{"id":1713,"date":"2021-07-23T20:24:03","date_gmt":"2021-07-23T20:24:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1713"},"modified":"2022-03-21T23:04:58","modified_gmt":"2022-03-21T23:04:58","slug":"comparison-test","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/comparison-test\/","title":{"raw":"Comparison Test","rendered":"Comparison Test"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the comparison test to test a series for convergence<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169739208245\"><span style=\"font-size: 1rem; text-align: initial;\">In the preceding two sections, we discussed two large classes of series: geometric series and <\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; text-align: initial;\">-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the <\/span><strong style=\"font-size: 1rem; text-align: initial;\">comparison test<\/strong><span style=\"font-size: 1rem; text-align: initial;\">.\u00a0 We begin by recalling a principle for comparing variable quantities in the denominator of a fraction.\r\n<\/span><\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Algebra of inequalities<\/h3>\r\nIf [latex] 0 &lt; a_n \\le b_n [\/latex], then [latex] \\frac{1}{a_n} \\ge \\frac{1}{b_n} [\/latex]\r\n\r\nIn other words, a larger denominator corresponds to a smaller fraction.\r\n\r\n<\/div>\r\n<section id=\"fs-id1169739336061\" data-depth=\"1\">\r\n<p id=\"fs-id1169739301994\">For example, consider the series<\/p>\r\n\r\n<div id=\"fs-id1169738985514\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739270411\">This series looks similar to the convergent series<\/p>\r\n\r\n<div id=\"fs-id1169739305240\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736654567\">Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing.<\/p>\r\n&nbsp;\r\n\r\nFurthermore, since\r\n<div id=\"fs-id1169739227434\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0&lt;\\frac{1}{{n}^{2}+1}&lt;\\frac{1}{{n}^{2}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739305297\">for all positive integers [latex]n[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}+1}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169736594931\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}+1}&lt;\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}&lt;\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739189474\">(See Figure 1 (a) and the table below.) Since the series on the right converges, the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded above. We conclude that [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges, and thus<\/p>\r\n\r\n<div id=\"fs-id1169739003848\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}+1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739029866\">converges.<\/p>\r\n<p id=\"fs-id1169739264296\">Similarly, consider the series<\/p>\r\n\r\n<div id=\"fs-id1169739015648\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n - \\frac{1}{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739242738\">This series looks similar to the divergent series<\/p>\r\n\r\n<div id=\"fs-id1169739225137\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739301476\">The sequence of partial sums for each series is monotone increasing and<\/p>\r\n\r\n<div id=\"fs-id1169739182905\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{n - \\frac{1}{2}}&gt;\\frac{1}{n}&gt;0[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739005914\">for every positive integer [latex]n[\/latex]. Therefore, the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n - \\frac{1}{2}}[\/latex] satisfies<\/p>\r\n\r\n<div id=\"fs-id1169738961628\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n - \\frac{1}{2}}&gt;\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739004447\">(See Figure 1 (b) and the following table.) Since the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges to infinity, the sequence of partial sums [latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}[\/latex] is unbounded. Consequently, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an unbounded sequence, and therefore diverges. We conclude that<\/p>\r\n\r\n<div id=\"fs-id1169738999447\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n - \\frac{1}{2}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739183589\">diverges.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_04_001\"><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\/11234356\/CNX_Calc_Figure_09_04_001.jpg\" alt=\"This shows two graphs side by side. The first shows plotted points for the partial sums for the sum of 1\/n^2 and the sum 1\/(n^2 + 1). Each of the partial sums for the latter is less than the corresponding partial sum for the former. The second shows plotted points for the partial sums for the sum of 1\/(n - 0.5) and the sum 1\/n. Each of the partial sums for the latter is less than the corresponding partial sum for the former.\" width=\"731\" height=\"363\" data-media-type=\"image\/jpeg\" \/> Figure 1. (a) Each of the partial sums for the given series is less than the corresponding partial sum for the converging [latex]p-\\text{series}[\/latex]. (b) Each of the partial sums for the given series is greater than the corresponding partial sum for the diverging harmonic series.