{"id":1719,"date":"2021-07-23T20:29:51","date_gmt":"2021-07-23T20:29:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1719"},"modified":"2022-03-21T23:06:03","modified_gmt":"2022-03-21T23:06:03","slug":"absolute-and-conditional-convergence","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/absolute-and-conditional-convergence\/","title":{"raw":"Absolute and Conditional Convergence","rendered":"Absolute and Conditional Convergence"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning OutcomeS<\/h3>\r\n<ul>\r\n \t<li>Explain the meaning of absolute convergence and conditional convergence<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169737816423\">Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and the related series [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex]. Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}[\/latex]. The series whose terms are the absolute value of these terms is the harmonic series, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{{\\left(-1\\right)}^{n+1}}{n}|=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.<\/p>\r\n<p id=\"fs-id1169737947414\">By comparison, consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex]. The series whose terms are the absolute values of the terms of this series is the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. Since both of these series converge, we say the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex] exhibits absolute convergence.<\/p>\r\n\r\n<div id=\"fs-id1169737232513\" 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-id1169737232517\">A series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits <strong>absolute convergence<\/strong> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges. A series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits <span data-type=\"term\">conditional convergence<\/span> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges but [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] diverges.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169737931155\">As shown by the alternating harmonic series, a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] may converge, but [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] may diverge. In the following theorem, however, we show that if [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\r\n\r\n<div id=\"fs-id1169737226135\" 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: Absolute Convergence Implies Convergence<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169737212169\">If [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1169738115704\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\r\n<p id=\"fs-id1169737264006\">Suppose that [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges. We show this by using the fact that [latex]{a}_{n}=|{a}_{n}|[\/latex] or [latex]{a}_{n}=\\text{-}|{a}_{n}|[\/latex] and therefore [latex]|{a}_{n}|+{a}_{n}=2|{a}_{n}|[\/latex] or [latex]|{a}_{n}|+{a}_{n}=0[\/latex]. Therefore, [latex]0\\le |{a}_{n}|+{a}_{n}\\le 2|{a}_{n}|[\/latex]. Consequently, by the comparison test, since [latex]2\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, the series<\/p>\r\n\r\n<div id=\"fs-id1169738045626\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(|{a}_{n}|+{a}_{n}\\right)[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738167761\">converges. By using the algebraic properties for convergent series, we conclude that<\/p>\r\n\r\n<div id=\"fs-id1169738167764\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\displaystyle\\sum _{n=1}^{\\infty }\\left(|{a}_{n}|+{a}_{n}\\right)\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737297637\">converges.<\/p>\r\n<p id=\"fs-id1169737297640\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1169738153895\" data-type=\"example\">\r\n<div id=\"fs-id1169738153897\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738153899\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Absolute versus Conditional Convergence<\/h3>\r\n<div id=\"fs-id1169738153899\" data-type=\"problem\">\r\n<p id=\"fs-id1169738153905\">For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169738153909\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{\\left(3n+1\\right)}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos\\left(n\\right)}{{n}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1169737232565\" data-type=\"solution\">\r\n<ol id=\"fs-id1169737232568\" type=\"a\">\r\n \t<li>We can see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738128166\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{{\\left(-1\\right)}^{n+1}}{3n+1}|=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{3n+1}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\ndiverges by using the limit comparison test with the harmonic series. In fact,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737162005\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left(3n+1\\right)}}{\\frac{1}{n}}=\\frac{1}{3}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, the series does not converge absolutely. However, since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738239379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{3\\left(n+1\\right)+1}&lt;\\frac{1}{3n+1}\\text{and}\\frac{1}{3n+1}\\to 0[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nthe series converges. We can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{\\left(3n+1\\right)}[\/latex] converges conditionally.