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.XM_IGNORE/remainders1172/digInAlternatingSeriesTestRemainders.tex

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+\documentclass{ximera}
+
+\input{../preamble.tex}
+
+\outcome{Determine if a series converges using the alternating series test.}
+
+\title[Dig-In:]{Remainders for alternating series}
+\author{Jim Talamo}
+
+\begin{document}
+\begin{abstract}
+There is a nice result for approximating the remainder of convergent alternating series. 
+\end{abstract}
+\maketitle
+
+\section{Introduction}
+Recall that if $\{a_n\}_{n=n_0}$ is a sequence of positive terms, we say the series $\sum_{k=n_0}^{\infty} (-1)^k a_k$ and $\sum_{k=n_0}^{\infty} (-1)^{k+1} a_k$ are \emph{alternating series}, and we have a nice result to test these for convergence, which is reprinted below.
+
+\begin{theorem}[Alternating Series Test]
+Let $\{a_n\}_{n=n_0}$ be a sequence whose terms are eventually positive and decreasing and
+$\lim_{n\to\infty}a_n=0$. Then, the series 
+\[
+\sum_{k=n_0}^\infty (-1)^{k}a_k \qquad \text{and}\qquad \sum_{k=n_0}^\infty (-1)^{k+1}a_k 
+\]
+both converge.
+\end{theorem}
+
+
+Note that this test gives us a way to determine if many alternating series \emph{must} converge, but it does not give us information about the value to which the series converges.  If we want to obtain insight as to what the approximate value of the convergent series is, we must study remainders.
+
+  
+
+\begin{image}
+\begin{tikzpicture}
+\draw[densely dashed] (-1,0)--(10,0);
+\node at (-1.2, 0) {$S$};
+\draw[fill=black] (0,5) circle (3pt) node[below,left] {$s_1$};
+\draw[fill=black] (1, -3.8) circle(3pt) node[below,left] {$s_2$};
+\draw[fill=black] (2, 2.8) circle(3pt) node[below,left] {$s_3$};
+\draw[fill=black] (4.6, 2) circle (1pt);
+\draw[fill=black] (5, 2) circle (1pt);
+\draw[fill=black] (5.4, 2) circle (1pt);
+\draw[fill=black] (7, -2.1) circle (3pt) node[below,left] {$s_n$};
+\draw[fill=black] (8, 1.6) circle (3pt) node[below, left] {$s_{n+1}$};
+\draw[fill=black] (9, -1.2) circle (3pt) node[below,left] {$s_{n+2}$};
+\draw[<->] (7, -0.1) -- (7, -1.9) node[pos=0.5, left] {$r_n$};
+\draw[<->] (7.5, -1.9) -- (7.5, 1.45) node[pos=0.4, right] {$a_{n+1}$};
+\end{tikzpicture}
+\end{image}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\section{Remainders for Alternating Series}
+As usual we must establish that a series converges first before we begin to think about remainders.  Once we have established that an alternating series $\sum_{k=1}^{\infty} (-1)^k a_k$ converges, we recall the important result
+
+\begin{image}
+  \begin{tikzpicture}
+        \node at (0,0) {
+          $ \sum_{k=1}^{\infty} (-1)^k a_k=\underbrace{\sum_{k=1}^{n}(-1)^k a_k}+ \underbrace{\sum_{k=n+1}^\infty (-1)^k a_k} = s_n+r_n$.};
+        \node at (1.5,-1) { This is $r_n$.};
+       
+        \node at (-1,-1) { This is $s_n$. };
+        
+      \end{tikzpicture}
+  \end{image}
+
+While we cannot find an explicit formula for $s_n$ here, we have a good way to establish bounds on the error made when approximating $ \sum_{k=1}^{\infty} (-1)^k a_k$ by the finite sum $s_n= \sum_{k=1}^{n} (-1)^k a_k$.   We'll state the test, then explore how it works in the context of an example.
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\begin{theorem}[Alternating Series Remainder Estimates]\index{alternating series remainder estimates}
+Let $\{a_n\}_{n=n_0}$ be a sequence whose terms are positive and decreasing  and
+$\lim_{n\to\infty}a_n=0$. Then,  
+\[
+\big| r_n \big| \leq a_{n+1} \qquad \textrm{ for all } n \geq n_0,
+\]
+where $r_n = \sum_{k=n+1}^{\infty} a_k$.
+\end{theorem}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+Note that the above guarantees that for any integer $N>n_0$, the term $s_N$ will approximate the series $\sum_{k=1}^{\infty} a_k$ with an error no greater than $a_{n+1}$.  Before we work an example, a few comments are in order.
+
+\begin{itemize}
+\item[1.] In most cases, we will have $n_0 = 0$ or $n_0 =1$.  As a technical point, we have to make the specification of when the sequence $\{a_n\}$ is actually both decreasing and contains only positive terms before we can use the result.  