[\/caption]<\/figure>\r\n<table id=\"fs-id1169738963340\" summary=\"This is a table with three rows and nine columns. The first row contains the label \"><caption><span data-type=\"title\">Comparing a series with a <em data-effect=\"italics\">p<\/em>-series (<em data-effect=\"italics\">p<\/em> = 2)<\/span><\/caption>\r\n<thead>\r\n<tr valign=\"top\">\r\n<th data-align=\"left\">[latex]k[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]1[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]2[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]3[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]4[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]5[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]6[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]7[\/latex]<\/th>\r\n<th data-align=\"left\">[latex]8[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}+1}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.5[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.7[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.8[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.8588[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.8973[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.9243[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.9443[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]0.9597[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]1[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]\\begin{array}{l}1.25\\hfill \\end{array}[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]1.3611[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]1.4236[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]1.4636[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]1.4914[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]1.5118[\/latex]<\/td>\r\n<td data-align=\"left\">[latex]1.5274[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<table id=\"fs-id1169739044243\" style=\"height: 98px;\" summary=\"This shows a table with three rows and nine columns. The first row contains the label \"><caption><span data-type=\"title\">Comparing a series with the harmonic series<\/span><\/caption>\r\n<thead>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<th style=\"height: 14px; width: 131.609px;\" data-align=\"left\">[latex]k[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 86.0781px;\" data-align=\"left\">[latex]1[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]2[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]3[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]4[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]5[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]6[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]7[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]8[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 42px;\" valign=\"top\">\r\n<td style=\"height: 42px; width: 131.609px;\" data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n - \\frac{1}{2}}[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 86.0781px;\" data-align=\"left\">[latex]2[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.6667[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.0667[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.3524[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.5746[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.7564[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.9103[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]4.0436[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 42px;\" valign=\"top\">\r\n<td style=\"height: 42px; width: 131.609px;\" data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 86.0781px;\" data-align=\"left\">[latex]1[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]1.5[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]\\begin{array}{l}1.8333\\hfill \\end{array}[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.0933[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.2833[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.45[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.5929[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.7179[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1169738915775\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 data-type=\"title\">Theorem: Comparison Test<\/h3>\r\n<ol id=\"fs-id1169738951381\" type=\"i\">\r\n \t<li>Suppose there exists an integer [latex]N[\/latex] such that [latex]0\\le {a}_{n}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex]. If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/li>\r\n \t<li>Suppose there exists an integer [latex]N[\/latex] such that [latex]{a}_{n}\\ge {b}_{n}\\ge 0[\/latex] for all [latex]n\\ge N[\/latex]. If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1169739035335\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\r\n<p id=\"fs-id1169739273903\">We prove part i. The proof of part ii. is the contrapositive of part i. Let [latex]\\left\\{{S}_{k}\\right\\}[\/latex] be the sequence of partial sums associated with [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], and let [latex]L=\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. Since the terms [latex]{a}_{n}\\ge 0[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169739189340\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={a}_{1}+{a}_{2}+\\cdots +{a}_{k}\\le {a}_{1}+{a}_{2}+\\cdots +{a}_{k}+{a}_{k+1}={S}_{k+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739064845\">Therefore, the sequence of partial sums is increasing. Further, since [latex]{a}_{n}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex], then<\/p>\r\n\r\n<div id=\"fs-id1169739300032\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=N}^{k}{a}_{n}\\le \\displaystyle\\sum _{n=N}^{k}{b}_{n}\\le \\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}=L[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739013220\">Therefore, for all [latex]k\\ge 