{"id":1768,"date":"2021-07-26T21:33:46","date_gmt":"2021-07-26T21:33:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1768"},"modified":"2022-03-21T23:04:06","modified_gmt":"2022-03-21T23:04:06","slug":"divergence-and-integral-tests","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/divergence-and-integral-tests\/","title":{"raw":"Divergence and Integral Tests","rendered":"Divergence and Integral Tests"},"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 divergence test to determine whether a series converges or diverges<\/li>\r\n \t<li>Use the integral test to determine the convergence of a series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Divergence Test<\/h2>\r\n<p id=\"fs-id1169737806278\">For a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] to converge, the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] must satisfy [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex].<\/p>\r\n<p id=\"fs-id1169737807028\">Therefore, from the algebraic limit properties of sequences,<\/p>\r\n\r\n<div id=\"fs-id1169737836979\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{k\\to \\infty }{\\text{lim}}{a}_{k}=\\underset{k\\to \\infty }{\\text{lim}}\\left({S}_{k}-{S}_{k - 1}\\right)=\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}-\\underset{k\\to \\infty }{\\text{lim}}{S}_{k - 1}=S-S=0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738184603\">Therefore, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges, the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex]. An important consequence of this fact is the following statement:<\/p>\r\n\r\n<div id=\"fs-id1169738046165\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\text{If }{a}_{n}\\nrightarrow 0\\text{ as }n\\to \\infty ,\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}\\text{ diverges}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738059517\">This test is known as the <span data-type=\"term\">divergence test<\/span> because it provides a way of proving that a series diverges.<\/p>\r\n\r\n<div id=\"fs-id1169737741998\" 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: Divergence Test<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169738046564\">If [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=c\\ne 0[\/latex] or [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist, then the series [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-id1169737906569\">It is important to note that the converse of this theorem is not true. That is, if [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex], we cannot make any conclusion about the convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. For example, [latex]\\underset{n\\to 0}{\\text{lim}}\\left(\\frac{1}{n}\\right)=0[\/latex], but the harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if [latex]{a}_{n}\\to 0[\/latex], the divergence test is inconclusive.<\/p>\r\n\r\n<div id=\"fs-id1169738054624\" data-type=\"example\">\r\n<div id=\"fs-id1169737739872\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738057974\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Using the divergence test<\/h3>\r\n<div id=\"fs-id1169738057974\" data-type=\"problem\">\r\n<p id=\"fs-id1169738187459\">For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.<\/p>\r\n\r\n<ol id=\"fs-id1169738212468\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3n - 1}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{\\frac{1}{{n}^{2}}}[\/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-id1169737817290\" data-type=\"solution\">\r\n<ol id=\"fs-id1169737795714\" type=\"a\">\r\n \t<li>Since [latex]\\frac{n}{\\left(3n - 1\\right)}\\to \\frac{1}{3}\\ne 0[\/latex], by the divergence test, we can conclude that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737950574\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3n - 1}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\ndiverges.<\/li>\r\n \t<li>Since [latex]\\frac{1}{{n}^{3}}\\to 0[\/latex], the divergence test is inconclusive.<\/li>\r\n \t<li>Since [latex]{e}^{\\frac{1}{{n}^{2}}}\\to 1\\ne 0[\/latex], by the divergence test, the series<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738227584\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{\\frac{1}{{n}^{2}}}[\/latex]<\/div>\r\ndiverges.<\/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-id1169738189136\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738044696\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737949827\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169737949827\" data-type=\"problem\">\r\n<p id=\"fs-id1169738149942\">What does the divergence test tell us about the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\cos\\left(\\frac{1}{{n}^{2}}\\right)\\text{?}[\/latex]<\/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-id1169738226267\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737946990\">Look at [latex]\\underset{n\\to \\infty }{\\text{lim}}\\cos\\left(\\frac{1}{{n}^{2}}\\right)[\/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-id1169738186762\" data-type=\"solution\">\r\n<p id=\"fs-id1169738093054\">The series diverges.<\/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\/-y848AvISyY?controls=0&amp;start=150&amp;end=186&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\/5.3.1_150to186_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3.1\" here (opens in new window)<\/a>.\r\n\r\n<section id=\"fs-id1169738186085\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Integral Test<\/h2>\r\n<p id=\"fs-id1169738220596\">In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] and showing that [latex]{S}_{{2}^{k}}&gt;1+\\frac{k}{2}[\/latex] for all positive integers [latex]k[\/latex]. In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the <strong>integral test<\/strong>, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.