<\/li>\r\n \t<li>Noting that [latex]|\\cos{n}|\\le 1[\/latex], to determine whether the series converges absolutely, compare<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738223988\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{\\cos{n}}{{n}^{2}}|[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwith the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, by the comparison test, [latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{\\cos{n}}{{n}^{2}}|[\/latex] converges, and therefore [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos{n}}{{n}^{2}}[\/latex] converges absolutely.<\/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-id1169737262013\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169737262016\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737262018\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169737262018\" data-type=\"problem\">\r\n<p id=\"fs-id1169737262020\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{n}{\\left(2{n}^{3}+1\\right)}[\/latex] converges absolutely, converges conditionally, or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1169737161109\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737161116\">Check for absolute convergence first.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1169737161103\" data-type=\"solution\">\r\n<p id=\"fs-id1169737161105\">The series converges absolutely.<\/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\/OZ6z5F1c8HA?controls=0&amp;start=347&amp;end=462&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>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.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.5.3_347to462_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.5.3\" here (opens in new window)<\/a>.\r\n<p id=\"fs-id1169737161122\">To see the difference between absolute and conditional convergence, look at what happens when we <em data-effect=\"italics\">rearrange<\/em> the terms of the alternating harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}[\/latex]. We show that we can rearrange the terms so that the new series diverges. Certainly if we rearrange the terms of a finite sum, the sum does not change. When we work with an infinite sum, however, interesting things can happen.<\/p>\r\n<p id=\"fs-id1169737161174\">Begin by adding enough of the positive terms to produce a sum that is larger than some real number [latex]M&gt;0[\/latex]. For example, let [latex]M=10[\/latex], and find an integer [latex]k[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169738198575\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+\\frac{1}{3}+\\frac{1}{5}+\\cdots +\\frac{1}{2k - 1}&gt;10[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738198624\">(We can do this because the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\left(2n - 1\\right)}[\/latex] diverges to infinity.) Then subtract [latex]\\frac{1}{2}[\/latex]. Then add more positive terms until the sum reaches 100. That is, find another integer [latex]j&gt;k[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169737911250\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+\\frac{1}{3}+\\cdots +\\frac{1}{2k - 1}-\\frac{1}{2}+\\frac{1}{2k+1}+\\cdots +\\frac{1}{2j+1}&gt;100[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737297380\">Then subtract [latex]\\frac{1}{4}[\/latex]. Continuing in this way, we have found a way of rearranging the terms in the alternating harmonic series so that the sequence of partial sums for the rearranged series is unbounded and therefore diverges.<\/p>\r\n<p id=\"fs-id1169737297396\">The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In the next example, we show how to rearrange the terms to create a new series that converges to [latex]3\\text{ln}\\frac{\\left(2\\right)}{2}[\/latex]. We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number [latex]r[\/latex]; however, the proof of that fact is beyond the scope of this text.<\/p>\r\n<p id=\"fs-id1169737297434\">In general, any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] that converges absolutely, the value of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.<\/p>\r\n\r\n<div id=\"fs-id1169738079721\" data-type=\"example\">\r\n<div id=\"fs-id1169738079723\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738079725\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Rearranging Series<\/h3>\r\n<div id=\"fs-id1169738079725\" data-type=\"problem\">\r\n<p id=\"fs-id1169738079730\">Use the fact that<\/p>\r\n\r\n<div id=\"fs-id1169738079734\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\cdots =\\text{ln}2[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738079783\" style=\"text-align: left;\">to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is [latex]\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1169737832891\" data-type=\"solution\">\r\n<p id=\"fs-id1169737832893\">Let<\/p>\r\n\r\n<div id=\"fs-id1169737832896\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\frac{1}{6}+\\frac{1}{7}-\\frac{1}{8}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738211778\">Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\text{ln}\\left(2\\right)[\/latex], by