{"id":1718,"date":"2021-07-23T20:29:55","date_gmt":"2021-07-23T20:29:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1718"},"modified":"2022-03-21T23:05:43","modified_gmt":"2022-03-21T23:05:43","slug":"the-alternating-series-test","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/the-alternating-series-test\/","title":{"raw":"The Alternating Series Test","rendered":"The Alternating Series Test"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the alternating series test to test an alternating series for convergence<\/li>\r\n \t<li>Estimate the sum of an alternating series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1169738007132\" data-depth=\"1\">\r\n<p id=\"fs-id1169738227030\">A series whose terms alternate between positive and negative values is an <strong>alternating series<\/strong>. For example, the series<\/p>\r\n\r\n<div id=\"fs-id1169737843037\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-\\frac{1}{2}\\right)}^{n}=-\\frac{1}{2}+\\frac{1}{4}-\\frac{1}{8}+\\frac{1}{16}-\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737783896\">and<\/p>\r\n\r\n<div id=\"fs-id1169738026978\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737760417\">are both alternating series.<\/p>\r\n\r\n<div id=\"fs-id1169738189084\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169737196494\">Any series whose terms alternate between positive and negative values is called an alternating series. An alternating series can be written in the form<\/p>\r\n\r\n<div id=\"fs-id1169737355805\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}={b}_{1}-{b}_{2}+{b}_{3}-{b}_{4}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738170113\">or<\/p>\r\n\r\n<div id=\"fs-id1169738222645\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n - 1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}=\\text{-}{b}_{1}+{b}_{2}-{b}_{3}+{b}_{4}-\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737948416\">Where [latex]{b}_{n}\\ge 0[\/latex] for all positive integers <em data-effect=\"italics\">n<\/em>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169737895556\">Series (1), shown as the first alternating series example, is a geometric series. Since [latex]|r|=|\\text{-}\\frac{1}{2}|&lt;1[\/latex], the series converges. Series (2), shown as the second alternating series example, is called the alternating harmonic series. We will show that whereas the harmonic series diverges, the alternating harmonic series converges.<\/p>\r\n<p id=\"fs-id1169737849876\">To prove this, we look at the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] (Figure 1).<\/p>\r\n\r\n<section id=\"fs-id1169737168496\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\r\n<p id=\"fs-id1169737779407\">Consider the odd terms [latex]{S}_{2k+1}[\/latex] for [latex]k\\ge 0[\/latex]. Since [latex]\\frac{1}{\\left(2k+1\\right)}&lt;\\frac{1}{2k}[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169738187484\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k+1}={S}_{2k - 1}-\\frac{1}{2k}+\\frac{1}{2k+1}&lt;{S}_{2k - 1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738217241\">Therefore, [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] is a decreasing sequence. Also,<\/p>\r\n\r\n<div id=\"fs-id1169737950278\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k+1}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots +\\left(\\frac{1}{2k - 1}-\\frac{1}{2k}\\right)+\\frac{1}{2k+1}&gt;0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738059585\">Therefore, [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] is bounded below. Since [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] is a decreasing sequence that is bounded below, by the Monotone Convergence Theorem, [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] converges. Similarly, the even terms [latex]\\left\\{{S}_{2k}\\right\\}[\/latex] form an increasing sequence that is bounded above because<\/p>\r\n\r\n<div id=\"fs-id1169737955583\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k}={S}_{2k - 2}+\\frac{1}{2k - 1}-\\frac{1}{2k}&gt;{S}_{2k - 2}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737822225\">and<\/p>\r\n\r\n<div id=\"fs-id1169737837204\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k}=1+\\left(-\\frac{1}{2}+\\frac{1}{3}\\right)+\\cdots +\\left(-\\frac{1}{2k - 2}+\\frac{1}{2k - 1}\\right)-\\frac{1}{2k}&lt;1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737770954\">Therefore, by the Monotone Convergence Theorem, the sequence [latex]\\left\\{{S}_{2k}\\right\\}[\/latex] also converges. Since<\/p>\r\n\r\n<div id=\"fs-id1169738064210\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k+1}={S}_{2k}+\\frac{1}{2k+1}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737845579\">we know that<\/p>\r\n\r\n<div id=\"fs-id1169737772510\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k+1}=\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k}+\\underset{k\\to \\infty }{\\text{lim}}\\frac{1}{2k+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737770972\">Letting [latex]S=\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k+1}[\/latex] and using the fact that [latex]\\frac{1}{\\left(2k+1\\right)}\\to 0[\/latex], we conclude that [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k}=S[\/latex]. Since the odd terms and the even terms in the sequence of partial sums converge to the same limit [latex]S[\/latex], it can be shown that the sequence of partial sums converges to [latex]S[\/latex], and therefore the alternating harmonic series converges to [latex]S[\/latex].