<\/p>\r\n<p id=\"fs-id1169737790154\">To illustrate how the integral test works, use the harmonic series as an example. In Figure 1, we depict the harmonic series by sketching a sequence of rectangles with areas [latex]1,\\frac{1}{2},\\frac{1}{3},\\frac{1}{4},\\ldots [\/latex] along with the function [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]. From the graph, we see that<\/p>\r\n\r\n<div id=\"fs-id1169737820611\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots +\\frac{1}{k}&gt;{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737796301\">Therefore, for each [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169738142452\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}&gt;{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx=\\text{ln}x{|}_{1}^{k+1}=\\text{ln}\\left(k+1\\right)-\\text{ln}\\left(1\\right)=\\text{ln}\\left(k+1\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738183956\">Since [latex]\\underset{k\\to \\infty }{\\text{lim}}\\text{ln}\\left(k+1\\right)=\\infty [\/latex], we see that the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded. Therefore, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] diverges, and, consequently, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] also diverges.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_03_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\/11234342\/CNX_Calc_Figure_09_03_001.jpg\" alt=\"This is a graph in quadrant 1 of a decreasing concave up curve approaching the x-axis \u2013 f(x) = 1\/x. Five rectangles are drawn with base 1 over the interval [1, 6]. The height of each rectangle is determined by the value of the function at the left endpoint of the rectangle\u2019s base. The areas for each are marked: 1, 1\/2, 1\/3, 1\/4, and 1\/5.\" width=\"325\" height=\"201\" data-media-type=\"image\/jpeg\" \/> Figure 1. The sum of the areas of the rectangles is greater than the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is infinite (as calculated by an improper integral), the sum of the areas of the rectangles is also infinite.[\/caption]<\/figure>\r\n<p id=\"fs-id1169738095548\">Now consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. We show how an integral can be used to prove that this series converges. In Figure 2, we sketch a sequence of rectangles with areas [latex]1,\\frac{1}{{2}^{2}},\\frac{1}{{3}^{2}},\\ldots [\/latex] along with the function [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]. From the graph we see that<\/p>\r\n\r\n<div id=\"fs-id1169737815648\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}=1+\\frac{1}{{2}^{2}}+\\frac{1}{{3}^{2}}+\\cdots +\\frac{1}{{k}^{2}}&lt;1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738005573\">Therefore, for each [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169738223359\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}&lt;1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx=1-{\\frac{1}{x}|}_{1}^{k}=1-\\frac{1}{k}+1=2-\\frac{1}{k}&lt;2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738149699\">We conclude that the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded. We also see that [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an increasing sequence:<\/p>\r\n\r\n<div id=\"fs-id1169737777076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={S}_{k - 1}+\\frac{1}{{k}^{2}}\\text{for}k\\ge 2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738187869\">Since [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_03_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"263\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234344\/CNX_Calc_Figure_09_03_004.jpg\" alt=\"This is a graph in quadrant 1 of the decreasing concave up curve f(x) = 1\/(x^2), which approaches the x axis. Rectangles of base 1 are drawn over the interval [0, 5]. The height of each rectangle is determined by the value of the function at the right endpoint of its base. The areas of each are marked: 1, 1\/(2^2), 1\/(3^2), 1\/(4^2) and 1\/(5^2).\" width=\"263\" height=\"278\" data-media-type=\"image\/jpeg\" \/> Figure 2. The sum of the areas of the rectangles is less than the sum of the area of the first rectangle and the area between the curve [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is finite, the sum of the areas of the rectangles is also finite.[\/caption]<\/figure>\r\n<p id=\"fs-id1169737838692\">We can extend this idea to prove convergence or divergence for many different series. Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a series with positive terms [latex]{a}_{n}[\/latex] such that there exists a continuous, positive, decreasing function [latex]f[\/latex] where [latex]f\\left(n\\right)={a}_{n}[\/latex] for all positive integers. Then, as in Figure 3 (a), for any integer [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169738063800\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots +{a}_{k}&lt;{a}_{1}+{\\displaystyle\\int }_{1}^{k}f\\left(x\\right)dx&lt;1+{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738159368\">Therefore, if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] converges, then the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded. Since [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] converges, then the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] also converges. On the other hand, from Figure 3 (b), for any integer [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169738116001\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots +{a}_{k}&gt;{\\displaystyle\\int }_{1}^{k+1}f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737167942\">If [latex]\\underset{k\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{k+1}f\\left(x\\right)dx=\\infty [\/latex], then [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an unbounded sequence and therefore diverges. As a result, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] also diverges. Since [latex]f[\/latex] is a positive function, if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] diverges, then [latex]\\underset{k\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{k+1}f\\left(x\\right)dx=\\infty [\/latex]. We conclude that if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_03_003\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234347\/CNX_Calc_Figure_09_03_002.jpg\" alt=\"This shows two graphs side by side of the same function y = f(x), a decreasing concave up curve approaching the x axis. Rectangles are drawn with base 1 over the intervals [0, 6] and [1, 6]. For the graph on the left, the height of each rectangle is determined by the value of the function at the right endpoint of its base. For the graph on the right, the height of each rectangle is determined by the value of the function at the left endpoint of its base. Areas a_1 through a_6 are marked in the graph on the left, and the same for a_1 to a_5 on the right.\" width=\"731\" height=\"313\" data-media-type=\"image\/jpeg\" \/> Figure 3. (a) If we can inscribe rectangles inside a region bounded by a curve [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] -axis, and the area bounded by those curves for [latex]x\\ge 1[\/latex] is finite, then the sum of the areas of the rectangles is also finite. (b) If a set of rectangles circumscribes the region bounded by [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] axis for [latex]x\\ge 1[\/latex] and the region has infinite area, then the sum of the areas of the rectangles is also infinite.[\/caption]<\/figure>\r\n<div id=\"fs-id1169737847749\" 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: Integral Test<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169738149739\">Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a series with positive terms [latex]{a}_{n}[\/latex]. Suppose there exists a function [latex]f[\/latex] and a positive integer [latex]N[\/latex] such that the following three conditions are satisfied:<\/p>\r\n\r\n<ol id=\"fs-id1169737848192\" type=\"i\">\r\n \t<li>[latex]f[\/latex] is continuous,<\/li>\r\n \t<li>[latex]f[\/latex] is decreasing, and<\/li>\r\n \t<li>[latex]f\\left(n\\right)={a}_{n}[\/latex] for all integers [latex]n\\ge N[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\nThen [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}\\text{ and }{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex]both converge or both diverge (see Figure 3).<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>&nbsp;\r\n<p id=\"fs-id1169737830929\">Although convergence of [latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex] implies convergence of the related series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,<\/p>\r\n\r\n<div id=\"fs-id1169738198560\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{n}=\\frac{1}{e}+{\\left(\\frac{1}{e}\\right)}^{2}+{\\left(\\frac{1}{e}\\right)}^{3}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737998021\">is a geometric series with initial term [latex]a=\\frac{1}{e}[\/latex] and ratio [latex]r=\\frac{1}{e}[\/latex], which converges to<\/p>\r\n\r\n<div id=\"fs-id1169737168794\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{e}}{1-\\left(\\frac{1}{e}\\right)}=\\frac{\\frac{1}{e}}{\\frac{\\left(e - 1\\right)}{e}}=\\frac{1}{e - 1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737297607\">However, the related integral [latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169737232835\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx={\\displaystyle\\int }_{1}^{\\infty }{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}-{e}^{\\text{-}x}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\left[\\text{-}{e}^{\\text{-}b}+{e}^{-1}\\right]=\\frac{1}{e}[\/latex].<\/div>\r\n<div data-type=\"equation\" data-label=\"\"><\/div>\r\nIn the following examples, we explore how to use the integral test.\u00a0 Before doing so, we should note that since one of the conditions of the test is that the function is decreasing, we can use calculus to verify that this condition is met.\r\n<div id=\"fs-id1169737903729\" data-type=\"example\">\r\n<div id=\"fs-id1169737903731\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737264790\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Showing that a function is decreasing<\/h3>\r\nA differentiable function [latex] f(x) [\/latex] is decreasing on an interval\u00a0[latex] \\left( a,b \\right) [\/latex] if [latex] f'(x) &lt; 0 [\/latex] for all [latex] x\\in\\left( a,b \\right) [\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Integral Test<\/h3>\r\n<div id=\"fs-id1169737264790\" data-type=\"problem\">\r\n<p id=\"fs-id1169737264795\">For each of the following series, use the integral test to determine whether the series converges or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169738048921\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1169738115791\" data-type=\"solution\">\r\n<ol id=\"fs-id1169738115793\" type=\"a\">\r\n \t<li>Compare\r\n<div id=\"fs-id1169738219658\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}\\text{and}{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nWe have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738155194\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{{x}^{3}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\left[-\\frac{1}{2{x}^{2}}{|}_{1}^{b}\\right]=\\underset{b\\to \\infty }{\\text{lim}}\\left[-\\frac{1}{2{b}^{2}}+\\frac{1}{2}\\right]=\\frac{1}{2}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThus the integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex] converges, and therefore so does the series<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737966896\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex].<\/div>\r\n&nbsp;\r\n<ol>\r\n \t<li>Compare<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ol>\r\n&nbsp;\r\n<div id=\"fs-id1169737981493\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}\\text{ and }{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738125948\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx&amp; =\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{\\sqrt{2x - 1}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\sqrt{2x - 1}{|}_{1}^{b}\\hfill \\\\ &amp; =\\underset{b\\to \\infty }{\\text{lim}}\\left[\\sqrt{2b - 1}-1\\right]=\\infty ,\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nthe integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx[\/latex] diverges, and therefore<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737169332\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\ndiverges.\r\n\r\n[\/hidden-answer]\r\n\r\n<span data-type=\"newline\">\u00a0<\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1169737284928\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169737284932\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738164454\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169738164454\" data-type=\"problem\">\r\n<p id=\"fs-id1169738164456\">Use the integral test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3{n}^{2}+1}[\/latex] converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1169738249441\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169738250182\">Compare to the integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{x}{3{x}^{2}+1}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169738068120\" data-type=\"solution\">\r\n<p id=\"fs-id1169738068122\">The series diverges.<\/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\/_zV7bzn3jUE?controls=0&amp;start=235&amp;end=342&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\/5.3.2_235to342_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3.2\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]20296[\/ohm_question]\r\n\r\n<\/div>\r\n<section id=\"fs-id1169737363877\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the divergence test to determine whether a series converges or diverges<\/li>\n<li>Use the integral test to determine the convergence of a series<\/li>\n<\/ul>\n<\/div>\n<h2>Divergence Test<\/h2>\n<p id=\"fs-id1169737806278\">For a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] to converge, the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}[\/latex] must satisfy [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex].<\/p>\n<p id=\"fs-id1169737807028\">Therefore, from the algebraic limit properties of sequences,<\/p>\n<div id=\"fs-id1169737836979\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{k\\to \\infty }{\\text{lim}}{a}_{k}=\\underset{k\\to \\infty }{\\text{lim}}\\left({S}_{k}-{S}_{k - 1}\\right)=\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}-\\underset{k\\to \\infty }{\\text{lim}}{S}_{k - 1}=S-S=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738184603\">Therefore, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges, the [latex]n\\text{th}[\/latex] term [latex]{a}_{n}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex]. An important consequence of this fact is the following statement:<\/p>\n<div id=\"fs-id1169738046165\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\text{If }{a}_{n}\\nrightarrow 0\\text{ as }n\\to \\infty ,\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}\\text{ diverges}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738059517\">This test is known as the <span data-type=\"term\">divergence test<\/span> because it provides a way of proving that a series diverges.<\/p>\n<div id=\"fs-id1169737741998\" 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: Divergence Test<\/h3>\n<hr \/>\n<p id=\"fs-id1169738046564\">If [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=c\\ne 0[\/latex] or [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex] does not exist, then the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169737906569\">It is important to note that the converse of this theorem is not true. That is, if [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex], we cannot make any conclusion about the convergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. For example, [latex]\\underset{n\\to 0}{\\text{lim}}\\left(\\frac{1}{n}\\right)=0[\/latex], but the harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if [latex]{a}_{n}\\to 0[\/latex], the divergence test is inconclusive.<\/p>\n<div id=\"fs-id1169738054624\" data-type=\"example\">\n<div id=\"fs-id1169737739872\" data-type=\"exercise\">\n<div id=\"fs-id1169738057974\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Using the divergence test<\/h3>\n<div id=\"fs-id1169738057974\" data-type=\"problem\">\n<p id=\"fs-id1169738187459\">For each of the following series, apply the divergence test. If the divergence test proves that the series diverges, state so. Otherwise, indicate that the divergence test is inconclusive.<\/p>\n<ol id=\"fs-id1169738212468\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3n - 1}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{\\frac{1}{{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=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737817290\" data-type=\"solution\">\n<ol id=\"fs-id1169737795714\" type=\"a\">\n<li>Since [latex]\\frac{n}{\\left(3n - 1\\right)}\\to \\frac{1}{3}\\ne 0[\/latex], by the divergence test, we can conclude that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737950574\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3n - 1}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\ndiverges.