the algebraic properties of convergent series,<\/p>\r\n\r\n<div id=\"fs-id1169738211821\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2}{a}_{n}=\\frac{1}{2}-\\frac{1}{4}+\\frac{1}{6}-\\frac{1}{8}+\\cdots =\\frac{1}{2}\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\frac{\\text{ln}2}{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737933401\">Now introduce the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] such that for all [latex]n\\ge 1[\/latex], [latex]{b}_{2n - 1}=0[\/latex] and [latex]{b}_{2n}=\\frac{{a}_{n}}{2}[\/latex]. Then<\/p>\r\n\r\n<div id=\"fs-id1169738080215\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}=0+\\frac{1}{2}+0-\\frac{1}{4}+0+\\frac{1}{6}+0-\\frac{1}{8}+\\cdots =\\frac{\\text{ln}2}{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737934699\">Then using the algebraic limit properties of convergent series, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converge, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)[\/latex] converges and<\/p>\r\n\r\n<div id=\"fs-id1169737934790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}=\\text{ln}2+\\frac{\\text{ln}2}{2}=\\frac{3\\text{ln}2}{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737254437\">Now adding the corresponding terms, [latex]{a}_{n}[\/latex] and [latex]{b}_{n}[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1169738228421\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)&amp; =\\left(1+0\\right)+\\left(-\\frac{1}{2}+\\frac{1}{2}\\right)+\\left(\\frac{1}{3}+0\\right)+\\left(-\\frac{1}{4}-\\frac{1}{4}\\right)+\\left(\\frac{1}{5}+0\\right)+\\left(-\\frac{1}{6}+\\frac{1}{6}\\right)\\hfill \\\\ \\\\ &amp; +\\left(\\frac{1}{7}+0\\right)+\\left(\\frac{1}{8}-\\frac{1}{8}\\right)+\\cdots \\hfill \\\\ &amp; =1+\\frac{1}{3}-\\frac{1}{2}+\\frac{1}{5}+\\frac{1}{7}-\\frac{1}{4}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738068068\">We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex], we conclude that<\/p>\r\n\r\n<div id=\"fs-id1169737162164\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+\\frac{1}{3}-\\frac{1}{2}+\\frac{1}{5}+\\frac{1}{7}-\\frac{1}{4}+\\cdots =\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737162236\">Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]230805[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1169737162243\" class=\"key-concepts\" data-depth=\"3\"><\/section><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning OutcomeS<\/h3>\n<ul>\n<li>Explain the meaning of absolute convergence and conditional convergence<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169737816423\">Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and the related series [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex]. Here we discuss possibilities for the relationship between the convergence of these two series. For example, consider the alternating harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}[\/latex]. The series whose terms are the absolute value of these terms is the harmonic series, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{{\\left(-1\\right)}^{n+1}}{n}|=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.<\/p>\n<p id=\"fs-id1169737947414\">By comparison, consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex]. The series whose terms are the absolute values of the terms of this series is the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. Since both of these series converge, we say the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex] exhibits absolute convergence.<\/p>\n<div id=\"fs-id1169737232513\" 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-id1169737232517\">A series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits <strong>absolute convergence<\/strong> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges. A series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits <span data-type=\"term\">conditional convergence<\/span> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges but [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] diverges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169737931155\">As shown by the alternating harmonic series, a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] may converge, but [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] may diverge. In the following theorem, however, we show that if [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\n<div id=\"fs-id1169737226135\" 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: Absolute Convergence Implies Convergence<\/h3>\n<hr \/>\n<p id=\"fs-id1169737212169\">If [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169738115704\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\n<p id=\"fs-id1169737264006\">Suppose that [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges. We show this by using the fact that [latex]{a}_{n}=|{a}_{n}|[\/latex] or [latex]{a}_{n}=\\text{-}|{a}_{n}|[\/latex] and therefore [latex]|{a}_{n}|+{a}_{n}=2|{a}_{n}|[\/latex] or [latex]|{a}_{n}|+{a}_{n}=0[\/latex]. Therefore, [latex]0\\le |{a}_{n}|+{a}_{n}\\le 2|{a}_{n}|[\/latex]. Consequently, by the comparison test, since [latex]2\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, the series<\/p>\n<div id=\"fs-id1169738045626\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left(|{a}_{n}|+{a}_{n}\\right)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738167761\">converges. By using the algebraic properties for convergent series, we conclude that<\/p>\n<div id=\"fs-id1169738167764\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\displaystyle\\sum _{n=1}^{\\infty }\\left(|{a}_{n}|+{a}_{n}\\right)\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737297637\">converges.