<\/p>\r\n<p id=\"fs-id1169738184603\">It can also be shown that [latex]S=\\text{ln}2[\/latex], and we can write<\/p>\r\n\r\n<div id=\"fs-id1169738249486\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots =\\text{ln}\\left(2\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<figure id=\"CNX_Calc_Figure_09_05_001\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234401\/CNX_Calc_Figure_09_05_001.jpg\" alt=\"This graph demonstrates the alternating hamanic series in the first quadrant. The highest line 1 is drawn to S1, the next line -1\/2 is drawn to S2, the next line +1\/3 is drawn to S3, the line -1\/4 is drawn to S4, and the last line +1\/5 is drawn to S5. The odd terms are decreasing and bounded below, and the even terms are increasing and bounded above. It seems to be converging to S, which is in the middle of S2, S4 and S5, S3, S1.\" width=\"325\" height=\"205\" data-media-type=\"image\/jpeg\" \/> Figure 1. For the alternating harmonic series, the odd terms [latex]{S}_{2k+1}[\/latex] in the sequence of partial sums are decreasing and bounded below. The even terms [latex]{S}_{2k}[\/latex] are increasing and bounded above.[\/caption]<\/figure>\r\n<p id=\"fs-id1169737844496\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1169737953787\">More generally, any alternating series of form (3) or (4) (see the definition) converges as long as [latex]{b}_{1}\\ge {b}_{2}\\ge {b}_{3}\\ge \\cdots [\/latex] and [latex]{b}_{n}\\to 0[\/latex] (Figure 2). The proof is similar to the proof for the alternating harmonic series.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_05_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234406\/CNX_Calc_Figure_09_05_002.jpg\" alt=\"This diagram illustrates an alternating series in quadrant 1. The highest line b1 is drawn out to S1, the next line \u2013b2 is drawn back to S2, the next line b3 is drawn out to S3, the next line \u2013b4 is drawn back to S4, and the last line is drawn out to S5. It seems to be converging to S, which is in between S2, S4 and S5, S3, and S1. The odd terms are decreasing and bounded below. The even terms are increasing and bounded above.\" width=\"325\" height=\"206\" data-media-type=\"image\/jpeg\" \/> Figure 2. For an alternating series [latex]{b}_{1}-{b}_{2}+{b}_{3}-\\cdots [\/latex] in which [latex]{b}_{1}&gt;{b}_{2}&gt;{b}_{3}&gt;\\cdots [\/latex], the odd terms [latex]{S}_{2k+1}[\/latex] in the sequence of partial sums are decreasing and bounded below. The even terms [latex]{S}_{2k}[\/latex] are increasing and bounded above.[\/caption]<\/figure>\r\n<div id=\"fs-id1169738154201\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Alternating Series Test<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169738059080\">An alternating series of the form<\/p>\r\n\r\n<div id=\"fs-id1169738101714\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}\\text{or}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737839161\">converges if<\/p>\r\n\r\n<ol id=\"fs-id1169738187898\" type=\"i\">\r\n \t<li>[latex]0\\le {b}_{n+1}\\le {b}_{n}[\/latex] for all [latex]n\\ge 1[\/latex] and<\/li>\r\n \t<li>[latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1169738152556\">This is known as the <span data-type=\"term\">alternating series test<\/span>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738142452\">We remark that this theorem is true more generally as long as there exists some integer [latex]N[\/latex] such that [latex]0\\le {b}_{n+1}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169737819249\" data-type=\"example\">\r\n<div id=\"fs-id1169737949836\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737788765\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Convergence of Alternating Series<\/h3>\r\n<div id=\"fs-id1169737788765\" data-type=\"problem\">\r\n<p id=\"fs-id1169737160608\">For each of the following alternating series, determine whether the series converges or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169738018992\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}n}{\\left(n+1\\right)}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1169737952092\" data-type=\"solution\">\r\n<ol id=\"fs-id1169737742192\" type=\"a\">\r\n \t<li>Since<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\frac{1}{{\\left(n+1\\right)}^{2}}&lt;\\frac{1}{{n}^{2}}\\text{ and }\\frac{1}{{n}^{2}}\\to 0[\/latex], <span data-type=\"newline\">\r\n<\/span>\r\nthe series converges.