1[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169739222211\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\left({a}_{1}+{a}_{2}+\\cdots +{a}_{N - 1}\\right)+\\displaystyle\\sum _{n=N}^{k}{a}_{n}\\le \\left({a}_{1}+{a}_{2}+\\cdots +{a}_{N - 1}\\right)+L[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738925226\">Since [latex]{a}_{1}+{a}_{2}+\\cdots +{a}_{N - 1}[\/latex] is a finite number, we conclude that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded above. Therefore, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges, and therefore the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\r\n<p id=\"fs-id1169739017799\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1169738938243\">To use the comparison test to determine the convergence or divergence of a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and <em data-effect=\"italics\">p<\/em>-series, these series are often used. If there exists an integer [latex]N[\/latex] such that for all [latex]n\\ge N[\/latex], each term [latex]{a}_{n}[\/latex] is less than each corresponding term of a known convergent series, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges. Similarly, if there exists an integer [latex]N[\/latex] such that for all [latex]n\\ge N[\/latex], each term [latex]{a}_{n}[\/latex] is greater than each corresponding term of a known divergent series, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/p>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169383[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738880258\" data-type=\"example\">\r\n<div id=\"fs-id1169738905678\" data-type=\"exercise\">\r\n<div id=\"fs-id1169739020328\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Using the Comparison Test<\/h3>\r\n<div id=\"fs-id1169739020328\" data-type=\"problem\">\r\n<p id=\"fs-id1169738994018\">For each of the following series, use the comparison test to determine whether the series converges or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169736613871\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}+3n+1}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{2}^{n}+1}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\text{ln}\\left(n\\right)}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1169738895467\" data-type=\"solution\">\r\n<ol id=\"fs-id1169738938121\" type=\"a\">\r\n \t<li>Compare to [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] is a <em data-effect=\"italics\">p<\/em>-series with [latex]p=3[\/latex], it converges. Further,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739298200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{n}^{3}+3n+1}&lt;\\frac{1}{{n}^{3}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor every positive integer [latex]n[\/latex]. Therefore, we can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}+3n+1}[\/latex] converges.<\/li>\r\n \t<li>Compare to [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n}[\/latex]. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n}[\/latex] is a geometric series with [latex]r=\\frac{1}{2}[\/latex] and [latex]|\\frac{1}{2}|&lt;1[\/latex], it converges. Also,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739039859\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{2}^{n}+1}&lt;\\frac{1}{{2}^{n}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor every positive integer [latex]n[\/latex]. Therefore, we see that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{2}^{n}+1}[\/latex] converges.<\/li>\r\n \t<li>Compare to [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n}[\/latex]. Since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739029860\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\text{ln}\\left(n\\right)}&gt;\\frac{1}{n}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nfor every integer [latex]n\\ge 2[\/latex] and [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n}[\/latex] diverges, we have that [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\text{ln}\\left(n\\right)}[\/latex] diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739333977\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169739096544\" data-type=\"exercise\">\r\n<div id=\"fs-id1169739269621\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169739269621\" data-type=\"problem\">\r\n<p id=\"fs-id1169739001213\">Use the comparison test to determine if the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{n}^{3}+n+1}[\/latex] converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1169738972533\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169736597699\">Find a value [latex]p[\/latex] such that [latex]\\frac{n}{{n}^{3}+n+1}\\le \\frac{1}{{n}^{p}}[\/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-id1169739034395\" data-type=\"solution\">\r\n<p id=\"fs-id1169739034397\">The series converges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/izlPnqyz9Y4?controls=0&amp;start=285&amp;end=375&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\/5.4.1_285to375_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.4.1\" here (opens in new window)<\/a>.\r\n\r\n<\/section><\/section><section id=\"fs-id1169736615268\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the comparison test to test a series for convergence<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169739208245\"><span style=\"font-size: 1rem; text-align: initial;\">In the preceding two sections, we discussed two large classes of series: geometric series and <\/span><em style=\"font-size: 1rem; text-align: initial;\" data-effect=\"italics\">p<\/em><span style=\"font-size: 1rem; text-align: initial;\">-series. We know exactly when these series converge and when they diverge. Here we show how to use the convergence or divergence of these series to prove convergence or divergence for other series, using a method called the <\/span><strong style=\"font-size: 1rem; text-align: initial;\">comparison test<\/strong><span style=\"font-size: 1rem; text-align: initial;\">.