+\item[2.] If the sequence $\{a_n\}_{n = n_0}$ meets the criteria, note that we can split up the series $\sum_{k=1}^{\infty} a_k$ as 
+
+\[ 
+\sum_{k=1}^{\infty} a_k = \sum_{k=1}^{n_0} a_k +\sum_{k=n_0+1}^{\infty} a_k 
+\]
+
+We can compute the first sum by hand since it requires adding finitely many terms, then use the remainder result on the infinite series.
+\end{itemize}
+
+Let's see how the test works in the context of an example.
+
+\begin{model}
+Consider the alternating harmonic series $\sum_{k=1}^{\infty} \frac{(-1)^k}{k}$.  Here, $a_n = \frac{1}{n}$, and the reader can verify that the series meets the hypothesis for the alternating series test.  To develop some intuition about the error made when we approximate the infinite series by adding only finitely many of its terms, let's write out some of its terms.
+
+\[
+\sum_{k=1}^{\infty} \frac{(-1)^k}{k} = -1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7} + \ldots
+\]
+
+Now, let's say we want to approximate the infinite series with $s_2$.
+
+\begin{image}
+  \begin{tikzpicture}
+        \node at (0,0) {
+          $\sum_{k=1}^{\infty} \frac{(-1)^k}{k}=\underbrace{-1+\frac{1}{2}}-\frac{1}{3}+ \underbrace{\left[\frac{1}{4}-\frac{1}{5}\right]}+\underbrace{\left[\frac{1}{6}-\frac{1}{7}\right]} + \ldots$};
+        \node at (.9,-1) { positive};
+         \node at (2.6,-1) { positive};      
+        \node at (-1.4,-.8) { $s_2$};
+        
+      \end{tikzpicture}
+  \end{image}
+  
+  Note that each pair of terms we've enclose in brackets is \wordChoice{\choice[correct]{positive}\choice{negative}}, which means that
+  
+  \[
+  \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \geq -1+\frac{1}{2}-\frac{1}{3} = s_2 -a_3
+  \]
+  
+  Similarly, had we used $s_3$ for the approximation, we could obtain an overestimate for the value of the series.
+  
+  \begin{image}
+  \begin{tikzpicture}
+        \node at (0,0) {
+          $\sum_{k=1}^{\infty} \frac{(-1)^k}{k}=\underbrace{-1+\frac{1}{2}-\frac{1}{3}}+ \frac{1}{4}+\underbrace{\left[-\frac{1}{5}+\frac{1}{6}\right]}+\underbrace{\left[-\frac{1}{7} +\frac{1}{8}\right]} + \ldots$};
+        \node at (1.1,-1) { negative};
+         \node at (3.1,-1) { negative};      
+        \node at (-1.7,-.8) { $s_3$};
+        
+      \end{tikzpicture}
+  \end{image}
+  
+    Note that each pair of terms we've enclose in brackets is \wordChoice{\choice{positive}\choice[correct]{negative}}, which means that
+  
+  \[
+  \sum_{k=1}^{\infty} \frac{(-1)^k}{k} \leq -1+\frac{1}{2}-\frac{1}{3} +\frac{1}{4} = s_3 -a_4
+  \]
+  
+Note that since we can split the convergent series up as
+
+\[
+\sum_{k=1}^{\infty} \frac{(-1)^k}{k} = s_n +r_n,
+\]
+We can see in both cases that the magnitude of the error is bounded above by the magnitude of the next term in the series; that is 
+
+\[\big| r_n \big| \leq a_{n+1}.\]
+
+We can repeat this construction by using $s_n$ for an unspecified value for $n$ and arrive at the conclusion above.  
+\end{model}
+
+We hope that the curious reader understands that a similar argument can be made for series that meet the assumptions of the alternating series remainder test.  In order to gain some practice, let's work some examples.
+
+\begin{example}
+Consider the alternating geometric series $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$.
+
+I. Compute $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$.
+
+\begin{explanation}
+First, note that by using laws of exponents, $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k = \sum_{k=0}^{\infty}  \left(\frac{-1}{2}\right)^k$.   
+Hence, the geometric series $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$ converges because $|r| = \frac{1}{2} <1$.
+
+
+When $r<1$, note that $\sum_{k=0}^{\infty} ar^k = \frac{a}{1-r}$.  Here, $r= -\frac{1}{2}$, so we find that $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k= \frac{2}{3}.$
+\end{explanation}
+
+II. Use the theorem above to obtain an estimate for the remainder if $s_4$ is used to approximate the series.  What do you notice?
+
+\begin{explanation}
+Since $\left\{\left(\frac{1}{2}\right)^k\right\}_{k=0}$ is decreasing and $\lim_{n \to \infty} \left(\frac{1}{2}\right)^n =0$, we can use the remainder results.  If we use $s_7$ to approximate  $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$, we have $n=7$, and conclude
+
+\[
+|r_4| < a_5 = \left(\frac{1}{2}\right)^5 = \frac{1}{32}.
+\]
+
+We can explicitly compute $s_4$ by using the formula for partial sums for geometric series or by hand.
+
+\[
+s_4 = 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16} = \frac{11}{16}.