<\/li>\n<li>Since [latex]\\frac{1}{{n}^{3}}\\to 0[\/latex], the divergence test is inconclusive.<\/li>\n<li>Since [latex]{e}^{\\frac{1}{{n}^{2}}}\\to 1\\ne 0[\/latex], by the divergence test, the series<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738227584\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{\\frac{1}{{n}^{2}}}[\/latex]<\/div>\n<p>diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738189136\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738044696\" data-type=\"exercise\">\n<div id=\"fs-id1169737949827\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169737949827\" data-type=\"problem\">\n<p id=\"fs-id1169738149942\">What does the divergence test tell us about the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\cos\\left(\\frac{1}{{n}^{2}}\\right)\\text{?}[\/latex]<\/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-id1169738226267\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737946990\">Look at [latex]\\underset{n\\to \\infty }{\\text{lim}}\\cos\\left(\\frac{1}{{n}^{2}}\\right)[\/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-id1169738186762\" data-type=\"solution\">\n<p id=\"fs-id1169738093054\">The series diverges.<\/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\/-y848AvISyY?controls=0&amp;start=150&amp;end=186&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\/5.3.1_150to186_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3.1&#8221; here (opens in new window)<\/a>.<\/p>\n<section id=\"fs-id1169738186085\" data-depth=\"1\">\n<h2 data-type=\"title\">Integral Test<\/h2>\n<p id=\"fs-id1169738220596\">In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] and showing that [latex]{S}_{{2}^{k}}>1+\\frac{k}{2}[\/latex] for all positive integers [latex]k[\/latex]. In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the <strong>integral test<\/strong>, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.<\/p>\n<p id=\"fs-id1169737790154\">To illustrate how the integral test works, use the harmonic series as an example. In Figure 1, we depict the harmonic series by sketching a sequence of rectangles with areas [latex]1,\\frac{1}{2},\\frac{1}{3},\\frac{1}{4},\\ldots[\/latex] along with the function [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]. From the graph, we see that<\/p>\n<div id=\"fs-id1169737820611\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots +\\frac{1}{k}>{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737796301\">Therefore, for each [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169738142452\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}>{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx=\\text{ln}x{|}_{1}^{k+1}=\\text{ln}\\left(k+1\\right)-\\text{ln}\\left(1\\right)=\\text{ln}\\left(k+1\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738183956\">Since [latex]\\underset{k\\to \\infty }{\\text{lim}}\\text{ln}\\left(k+1\\right)=\\infty[\/latex], we see that the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded. Therefore, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] diverges, and, consequently, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] also diverges.<\/p>\n<figure id=\"CNX_Calc_Figure_09_03_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\/11234342\/CNX_Calc_Figure_09_03_001.jpg\" alt=\"This is a graph in quadrant 1 of a decreasing concave up curve approaching the x-axis \u2013 f(x) = 1\/x. Five rectangles are drawn with base 1 over the interval [1, 6]. The height of each rectangle is determined by the value of the function at the left endpoint of the rectangle\u2019s base. The areas for each are marked: 1, 1\/2, 1\/3, 1\/4, and 1\/5.\" width=\"325\" height=\"201\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The sum of the areas of the rectangles is greater than the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is infinite (as calculated by an improper integral), the sum of the areas of the rectangles is also infinite.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1169738095548\">Now consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. We show how an integral can be used to prove that this series converges. In Figure 2, we sketch a sequence of rectangles with areas [latex]1,\\frac{1}{{2}^{2}},\\frac{1}{{3}^{2}},\\ldots[\/latex] along with the function [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex]. From the graph we see that<\/p>\n<div id=\"fs-id1169737815648\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}=1+\\frac{1}{{2}^{2}}+\\frac{1}{{3}^{2}}+\\cdots +\\frac{1}{{k}^{2}}<1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738005573\">Therefore, for each [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169738223359\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}<1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx=1-{\\frac{1}{x}|}_{1}^{k}=1-\\frac{1}{k}+1=2-\\frac{1}{k}<2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738149699\">We conclude that the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded. We also see that [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an increasing sequence:<\/p>\n<div id=\"fs-id1169737777076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={S}_{k - 1}+\\frac{1}{{k}^{2}}\\text{for}k\\ge 2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738187869\">Since [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges.<\/p>\n<figure id=\"CNX_Calc_Figure_09_03_002\"><figcaption><\/figcaption><div style=\"width: 273px\" 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\/11234344\/CNX_Calc_Figure_09_03_004.jpg\" alt=\"This is a graph in quadrant 1 of the decreasing concave up curve f(x) = 1\/(x^2), which approaches the x axis. Rectangles of base 1 are drawn over the interval [0, 5]. The height of each rectangle is determined by the value of the function at the right endpoint of its base. The areas of each are marked: 1, 1\/(2^2), 1\/(3^2), 1\/(4^2) and 1\/(5^2).\" width=\"263\" height=\"278\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The sum of the areas of the rectangles is less than the sum of the area of the first rectangle and the area between the curve [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is finite, the sum of the areas of the rectangles is also finite.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1169737838692\">We can extend this idea to prove convergence or divergence for many different series. Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a series with positive terms [latex]{a}_{n}[\/latex] such that there exists a continuous, positive, decreasing function [latex]f[\/latex] where [latex]f\\left(n\\right)={a}_{n}[\/latex] for all positive integers. Then, as in Figure 3 (a), for any integer [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169738063800\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots +{a}_{k}<{a}_{1}+{\\displaystyle\\int }_{1}^{k}f\\left(x\\right)dx<1+{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738159368\">Therefore, if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] converges, then the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded. Since [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] converges, then the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] also converges. On the other hand, from Figure 3 (b), for any integer [latex]k[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169738116001\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots +{a}_{k}>{\\displaystyle\\int }_{1}^{k+1}f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737167942\">If [latex]\\underset{k\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{k+1}f\\left(x\\right)dx=\\infty[\/latex], then [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an unbounded sequence and therefore diverges. As a result, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] also diverges. Since [latex]f[\/latex] is a positive function, if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] diverges, then [latex]\\underset{k\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{k+1}f\\left(x\\right)dx=\\infty[\/latex]. We conclude that if [latex]{\\displaystyle\\int }_{1}^{\\infty }f\\left(x\\right)dx[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/p>\n<figure id=\"CNX_Calc_Figure_09_03_003\"><figcaption><\/figcaption><div style=\"width: 741px\" 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\/11234347\/CNX_Calc_Figure_09_03_002.jpg\" alt=\"This shows two graphs side by side of the same function y = f(x), a decreasing concave up curve approaching the x axis. Rectangles are drawn with base 1 over the intervals [0, 6] and [1, 6]. For the graph on the left, the height of each rectangle is determined by the value of the function at the right endpoint of its base. For the graph on the right, the height of each rectangle is determined by the value of the function at the left endpoint of its base. Areas a_1 through a_6 are marked in the graph on the left, and the same for a_1 to a_5 on the right.\" width=\"731\" height=\"313\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. (a) If we can inscribe rectangles inside a region bounded by a curve [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] -axis, and the area bounded by those curves for [latex]x\\ge 1[\/latex] is finite, then the sum of the areas of the rectangles is also finite. (b) If a set of rectangles circumscribes the region bounded by [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] axis for [latex]x\\ge 1[\/latex] and the region has infinite area, then the sum of the areas of the rectangles is also infinite.<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1169737847749\" 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: Integral Test<\/h3>\n<hr \/>\n<p id=\"fs-id1169738149739\">Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a series with positive terms [latex]{a}_{n}[\/latex]. Suppose there exists a function [latex]f[\/latex] and a positive integer [latex]N[\/latex] such that the following three conditions are satisfied:<\/p>\n<ol id=\"fs-id1169737848192\" type=\"i\">\n<li>[latex]f[\/latex] is continuous,<\/li>\n<li>[latex]f[\/latex] is decreasing, and<\/li>\n<li>[latex]f\\left(n\\right)={a}_{n}[\/latex] for all integers [latex]n\\ge N[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThen [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}\\text{ and }{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex]both converge or both diverge (see Figure 3).<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737830929\">Although convergence of [latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex] implies convergence of the related series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,<\/p>\n<div id=\"fs-id1169738198560\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{n}=\\frac{1}{e}+{\\left(\\frac{1}{e}\\right)}^{2}+{\\left(\\frac{1}{e}\\right)}^{3}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737998021\">is a geometric series with initial term [latex]a=\\frac{1}{e}[\/latex] and ratio [latex]r=\\frac{1}{e}[\/latex], which converges to<\/p>\n<div id=\"fs-id1169737168794\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{e}}{1-\\left(\\frac{1}{e}\\right)}=\\frac{\\frac{1}{e}}{\\frac{\\left(e - 1\\right)}{e}}=\\frac{1}{e - 1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737297607\">However, the related integral [latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx[\/latex] satisfies<\/p>\n<div id=\"fs-id1169737232835\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx={\\displaystyle\\int }_{1}^{\\infty }{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}-{e}^{\\text{-}x}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\left[\\text{-}{e}^{\\text{-}b}+{e}^{-1}\\right]=\\frac{1}{e}[\/latex].