<\/p>\n<p id=\"fs-id1169737297640\">[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1169738153895\" data-type=\"example\">\n<div id=\"fs-id1169738153897\" data-type=\"exercise\">\n<div id=\"fs-id1169738153899\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Absolute versus Conditional Convergence<\/h3>\n<div id=\"fs-id1169738153899\" data-type=\"problem\">\n<p id=\"fs-id1169738153905\">For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.<\/p>\n<ol id=\"fs-id1169738153909\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{\\left(3n+1\\right)}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos\\left(n\\right)}{{n}^{2}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737232565\" data-type=\"solution\">\n<ol id=\"fs-id1169737232568\" type=\"a\">\n<li>We can see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738128166\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{{\\left(-1\\right)}^{n+1}}{3n+1}|=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{3n+1}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\ndiverges by using the limit comparison test with the harmonic series. In fact,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737162005\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left(3n+1\\right)}}{\\frac{1}{n}}=\\frac{1}{3}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, the series does not converge absolutely. However, since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738239379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{3\\left(n+1\\right)+1}<\\frac{1}{3n+1}\\text{and}\\frac{1}{3n+1}\\to 0[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nthe series converges. We can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{\\left(3n+1\\right)}[\/latex] converges conditionally.<\/li>\n<li>Noting that [latex]|\\cos{n}|\\le 1[\/latex], to determine whether the series converges absolutely, compare<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738223988\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{\\cos{n}}{{n}^{2}}|[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwith the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, by the comparison test, [latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{\\cos{n}}{{n}^{2}}|[\/latex] converges, and therefore [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos{n}}{{n}^{2}}[\/latex] converges absolutely.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737262013\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169737262016\" data-type=\"exercise\">\n<div id=\"fs-id1169737262018\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169737262018\" data-type=\"problem\">\n<p id=\"fs-id1169737262020\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{n}{\\left(2{n}^{3}+1\\right)}[\/latex] converges absolutely, converges conditionally, or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Hint<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737161109\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737161116\">Check for absolute convergence first.<\/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=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737161103\" data-type=\"solution\">\n<p id=\"fs-id1169737161105\">The series converges absolutely.<\/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\/OZ6z5F1c8HA?controls=0&amp;start=347&amp;end=462&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>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\/5.5.3_347to462_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.5.3&#8221; here (opens in new window)<\/a>.<\/p>\n<p id=\"fs-id1169737161122\">To see the difference between absolute and conditional convergence, look at what happens when we <em data-effect=\"italics\">rearrange<\/em> the terms of the alternating harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{n}[\/latex]. We show that we can rearrange the terms so that the new series diverges. Certainly if we rearrange the terms of a finite sum, the sum does not change. When we work with an infinite sum, however, interesting things can happen.