<\/li>\r\n \t<li>Since [latex]\\frac{n}{\\left(n+1\\right)}\\nrightarrow 0[\/latex] as [latex]n\\to \\infty [\/latex], we cannot apply the alternating series test. Instead, we use the <em data-effect=\"italics\">n<\/em>th term test for divergence. Since<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{\\left(-1\\right)}^{n+1}n}{n+1}\\ne 0[\/latex], <span data-type=\"newline\">\r\n<\/span>\r\nthe series diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738143988\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169737814736\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737745304\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169737745304\" data-type=\"problem\">\r\n<p id=\"fs-id1169738109091\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}n}{{2}^{n}}[\/latex] converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1169738000344\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737784927\">Is [latex]\\left\\{\\frac{n}{{2}^{n}}\\right\\}[\/latex] decreasing? What is [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{n}{{2}^{n}}\\text{?}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1169737825467\" data-type=\"solution\">\r\n<p id=\"fs-id1169737790551\">The series converges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/isJ2P7efSEM?controls=0&amp;start=295&amp;end=507&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.5.1_295to507_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.5.1\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question] 230800[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><\/section><section id=\"fs-id1169737209190\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Remainder of an Alternating Series<\/h2>\r\n<p id=\"fs-id1169738214451\">It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series<\/p>\r\n\r\n<div id=\"fs-id1169296846669\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738114153\">satisfying the hypotheses of the alternating series test. Let [latex]S[\/latex] denote the sum of this series and [latex]\\left\\{{S}_{k}\\right\\}[\/latex] be the corresponding sequence of partial sums. From Figure 2, we see that for any integer [latex]N\\ge 1[\/latex], the remainder [latex]{R}_{N}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169737790066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{N}|=|S-{S}_{N}|\\le |{S}_{N+1}-{S}_{N}|={b}_{n+1}[\/latex].<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1169738217519\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Remainders in Alternating Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169737806314\">Consider an alternating series of the form<\/p>\r\n\r\n<div id=\"fs-id1169738045618\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}\\text{or}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737743754\">that satisfies the hypotheses of the alternating series test. Let [latex]S[\/latex] denote the sum of the series and [latex]{S}_{N}[\/latex] denote the [latex]N\\text{th}[\/latex] partial sum. For any integer [latex]N\\ge 1[\/latex], the remainder [latex]{R}_{N}=S-{S}_{N}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169737725569\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{N}|\\le {b}_{N+1}[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169737842987\">In other words, if the conditions of the alternating series test apply, then the error in approximating the infinite series by the [latex]N\\text{th}[\/latex] partial sum [latex]{S}_{N}[\/latex] is in magnitude at most the size of the next term [latex]{b}_{N+1}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169737939430\" data-type=\"example\">\r\n<div id=\"fs-id1169737939432\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738153570\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Estimating the Remainder of an Alternating Series<\/h3>\r\n<div id=\"fs-id1169738153570\" data-type=\"problem\">\r\n<p id=\"fs-id1169738153575\">Consider the alternating series<\/p>\r\n\r\n<div id=\"fs-id1169738249304\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738087686\">Use the remainder estimate to determine a bound on the error [latex]{R}_{10}[\/latex] if we approximate the sum of the series by the partial sum [latex]{S}_{10}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169738183341\" data-type=\"solution\">\r\n<p id=\"fs-id1169738183343\">From the theorem stated above,<\/p>\r\n<p id=\"fs-id1169737931110\" style=\"text-align: center;\">[latex]|{R}_{10}|\\le {b}_{11}=\\frac{1}{{11}^{2}}\\approx 0.008265[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738057293\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738057296\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738154834\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169738154834\" data-type=\"problem\">\r\n<p id=\"fs-id1169738154836\">Find a bound for [latex]{R}_{20}[\/latex] when approximating [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}[\/latex] by [latex]{S}_{20}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1169738190280\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169738186221\">[latex]|{R}_{20}|\\le {b}_{21}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1169738111584\" data-type=\"solution\">\r\n<p id=\"fs-id1169738111585\">[latex]0.04762[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1169737985058\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the alternating series test to test an alternating series for convergence<\/li>\n<li>Estimate the sum of an alternating series<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1169738007132\" data-depth=\"1\">\n<p id=\"fs-id1169738227030\">A series whose terms alternate between positive and negative values is an <strong>alternating series<\/strong>. For example, the series<\/p>\n<div id=\"fs-id1169737843037\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-\\frac{1}{2}\\right)}^{n}=-\\frac{1}{2}+\\frac{1}{4}-\\frac{1}{8}+\\frac{1}{16}-\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737783896\">and<\/p>\n<div id=\"fs-id1169738026978\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737760417\">are both alternating series.