\u00a0 We begin by recalling a principle for comparing variable quantities in the denominator of a fraction.<br \/>\n<\/span><\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Algebra of inequalities<\/h3>\n<p>If [latex]0 < a_n \\le b_n[\/latex], then [latex]\\frac{1}{a_n} \\ge \\frac{1}{b_n}[\/latex]\n\nIn other words, a larger denominator corresponds to a smaller fraction.\n\n<\/div>\n<section id=\"fs-id1169739336061\" data-depth=\"1\">\n<p id=\"fs-id1169739301994\">For example, consider the series<\/p>\n<div id=\"fs-id1169738985514\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739270411\">This series looks similar to the convergent series<\/p>\n<div id=\"fs-id1169739305240\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736654567\">Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing.<\/p>\n<p>&nbsp;<\/p>\n<p>Furthermore, since<\/p>\n<div id=\"fs-id1169739227434\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0<\\frac{1}{{n}^{2}+1}<\\frac{1}{{n}^{2}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739305297\">for all positive integers [latex]n[\/latex], the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}+1}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169736594931\" 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\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}<\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739189474\">(See Figure 1 (a) and the table below.) Since the series on the right converges, the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded above. We conclude that [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is a monotone increasing sequence that is bounded above. Therefore, by the Monotone Convergence Theorem, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges, and thus<\/p>\n<div id=\"fs-id1169739003848\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}+1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739029866\">converges.<\/p>\n<p id=\"fs-id1169739264296\">Similarly, consider the series<\/p>\n<div id=\"fs-id1169739015648\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n - \\frac{1}{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739242738\">This series looks similar to the divergent series<\/p>\n<div id=\"fs-id1169739225137\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739301476\">The sequence of partial sums for each series is monotone increasing and<\/p>\n<div id=\"fs-id1169739182905\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{n - \\frac{1}{2}}>\\frac{1}{n}>0[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739005914\">for every positive integer [latex]n[\/latex]. Therefore, the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}[\/latex] of [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n - \\frac{1}{2}}[\/latex] satisfies<\/p>\n<div id=\"fs-id1169738961628\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n - \\frac{1}{2}}>\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739004447\">(See Figure 1 (b) and the following table.) Since the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges to infinity, the sequence of partial sums [latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}[\/latex] is unbounded. Consequently, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an unbounded sequence, and therefore diverges. We conclude that<\/p>\n<div id=\"fs-id1169738999447\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n - \\frac{1}{2}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739183589\">diverges.<\/p>\n<figure id=\"CNX_Calc_Figure_09_04_001\"><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\/11234356\/CNX_Calc_Figure_09_04_001.jpg\" alt=\"This shows two graphs side by side. The first shows plotted points for the partial sums for the sum of 1\/n^2 and the sum 1\/(n^2 + 1). Each of the partial sums for the latter is less than the corresponding partial sum for the former. The second shows plotted points for the partial sums for the sum of 1\/(n - 0.5) and the sum 1\/n. Each of the partial sums for the latter is less than the corresponding partial sum for the former.\" width=\"731\" height=\"363\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. (a) Each of the partial sums for the given series is less than the corresponding partial sum for the converging [latex]p-\\text{series}[\/latex]. (b) Each of the partial sums for the given series is greater than the corresponding partial sum for the diverging harmonic series.<\/p>\n<\/div>\n<\/figure>\n<table id=\"fs-id1169738963340\" summary=\"This is a table with three rows and nine columns. The first row contains the label\">\n<caption><span data-type=\"title\">Comparing a series with a <em data-effect=\"italics\">p<\/em>-series (<em data-effect=\"italics\">p<\/em> = 2)<\/span><\/caption>\n<thead>\n<tr valign=\"top\">\n<th data-align=\"left\">[latex]k[\/latex]<\/th>\n<th data-align=\"left\">[latex]1[\/latex]<\/th>\n<th data-align=\"left\">[latex]2[\/latex]<\/th>\n<th data-align=\"left\">[latex]3[\/latex]<\/th>\n<th data-align=\"left\">[latex]4[\/latex]<\/th>\n<th data-align=\"left\">[latex]5[\/latex]<\/th>\n<th data-align=\"left\">[latex]6[\/latex]<\/th>\n<th data-align=\"left\">[latex]7[\/latex]<\/th>\n<th