+\]
+
+We note that to five decimal places, we have the following.
+
+\begin{itemize}
+\item The error bound from the theorem is $|r_4| < \frac{1}{32} = .03125$.
+\item The approximate value of the series $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$ is $s_4 = \frac{11}{16} = .68750$.
+\item The actual value of the series $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$ is $\frac{2}{3} \approx .66667$.
+\end{itemize}
+
+Thus, $s_4$ is actually within $.02083$ of the actual value of $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$; $s_4$ is thus within the $.03125$ of the actual value of $\sum_{k=0}^{\infty} (-1)^k \left(\frac{1}{2}\right)^k$ guaranteed by the remainder result.
+
+\end{explanation}
+\end{example}
+
+In the above example, we could compute the exact value of the series in question, so the remainder result could be verified explicitly.  In the next example, the exact value of the series is not known, but we can approximate as closely as we want.
+
+\begin{example}
+Consider the alternating series $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$.
+
+\begin{itemize}
+\item[I.] Find a value for $N$ so $s_N$ approximates $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$ to within $.0001$ of its actual value.
+\item[II.] Using technology, compute $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$ to within $.0001$.
+\item[III.] Using technology, it can be shown that $\sum_{k=1}^{100000} \frac{(-1)^k}{k^3} = -.9015426774$ to ten decimal places.  Using the remainder result, how far off from the exact value of the series $\sum_{k=1}^{100000} \frac{(-1)^k}{k^3}$ could this be?
+\end{itemize}
+
+\begin{explanation}
+Check for yourself that $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$ satisfies the criteria for the remainder result for alternating series to hold before reading on.
+
+\begin{itemize}
+\item[I.] To find a value for $N$ so $s_N$ approximates $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$ to within $.0001$ of its actual value, we can find $N$ so $\big|r_N\big| < .0001$.  Since we have an explicit formula $a_n = \frac{1}{n^3}$ and that $\big|r_N\big| < a_{N+1}$, we can set $a_{N+1} < .0001$ (so $|r_N|$ will be less than $.0001$ by transitivity).
+
+\begin{align*}
+a_{N+1} = \frac{1}{\answer{\left(N+1\right)^3}} &< .0001 \\
+1 &<  .0001 (N+1)^3 \\
+\answer{10000} &< (N+1)^3 \\
+21.54 &< N+1 \qquad  \textrm{ (to two decimal places) }\\
+20.54 &< N \\
+\end{align*}
+
+We should thus use $N =$  \wordChoice{\choice{$20$}\choice[correct]{$21$}}.
+
+\item[II.] We can thus approximate $\sum_{k=1}^{\infty} \frac{(-1)^k}{k^3}$ by using \wordChoice{\choice{$s_20$}\choice[correct]{$s_21$}}.  While this would be tedious to do by hand, technology can compute this sum quickly to find 
+
+\[
+\sum_{k=1}^{21} \frac{(-1)^k}{k^3} \approx -.90159.
+\]
+
+Using this result above, we conclude that $-.90159-.0001 \leq \sum_{k=1}^{\infty} \frac{(-1)^k}{k^3} \leq -.90159+.0001$ or
+
+\[
+-.90169 \leq \sum_{k=1}^{\infty} \frac{(-1)^k}{k^3} \leq -.90149
+\]
+\item[III.] Here, we have used $s_{100000}$ to approximate the value of the series.  The remainder result tells us that the error $r_{100000}$ made in this approximation can be no worse than $a_{100001}$; i.e.
+
+\[
+\big|r_{100000}\big| < \frac{1}{(100001)^3}. 
+\]
+
+Noting that $\frac{1}{(100001)^3} = 9.9997 \times 10^{-16}$, we see that the approximation given above corresponds with the actual value of the series to the ten decimal places listed.
+
+\end{itemize}
+\end{explanation}
+\end{example}
+
+We conclude this section with a question.
+\begin{question}
+Determine if there is a value for $N$ so $\sum_{k=1}^{N} (-1)^k \cdot \frac{k+2}{2k+1}$ is within $.01$ of the value of $\sum_{k=1}^{N} (-1)^k \cdot \frac{k+2}{2k+1}$.  If there is such a value, give one possibility for it.
+
+\begin{multipleChoice}
+\choice[correct]{There is no such value for $N$ since the series $\sum_{k=1}^{N} (-1)^k \cdot \frac{k+2}{2k+1}$ diverges.}
+\choice{There is such a value for $N$.}
+\end{multipleChoice}
+
+\begin{feedback}
+Note that before we try to approximate the value of a series, we must first determine that it converges (so it actually \emph{has} a value).  Here, note that $\lim_{n \to \infty} \frac{n+2}{2n+1} = \frac{1}{2}$, so $\lim_{n \to \infty} (-1)^n \cdot \frac{n+2}{2n+1}$ does not exist, and hence $\sum_{k=1}^{N} (-1)^k \cdot \frac{k+2}{2k+1}$ diverges by the divergence test.
+\end{feedback}
+\end{question}
+
+
+\end{document}