<\/p>\n<div id=\"fs-id1169738189084\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1169737196494\">Any series whose terms alternate between positive and negative values is called an alternating series. An alternating series can be written in the form<\/p>\n<div id=\"fs-id1169737355805\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}={b}_{1}-{b}_{2}+{b}_{3}-{b}_{4}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738170113\">or<\/p>\n<div id=\"fs-id1169738222645\" class=\"numbered\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n - 1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}=\\text{-}{b}_{1}+{b}_{2}-{b}_{3}+{b}_{4}-\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737948416\">Where [latex]{b}_{n}\\ge 0[\/latex] for all positive integers <em data-effect=\"italics\">n<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169737895556\">Series (1), shown as the first alternating series example, is a geometric series. Since [latex]|r|=|\\text{-}\\frac{1}{2}|<1[\/latex], the series converges. Series (2), shown as the second alternating series example, is called the alternating harmonic series. We will show that whereas the harmonic series diverges, the alternating harmonic series converges.<\/p>\n<p id=\"fs-id1169737849876\">To prove this, we look at the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] (Figure 1).<\/p>\n<section id=\"fs-id1169737168496\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\n<p id=\"fs-id1169737779407\">Consider the odd terms [latex]{S}_{2k+1}[\/latex] for [latex]k\\ge 0[\/latex]. Since [latex]\\frac{1}{\\left(2k+1\\right)}<\\frac{1}{2k}[\/latex],<\/p>\n<div id=\"fs-id1169738187484\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k+1}={S}_{2k - 1}-\\frac{1}{2k}+\\frac{1}{2k+1}<{S}_{2k - 1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738217241\">Therefore, [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] is a decreasing sequence. Also,<\/p>\n<div id=\"fs-id1169737950278\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k+1}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots +\\left(\\frac{1}{2k - 1}-\\frac{1}{2k}\\right)+\\frac{1}{2k+1}>0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738059585\">Therefore, [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] is bounded below. Since [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] is a decreasing sequence that is bounded below, by the Monotone Convergence Theorem, [latex]\\left\\{{S}_{2k+1}\\right\\}[\/latex] converges. Similarly, the even terms [latex]\\left\\{{S}_{2k}\\right\\}[\/latex] form an increasing sequence that is bounded above because<\/p>\n<div id=\"fs-id1169737955583\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k}={S}_{2k - 2}+\\frac{1}{2k - 1}-\\frac{1}{2k}>{S}_{2k - 2}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737822225\">and<\/p>\n<div id=\"fs-id1169737837204\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k}=1+\\left(-\\frac{1}{2}+\\frac{1}{3}\\right)+\\cdots +\\left(-\\frac{1}{2k - 2}+\\frac{1}{2k - 1}\\right)-\\frac{1}{2k}<1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737770954\">Therefore, by the Monotone Convergence Theorem, the sequence [latex]\\left\\{{S}_{2k}\\right\\}[\/latex] also converges. Since<\/p>\n<div id=\"fs-id1169738064210\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{2k+1}={S}_{2k}+\\frac{1}{2k+1}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737845579\">we know that<\/p>\n<div id=\"fs-id1169737772510\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k+1}=\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k}+\\underset{k\\to \\infty }{\\text{lim}}\\frac{1}{2k+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737770972\">Letting [latex]S=\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k+1}[\/latex] and using the fact that [latex]\\frac{1}{\\left(2k+1\\right)}\\to 0[\/latex], we conclude that [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{2k}=S[\/latex]. Since the odd terms and the even terms in the sequence of partial sums converge to the same limit [latex]S[\/latex], it can be shown that the sequence of partial sums converges to [latex]S[\/latex], and therefore the alternating harmonic series converges to [latex]S[\/latex].<\/p>\n<p id=\"fs-id1169738184603\">It can also be shown that [latex]S=\\text{ln}2[\/latex], and we can write<\/p>\n<div id=\"fs-id1169738249486\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots =\\text{ln}\\left(2\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<figure id=\"CNX_Calc_Figure_09_05_001\"><figcaption><\/figcaption><div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234401\/CNX_Calc_Figure_09_05_001.jpg\" alt=\"This graph demonstrates the alternating hamanic series in the first quadrant. The highest line 1 is drawn to S1, the next line -1\/2 is drawn to S2, the next line +1\/3 is drawn to S3, the line -1\/4 is drawn to S4, and the last line +1\/5 is drawn to S5. The odd terms are decreasing and bounded below, and the even terms are increasing and bounded above. It seems to be converging to S, which is in the middle of S2, S4 and S5, S3, S1.