data-align=\"left\">[latex]8[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}+1}[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.5[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.7[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.8[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.8588[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.8973[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.9243[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.9443[\/latex]<\/td>\n<td data-align=\"left\">[latex]0.9597[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}[\/latex]<\/td>\n<td data-align=\"left\">[latex]1[\/latex]<\/td>\n<td data-align=\"left\">[latex]\\begin{array}{l}1.25\\hfill \\end{array}[\/latex]<\/td>\n<td data-align=\"left\">[latex]1.3611[\/latex]<\/td>\n<td data-align=\"left\">[latex]1.4236[\/latex]<\/td>\n<td data-align=\"left\">[latex]1.4636[\/latex]<\/td>\n<td data-align=\"left\">[latex]1.4914[\/latex]<\/td>\n<td data-align=\"left\">[latex]1.5118[\/latex]<\/td>\n<td data-align=\"left\">[latex]1.5274[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<table id=\"fs-id1169739044243\" style=\"height: 98px;\" summary=\"This shows a table with three rows and nine columns. The first row contains the label\">\n<caption><span data-type=\"title\">Comparing a series with the harmonic series<\/span><\/caption>\n<thead>\n<tr style=\"height: 14px;\" valign=\"top\">\n<th style=\"height: 14px; width: 131.609px;\" data-align=\"left\">[latex]k[\/latex]<\/th>\n<th style=\"height: 14px; width: 86.0781px;\" data-align=\"left\">[latex]1[\/latex]<\/th>\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]2[\/latex]<\/th>\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]3[\/latex]<\/th>\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]4[\/latex]<\/th>\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]5[\/latex]<\/th>\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]6[\/latex]<\/th>\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]7[\/latex]<\/th>\n<th style=\"height: 14px; width: 111.016px;\" data-align=\"left\">[latex]8[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 42px;\" valign=\"top\">\n<td style=\"height: 42px; width: 131.609px;\" data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n - \\frac{1}{2}}[\/latex]<\/td>\n<td style=\"height: 42px; width: 86.0781px;\" data-align=\"left\">[latex]2[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.6667[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.0667[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.3524[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.5746[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.7564[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]3.9103[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]4.0436[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 42px;\" valign=\"top\">\n<td style=\"height: 42px; width: 131.609px;\" data-align=\"left\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}[\/latex]<\/td>\n<td style=\"height: 42px; width: 86.0781px;\" data-align=\"left\">[latex]1[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]1.5[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]\\begin{array}{l}1.8333\\hfill \\end{array}[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.0933[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.2833[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.45[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.5929[\/latex]<\/td>\n<td style=\"height: 42px; width: 111.016px;\" data-align=\"left\">[latex]2.7179[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1169738915775\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 data-type=\"title\">Theorem: Comparison Test<\/h3>\n<ol id=\"fs-id1169738951381\" type=\"i\">\n<li>Suppose there exists an integer [latex]N[\/latex] such that [latex]0\\le {a}_{n}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex]. If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/li>\n<li>Suppose there exists an integer [latex]N[\/latex] such that [latex]{a}_{n}\\ge {b}_{n}\\ge 0[\/latex] for all [latex]n\\ge N[\/latex]. If [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169739035335\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\n<p id=\"fs-id1169739273903\">We prove part i. The proof of part ii. is the contrapositive of part i. Let [latex]\\left\\{{S}_{k}\\right\\}[\/latex] be the sequence of partial sums associated with [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], and let [latex]L=\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. Since the terms [latex]{a}_{n}\\ge 0[\/latex],<\/p>\n<div id=\"fs-id1169739189340\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={a}_{1}+{a}_{2}+\\cdots +{a}_{k}\\le {a}_{1}+{a}_{2}+\\cdots +{a}_{k}+{a}_{k+1}={S}_{k+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739064845\">Therefore, the sequence of partial sums is increasing. Further, since [latex]{a}_{n}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex], then<\/p>\n<div id=\"fs-id1169739300032\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=N}^{k}{a}_{n}\\le \\displaystyle\\sum _{n=N}^{k}{b}_{n}\\le \\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}=L[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739013220\">Therefore, for all [latex]k\\ge 1[\/latex],<\/p>\n<div id=\"fs-id1169739222211\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\left({a}_{1}+{a}_{2}+\\cdots +{a}_{N - 1}\\right)+\\displaystyle\\sum _{n=N}^{k}{a}_{n}\\le \\left({a}_{1}+{a}_{2}+\\cdots +{a}_{N - 1}\\right)+L[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738925226\">Since [latex]{a}_{1}+{a}_{2}+\\cdots +{a}_{N - 1}[\/latex] is a finite number, we conclude that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is