercise}
Compute the sum of the infinite series below.
\[ \sum_{n=1}^\infty \frac{2}{n(n+2)} = \answer{\frac{3}{2}} \]
\begin{hint}
Expand the terms using partial fractions and compute several partial sums by hand. What you get is something like a telescoping series, but cancellations occur in a slightly different way than usual.
\begin{hint}
\[ \frac{2}{n(n+2)} = \frac{1}{n} - \frac{1}{n+2}. \]
\begin{hint}
The formula for a general partial sum is
\[ \frac{2}{1 \cdot 3} + \cdots + \frac{2}{N \cdot (N+2)} = 1 + \frac{1}{2} - \frac{1}{N+1} - \frac{1}{N+2} \]
\end{hint}
\end{hint}
\end{hint}
\end{exercise}

\begin{exercise}
\begin{itemize}
\item Use your answer to the previous exercise to compute the sum of the series:
\[ \sum_{n=3}^\infty \frac{1}{n(n-2)} = \answer{\frac{3}{4}} \]
\item Use the answer you just found to compute the sum of the series:
\[ \sum_{n=4}^\infty \frac{1}{n(n-2)} = \answer{\frac{5}{12}} \]
\begin{hint}
To drop the $n=3$ term, you can just calculate it and then subtract it from the previous series.
\end{hint}
\end{itemize}
\end{exercise}



\section*{Sample Quiz Questions}

\begin{question}%%%%%[2019Telescope1]

Compute the exact value of the infinite series
\[ \sum_{n=1}^\infty \ln \left( \frac{1 + n^{-1}}{1 + (n+1)^{-1}} \right). \]
\begin{multiplechoice}
\choice[correct]{\(\ln 2\)}
\choice{\(\ln 3\)}
\choice{\(\ln 4\)}
\choice{\(\ln 5\)}
\choice{\(\ln 6\)}
\choice{\(\ln 7\)}
\end{multiplechoice}
\begin{feedback}
The series is not a geometric series or Taylor series, we compute the first few partial sums:
\[ 
\begin{aligned}
S_1 & = \ln \left( \frac{1 + 1}{1 + 2^{-1}} \right)  = \ln \left( \frac{2}{\frac{3}{2}} \right) \\
S_2 & = \ln \left( \frac{1 + 1}{1 + 2^{-1}} \right)  + \ln \left( \frac{1 + 2^{-1}}{1 + 3^{-1}} \right)  = \ln \left( \frac{1 + 1}{1 + 3^{-1}} \right) = \ln \left( \frac{2}{\frac{4}{3}} \right) \\
S_3 & = \ln \left( \frac{2}{1 + 3^{-1}} \right)  + \ln \left( \frac{1 + 3^{-1}}{1 + 4^{-1}} \right)  = \ln \left( \frac{2}{1 + 4^{-1}} \right) = \ln \left( \frac{2}{\frac{5}{4}} \right) \\
& \ \vdots  \\
S_n & = \ln \left( \frac{2}{1 + (n+1)^{-1}} \right).
 \end{aligned}
\]
In particular, writing the sum of logarithms as a logarithm of a product leads to substantial cancellation.  By letting \(n \rightarrow \infty\), we get \(S_n \rightarrow \ln 2\).
\end{feedback}

\end{question}




\end{document}
