{"id":1111,"date":"2025-06-30T16:34:45","date_gmt":"2025-06-30T16:34:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1111"},"modified":"2025-08-18T15:17:08","modified_gmt":"2025-08-18T15:17:08","slug":"alternating-series-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/alternating-series-learn-it-1\/","title":{"raw":"Alternating Series: Learn It 1","rendered":"Alternating Series: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the alternating series test to check if an alternating series converges<\/li>\r\n \t<li>Understand the difference between absolute and conditional convergence<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Alternating Series<\/h2>\r\nUp to this point, we've focused on series with positive terms. Now we'll explore <strong>alternating series<\/strong>\u2014series whose terms switch between positive and negative values.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>theorem: alternating series<\/h3>\r\n<p class=\"whitespace-normal break-words\">An alternating series has terms that alternate between positive and negative values. Any alternating series can be written as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^{n+1}b_n = b_1 - b_2 + b_3 - b_4 + \\cdots[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">or<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^n b_n = -b_1 + b_2 - b_3 + b_4 - \\cdots[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">where [latex]b_n \\geq 0[\/latex] for all positive integers [latex]n[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Let's look at two common examples:<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\"><strong>Example 1:<\/strong> [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]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Example 2:<\/strong> [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\cdots[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">The first series is a geometric series with [latex]r = -\\frac{1}{2}[\/latex]. Since [latex]|r| = \\left|-\\frac{1}{2}\\right| = \\frac{1}{2} &lt; 1[\/latex], this series converges.<\/p>\r\n<p class=\"whitespace-normal break-words\">The second series is the <strong>alternating harmonic series<\/strong>. While the regular harmonic series [latex]\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] diverges, we'll discover that this alternating version actually converges.<\/p>\r\n\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Alternating Series Test<\/h2>\r\n<p id=\"fs-id1169737849876\">To understand why the alternating harmonic series converges, we need to examine its sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] (Figure 1). The key insight is to analyze the odd and even partial sums separately, showing that both subsequences converge to the same limit.<\/p>\r\n\r\n<section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\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\">[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\r\n<\/section>\r\n<p id=\"fs-id1169737953787\">The proof above suggests a general pattern. Any alternating series converges under specific conditions, as stated in the following theorem.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_05_002\">[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<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>alternating series test<\/h3>\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] (the terms [latex]b_n[\/latex] are decreasing)<\/li>\r\n \t<li>[latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex] (the terms approach zero)<\/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<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">This theorem remains true even if the conditions only hold for [latex]n \\geq N[\/latex] for some integer [latex]N[\/latex]. The first few terms don't affect convergence, only the eventual behavior matters.<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]230800[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the alternating series test to check if an alternating series converges<\/li>\n<li>Understand the difference between absolute and conditional convergence<\/li>\n<\/ul>\n<\/section>\n<h2>Alternating Series<\/h2>\n<p>Up to this point, we&#8217;ve focused on series with positive terms. Now we&#8217;ll explore <strong>alternating series<\/strong>\u2014series whose terms switch between positive and negative values.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>theorem: alternating series<\/h3>\n<p class=\"whitespace-normal break-words\">An alternating series has terms that alternate between positive and negative values. Any alternating series can be written as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^{n+1}b_n = b_1 - b_2 + b_3 - b_4 + \\cdots[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">or<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^n b_n = -b_1 + b_2 - b_3 + b_4 - \\cdots[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">where [latex]b_n \\geq 0[\/latex] for all positive integers [latex]n[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Let&#8217;s look at two common examples:<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\"><strong>Example 1:<\/strong> [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]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Example 2:<\/strong> [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\cdots[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">The first series is a geometric series with [latex]r = -\\frac{1}{2}[\/latex]. Since [latex]|r| = \\left|-\\frac{1}{2}\\right| = \\frac{1}{2} < 1[\/latex], this series converges.<\/p>\n<p class=\"whitespace-normal break-words\">The second series is the <strong>alternating harmonic series<\/strong>. While the regular harmonic series [latex]\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] diverges, we&#8217;ll discover that this alternating version actually converges.<\/p>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Alternating Series Test<\/h2>\n<p id=\"fs-id1169737849876\">To understand why the alternating harmonic series converges, we need to examine its sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] (Figure 1). The key insight is to analyze the odd and even partial sums separately, showing that both subsequences converge to the same limit.<\/p>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\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\">\n<figure style=\"width: 325px\" 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\" \/><figcaption 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.<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1169737844496\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1169737953787\">The proof above suggests a general pattern. Any alternating series converges under specific conditions, as stated in the following theorem.<\/p>\n<figure id=\"CNX_Calc_Figure_09_05_002\">\n<figure style=\"width: 325px\" 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\" \/><figcaption 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.<\/figcaption><\/figure>\n<\/figure>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>alternating series test<\/h3>\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] (the terms [latex]b_n[\/latex] are decreasing)<\/li>\n<li>[latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}=0[\/latex] (the terms approach zero)<\/li>\n<\/ol>\n<p id=\"fs-id1169738152556\">This is known as the <span data-type=\"term\">alternating series test<\/span>.<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">This theorem remains true even if the conditions only hold for [latex]n \\geq N[\/latex] for some integer [latex]N[\/latex]. The first few terms don&#8217;t affect convergence, only the eventual behavior matters.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm230800\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=230800&theme=lumen&iframe_resize_id=ohm230800&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":28,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1111"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1111\/revisions"}],"predecessor-version":[{"id":1900,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1111\/revisions\/1900"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1111\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1111"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1111"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1111"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1111"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}