\" width=\"325\" height=\"205\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. For the alternating harmonic series, the odd terms [latex]{S}_{2k+1}[\/latex] in the sequence of partial sums are decreasing and bounded below. The even terms [latex]{S}_{2k}[\/latex] are increasing and bounded above.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1169737844496\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1169737953787\">More generally, any alternating series of form (3) or (4) (see the definition) converges as long as [latex]{b}_{1}\\ge {b}_{2}\\ge {b}_{3}\\ge \\cdots[\/latex] and [latex]{b}_{n}\\to 0[\/latex] (Figure 2). The proof is similar to the proof for the alternating harmonic series.<\/p>\n<figure id=\"CNX_Calc_Figure_09_05_002\"><figcaption><\/figcaption><div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234406\/CNX_Calc_Figure_09_05_002.jpg\" alt=\"This diagram illustrates an alternating series in quadrant 1. The highest line b1 is drawn out to S1, the next line \u2013b2 is drawn back to S2, the next line b3 is drawn out to S3, the next line \u2013b4 is drawn back to S4, and the last line is drawn out to S5. It seems to be converging to S, which is in between S2, S4 and S5, S3, and S1. The odd terms are decreasing and bounded below. The even terms are increasing and bounded above.\" width=\"325\" height=\"206\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. For an alternating series [latex]{b}_{1}-{b}_{2}+{b}_{3}-\\cdots [\/latex] in which [latex]{b}_{1}&gt;{b}_{2}&gt;{b}_{3}&gt;\\cdots [\/latex], the odd terms [latex]{S}_{2k+1}[\/latex] in the sequence of partial sums are decreasing and bounded below. The even terms [latex]{S}_{2k}[\/latex] are increasing and bounded above.<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1169738154201\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Alternating Series Test<\/h3>\n<hr \/>\n<p id=\"fs-id1169738059080\">An alternating series of the form<\/p>\n<div id=\"fs-id1169738101714\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}\\text{or}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737839161\">converges if<\/p>\n<ol id=\"fs-id1169738187898\" type=\"i\">\n<li>[latex]0\\le {b}_{n+1}\\le {b}_{n}[\/latex] for all [latex]n\\ge 1[\/latex] and<\/li>\n<li>[latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1169738152556\">This is known as the <span data-type=\"term\">alternating series test<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738142452\">We remark that this theorem is true more generally as long as there exists some integer [latex]N[\/latex] such that [latex]0\\le {b}_{n+1}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex].<\/p>\n<div id=\"fs-id1169737819249\" data-type=\"example\">\n<div id=\"fs-id1169737949836\" data-type=\"exercise\">\n<div id=\"fs-id1169737788765\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Convergence of Alternating Series<\/h3>\n<div id=\"fs-id1169737788765\" data-type=\"problem\">\n<p id=\"fs-id1169737160608\">For each of the following alternating series, determine whether the series converges or diverges.<\/p>\n<ol id=\"fs-id1169738018992\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}n}{\\left(n+1\\right)}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737952092\" data-type=\"solution\">\n<ol id=\"fs-id1169737742192\" type=\"a\">\n<li>Since<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\frac{1}{{\\left(n+1\\right)}^{2}}<\\frac{1}{{n}^{2}}\\text{ and }\\frac{1}{{n}^{2}}\\to 0[\/latex], <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nthe series converges.<\/li>\n<li>Since [latex]\\frac{n}{\\left(n+1\\right)}\\nrightarrow 0[\/latex] as [latex]n\\to \\infty[\/latex], we cannot apply the alternating series test. Instead, we use the <em data-effect=\"italics\">n<\/em>th term test for divergence. Since<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{\\left(-1\\right)}^{n+1}n}{n+1}\\ne 0[\/latex], <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nthe series diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738143988\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169737814736\" data-type=\"exercise\">\n<div id=\"fs-id1169737745304\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169737745304\" data-type=\"problem\">\n<p id=\"fs-id1169738109091\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}n}{{2}^{n}}[\/latex] converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738000344\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737784927\">Is [latex]\\left\\{\\frac{n}{{2}^{n}}\\right\\}[\/latex] decreasing? What is [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{n}{{2}^{n}}\\text{?