<\/p>\n<p id=\"fs-id1169737161174\">Begin by adding enough of the positive terms to produce a sum that is larger than some real number [latex]M>0[\/latex]. For example, let [latex]M=10[\/latex], and find an integer [latex]k[\/latex] such that<\/p>\n<div id=\"fs-id1169738198575\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+\\frac{1}{3}+\\frac{1}{5}+\\cdots +\\frac{1}{2k - 1}>10[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738198624\">(We can do this because the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\left(2n - 1\\right)}[\/latex] diverges to infinity.) Then subtract [latex]\\frac{1}{2}[\/latex]. Then add more positive terms until the sum reaches 100. That is, find another integer [latex]j>k[\/latex] such that<\/p>\n<div id=\"fs-id1169737911250\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+\\frac{1}{3}+\\cdots +\\frac{1}{2k - 1}-\\frac{1}{2}+\\frac{1}{2k+1}+\\cdots +\\frac{1}{2j+1}>100[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737297380\">Then subtract [latex]\\frac{1}{4}[\/latex]. Continuing in this way, we have found a way of rearranging the terms in the alternating harmonic series so that the sequence of partial sums for the rearranged series is unbounded and therefore diverges.<\/p>\n<p id=\"fs-id1169737297396\">The terms in the alternating harmonic series can also be rearranged so that the new series converges to a different value. In the next example, we show how to rearrange the terms to create a new series that converges to [latex]3\\text{ln}\\frac{\\left(2\\right)}{2}[\/latex]. We point out that the alternating harmonic series can be rearranged to create a series that converges to any real number [latex]r[\/latex]; however, the proof of that fact is beyond the scope of this text.<\/p>\n<p id=\"fs-id1169737297434\">In general, any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] that converges absolutely, the value of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.<\/p>\n<div id=\"fs-id1169738079721\" data-type=\"example\">\n<div id=\"fs-id1169738079723\" data-type=\"exercise\">\n<div id=\"fs-id1169738079725\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Rearranging Series<\/h3>\n<div id=\"fs-id1169738079725\" data-type=\"problem\">\n<p id=\"fs-id1169738079730\">Use the fact that<\/p>\n<div id=\"fs-id1169738079734\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\cdots =\\text{ln}2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738079783\" style=\"text-align: left;\">to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is [latex]\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Show Solution<\/span><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737832891\" data-type=\"solution\">\n<p id=\"fs-id1169737832893\">Let<\/p>\n<div id=\"fs-id1169737832896\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\frac{1}{6}+\\frac{1}{7}-\\frac{1}{8}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738211778\">Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\text{ln}\\left(2\\right)[\/latex], by the algebraic properties of convergent series,<\/p>\n<div id=\"fs-id1169738211821\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2}{a}_{n}=\\frac{1}{2}-\\frac{1}{4}+\\frac{1}{6}-\\frac{1}{8}+\\cdots =\\frac{1}{2}\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\frac{\\text{ln}2}{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737933401\">Now introduce the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] such that for all [latex]n\\ge 1[\/latex], [latex]{b}_{2n - 1}=0[\/latex] and [latex]{b}_{2n}=\\frac{{a}_{n}}{2}[\/latex]. Then<\/p>\n<div id=\"fs-id1169738080215\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}=0+\\frac{1}{2}+0-\\frac{1}{4}+0+\\frac{1}{6}+0-\\frac{1}{8}+\\cdots =\\frac{\\text{ln}2}{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737934699\">Then using the algebraic limit properties of convergent series, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converge, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)[\/latex] converges and<\/p>\n<div id=\"fs-id1169737934790\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}=\\text{ln}2+\\frac{\\text{ln}2}{2}=\\frac{3\\text{ln}2}{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737254437\">Now adding the corresponding terms, [latex]{a}_{n}[\/latex] and [latex]{b}_{n}[\/latex], we see that<\/p>\n<div id=\"fs-id1169738228421\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)& =\\left(1+0\\right)+\\left(-\\frac{1}{2}+\\frac{1}{2}\\right)+\\left(\\frac{1}{3}+0\\right)+\\left(-\\frac{1}{4}-\\frac{1}{4}\\right)+\\left(\\frac{1}{5}+0\\right)+\\left(-\\frac{1}{6}+\\frac{1}{6}\\right)\\hfill \\\\ \\\\ & +\\left(\\frac{1}{7}+0\\right)+\\left(\\frac{1}{8}-\\frac{1}{8}\\right)+\\cdots \\hfill \\\\ & =1+\\frac{1}{3}-\\frac{1}{2}+\\frac{1}{5}+\\frac{1}{7}-\\frac{1}{4}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738068068\">We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex], we conclude that<\/p>\n<div id=\"fs-id1169737162164\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]1+\\frac{1}{3}-\\frac{1}{2}+\\frac{1}{5}+\\frac{1}{7}-\\frac{1}{4}+\\cdots =\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737162236\">Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm230805\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=230805&theme=oea&iframe_resize_id=ohm230805&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169737162243\" class=\"key-concepts\" data-depth=\"3\"><\/section>\n<\/section>\n<div data-type=\"glossary\"><\/div>\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-1719\">\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>5.5.3. <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":22,"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\":\"5.5.3\",\"author\":\"Ryan 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