<\/div>\n<div data-type=\"equation\" data-label=\"\"><\/div>\n<p>In the following examples, we explore how to use the integral test.\u00a0 Before doing so, we should note that since one of the conditions of the test is that the function is decreasing, we can use calculus to verify that this condition is met.<\/p>\n<div id=\"fs-id1169737903729\" data-type=\"example\">\n<div id=\"fs-id1169737903731\" data-type=\"exercise\">\n<div id=\"fs-id1169737264790\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3>Recall: Showing that a function is decreasing<\/h3>\n<p>A differentiable function [latex]f(x)[\/latex] is decreasing on an interval\u00a0[latex]\\left( a,b \\right)[\/latex] if [latex]f'(x) < 0[\/latex] for all [latex]x\\in\\left( a,b \\right)[\/latex].\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Using the Integral Test<\/h3>\n<div id=\"fs-id1169737264790\" data-type=\"problem\">\n<p id=\"fs-id1169737264795\">For each of the following series, use the integral test to determine whether the series converges or diverges.<\/p>\n<ol id=\"fs-id1169738048921\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738115791\" data-type=\"solution\">\n<ol id=\"fs-id1169738115793\" type=\"a\">\n<li>Compare\n<div id=\"fs-id1169738219658\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}\\text{and}{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nWe have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738155194\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{{x}^{3}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\left[-\\frac{1}{2{x}^{2}}{|}_{1}^{b}\\right]=\\underset{b\\to \\infty }{\\text{lim}}\\left[-\\frac{1}{2{b}^{2}}+\\frac{1}{2}\\right]=\\frac{1}{2}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThus the integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex] converges, and therefore so does the series<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737966896\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<ol>\n<li>Compare<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169737981493\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}\\text{ and }{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738125948\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx& =\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{\\sqrt{2x - 1}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\sqrt{2x - 1}{|}_{1}^{b}\\hfill \\\\ & =\\underset{b\\to \\infty }{\\text{lim}}\\left[\\sqrt{2b - 1}-1\\right]=\\infty ,\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nthe integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx[\/latex] diverges, and therefore<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737169332\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\ndiverges.<\/p>\n<\/div>\n<\/div>\n<p><span data-type=\"newline\">\u00a0<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169737284928\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169737284932\" data-type=\"exercise\">\n<div id=\"fs-id1169738164454\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169738164454\" data-type=\"problem\">\n<p id=\"fs-id1169738164456\">Use the integral test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3{n}^{2}+1}[\/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=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738249441\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169738250182\">Compare to the integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{x}{3{x}^{2}+1}dx[\/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=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738068120\" data-type=\"solution\">\n<p id=\"fs-id1169738068122\">The series diverges.<\/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\/_zV7bzn3jUE?controls=0&amp;start=235&amp;end=342&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\/5.3.2_235to342_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3.2&#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=\"ohm20296\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20296&theme=oea&iframe_resize_id=ohm20296&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<section id=\"fs-id1169737363877\" 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-1768\">\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>5.3.1. <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><li>5.3.2. <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":13,"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\":\"5.3.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"5.3.2\",\"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-1768","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1768","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":9,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1768\/revisions"}],"predecessor-version":[{"id":2667,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1768\/revisions\/2667"}],"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\/1768\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1768"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1768"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1768"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1768"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}