}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737825467\" data-type=\"solution\">\n<p id=\"fs-id1169737790551\">The series converges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/isJ2P7efSEM?controls=0&amp;start=295&amp;end=507&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.5.1_295to507_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.5.1&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm230800\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=230800&theme=oea&iframe_resize_id=ohm230800&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<\/section>\n<section id=\"fs-id1169737209190\" data-depth=\"1\">\n<h2 data-type=\"title\">Remainder of an Alternating Series<\/h2>\n<p id=\"fs-id1169738214451\">It is difficult to explicitly calculate the sum of most alternating series, so typically the sum is approximated by using a partial sum. When doing so, we are interested in the amount of error in our approximation. Consider an alternating series<\/p>\n<div id=\"fs-id1169296846669\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738114153\">satisfying the hypotheses of the alternating series test. Let [latex]S[\/latex] denote the sum of this series and [latex]\\left\\{{S}_{k}\\right\\}[\/latex] be the corresponding sequence of partial sums. From Figure 2, we see that for any integer [latex]N\\ge 1[\/latex], the remainder [latex]{R}_{N}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169737790066\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{N}|=|S-{S}_{N}|\\le |{S}_{N+1}-{S}_{N}|={b}_{n+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169738217519\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Remainders in Alternating Series<\/h3>\n<hr \/>\n<p id=\"fs-id1169737806314\">Consider an alternating series of the form<\/p>\n<div id=\"fs-id1169738045618\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}\\text{or}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737743754\">that satisfies the hypotheses of the alternating series test. Let [latex]S[\/latex] denote the sum of the series and [latex]{S}_{N}[\/latex] denote the [latex]N\\text{th}[\/latex] partial sum. For any integer [latex]N\\ge 1[\/latex], the remainder [latex]{R}_{N}=S-{S}_{N}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169737725569\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{R}_{N}|\\le {b}_{N+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169737842987\">In other words, if the conditions of the alternating series test apply, then the error in approximating the infinite series by the [latex]N\\text{th}[\/latex] partial sum [latex]{S}_{N}[\/latex] is in magnitude at most the size of the next term [latex]{b}_{N+1}[\/latex].<\/p>\n<div id=\"fs-id1169737939430\" data-type=\"example\">\n<div id=\"fs-id1169737939432\" data-type=\"exercise\">\n<div id=\"fs-id1169738153570\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Estimating the Remainder of an Alternating Series<\/h3>\n<div id=\"fs-id1169738153570\" data-type=\"problem\">\n<p id=\"fs-id1169738153575\">Consider the alternating series<\/p>\n<div id=\"fs-id1169738249304\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738087686\">Use the remainder estimate to determine a bound on the error [latex]{R}_{10}[\/latex] if we approximate the sum of the series by the partial sum [latex]{S}_{10}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738183341\" data-type=\"solution\">\n<p id=\"fs-id1169738183343\">From the theorem stated above,<\/p>\n<p id=\"fs-id1169737931110\" style=\"text-align: center;\">[latex]|{R}_{10}|\\le {b}_{11}=\\frac{1}{{11}^{2}}\\approx 0.008265[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738057293\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738057296\" data-type=\"exercise\">\n<div id=\"fs-id1169738154834\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169738154834\" data-type=\"problem\">\n<p id=\"fs-id1169738154836\">Find a bound for [latex]{R}_{20}[\/latex] when approximating [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}[\/latex] by [latex]{S}_{20}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Hint<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738190280\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169738186221\">[latex]|{R}_{20}|\\le {b}_{21}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Show Solution<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738111584\" data-type=\"solution\">\n<p id=\"fs-id1169738111585\">[latex]0.04762[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1169737985058\" data-depth=\"1\"><\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1718\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>5.5.1. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":21,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.5.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1718","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1718","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1718\/revisions"}],"predecessor-version":[{"id":2334,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1718\/revisions\/2334"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1718\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1718"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1718"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1718"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1718"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}