bounded above. Therefore, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges, and therefore the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\n<p id=\"fs-id1169739017799\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1169738938243\">To use the comparison test to determine the convergence or divergence of a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and <em data-effect=\"italics\">p<\/em>-series, these series are often used. If there exists an integer [latex]N[\/latex] such that for all [latex]n\\ge N[\/latex], each term [latex]{a}_{n}[\/latex] is less than each corresponding term of a known convergent series, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges. Similarly, if there exists an integer [latex]N[\/latex] such that for all [latex]n\\ge N[\/latex], each term [latex]{a}_{n}[\/latex] is greater than each corresponding term of a known divergent series, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169383\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169383&theme=oea&iframe_resize_id=ohm169383&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1169738880258\" data-type=\"example\">\n<div id=\"fs-id1169738905678\" data-type=\"exercise\">\n<div id=\"fs-id1169739020328\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Using the Comparison Test<\/h3>\n<div id=\"fs-id1169739020328\" data-type=\"problem\">\n<p id=\"fs-id1169738994018\">For each of the following series, use the comparison test to determine whether the series converges or diverges.<\/p>\n<ol id=\"fs-id1169736613871\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}+3n+1}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{2}^{n}+1}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\text{ln}\\left(n\\right)}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738895467\" data-type=\"solution\">\n<ol id=\"fs-id1169738938121\" type=\"a\">\n<li>Compare to [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] is a <em data-effect=\"italics\">p<\/em>-series with [latex]p=3[\/latex], it converges. Further,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739298200\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{n}^{3}+3n+1}<\\frac{1}{{n}^{3}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor every positive integer [latex]n[\/latex]. Therefore, we can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}+3n+1}[\/latex] converges.<\/li>\n<li>Compare to [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n}[\/latex]. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n}[\/latex] is a geometric series with [latex]r=\\frac{1}{2}[\/latex] and [latex]|\\frac{1}{2}|<1[\/latex], it converges. Also,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739039859\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{2}^{n}+1}<\\frac{1}{{2}^{n}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor every positive integer [latex]n[\/latex]. Therefore, we see that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{2}^{n}+1}[\/latex] converges.<\/li>\n<li>Compare to [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n}[\/latex]. Since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739029860\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\text{ln}\\left(n\\right)}>\\frac{1}{n}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nfor every integer [latex]n\\ge 2[\/latex] and [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{n}[\/latex] diverges, we have that [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\text{ln}\\left(n\\right)}[\/latex] diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739333977\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169739096544\" data-type=\"exercise\">\n<div id=\"fs-id1169739269621\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169739269621\" data-type=\"problem\">\n<p id=\"fs-id1169739001213\">Use the comparison test to determine if the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{{n}^{3}+n+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=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738972533\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169736597699\">Find a value [latex]p[\/latex] such that [latex]\\frac{n}{{n}^{3}+n+1}\\le \\frac{1}{{n}^{p}}[\/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-id1169739034395\" data-type=\"solution\">\n<p id=\"fs-id1169739034397\">The series converges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/izlPnqyz9Y4?controls=0&amp;start=285&amp;end=375&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\/5.4.1_285to375_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.4.1&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<\/section>\n<section id=\"fs-id1169736615268\" 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-1713\">\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>5.4.1. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":17,"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\":\"5.4.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1713","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1713","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":8,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1713\/revisions"}],"predecessor-version":[{"id":2217,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1713\/revisions\/2217"}],"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\/1713\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1713"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1713"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1713"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1713"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}