{"id":895,"date":"2025-06-20T17:20:34","date_gmt":"2025-06-20T17:20:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=895"},"modified":"2025-09-11T16:18:59","modified_gmt":"2025-09-11T16:18:59","slug":"first-order-linear-equations-and-applications-learn-it-1-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/first-order-linear-equations-and-applications-learn-it-1-2\/","title":{"raw":"Comparison Tests: Learn It 1","rendered":"Comparison Tests: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Compare series to determine convergence using the comparison test<\/li>\r\n \t<li>Use the limit comparison test to figure out if a series converges<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">We've seen that the integral test determines convergence by comparing a series to a related improper integral. Now we'll learn how to determine convergence by comparing a series to another series whose behavior we already know. Comparison tests are particularly useful for series that resemble geometric series or <em>p<\/em>-series\u2014two types whose convergence patterns we understand completely.<\/p>\r\n\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Comparison Test<\/h2>\r\nIn previous sections, we studied geometric series and <em>p<\/em>-series and learned exactly when they converge or diverge. The <strong>comparison test<\/strong> lets us use the known behavior of these series to determine convergence for other, similar series.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\">Before diving into the test, recall this important inequality behavior:<\/p>\r\n<p style=\"text-align: center;\">If [latex] 0 &lt; a_n \\le b_n [\/latex], then [latex] \\frac{1}{a_n} \\ge \\frac{1}{b_n} [\/latex]<\/p>\r\nIn other words, a larger denominator corresponds to a smaller fraction.\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Comparing Similar Series<\/h3>\r\n<p class=\"whitespace-normal break-words\">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<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<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<p class=\"whitespace-normal break-words\">Notice that for all positive integers [latex]n[\/latex]:<\/p>\r\n\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<div data-type=\"equation\" data-label=\"\">This inequality holds because [latex]n^2 + 1 &gt; n^2[\/latex], making [latex]\\frac{1}{n^2 + 1}[\/latex] smaller than [latex]\\frac{1}{n^2}[\/latex].<\/div>\r\nTherefore, the [latex]k[\/latex]th partial sum [latex]S_k[\/latex] of [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2 + 1}[\/latex] satisfies:\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<p id=\"fs-id1169739189474\">(See Figure 1 (a) and the following table.)<\/p>\r\n\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<p class=\"whitespace-normal break-words\">Since the series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex] converges, the sequence [latex]{S_k}[\/latex] is bounded above. We now have a sequence that is both monotone increasing and bounded above.<\/p>\r\n<p class=\"whitespace-normal break-words\">By the Monotone Convergence Theorem, [latex]{S_k}[\/latex] must converge. Therefore, [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2 + 1}[\/latex] <strong>converges<\/strong>.<\/p>\r\nNow consider the series:\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<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<p id=\"fs-id1169739301476\">The sequence of partial sums for each series is monotone increasing, and we can compare the terms:<\/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<p class=\"whitespace-normal break-words\">This inequality holds because [latex]n - \\frac{1}{2} &lt; n[\/latex], making [latex]\\frac{1}{n - \\frac{1}{2}}[\/latex] larger than [latex]\\frac{1}{n}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, the [latex]k[\/latex]th 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<p id=\"fs-id1169739004447\">(See Figure 1 (a) and the following table.)<\/p>\r\n\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<p class=\"whitespace-normal break-words\">Since the harmonic 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.<\/p>\r\n<p class=\"whitespace-normal break-words\">Consequently, [latex]{S_k}[\/latex] is also unbounded and therefore diverges. We conclude that [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n - \\frac{1}{2}}[\/latex] <strong>diverges<\/strong>.<\/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<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>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].\r\nIf [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] <strong>converges<\/strong>, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] <strong>converges<\/strong>.<\/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].\r\nIf [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] <strong>diverges<\/strong>, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] <strong>diverges<\/strong>.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\r\n<p id=\"fs-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\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\"><strong>How to: Apply the Comparison Test<\/strong><\/p>\r\nTo use the comparison test for a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex], you need to find a suitable comparison series. Since we know the convergence properties of geometric series and [latex]p[\/latex]-series, these are the most commonly used comparison series.\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Strategy for convergence<\/strong>: If you can show that [latex]a_n \\leq b_n[\/latex] for all [latex]n \\geq N[\/latex] (for some integer [latex]N[\/latex]), where [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] is a known convergent series, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] converges.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Strategy for divergence<\/strong>: If you can show that [latex]a_n \\geq b_n[\/latex] for all [latex]n \\geq N[\/latex], where [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] is a known divergent series, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] diverges.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311398[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Compare series to determine convergence using the comparison test<\/li>\n<li>Use the limit comparison test to figure out if a series converges<\/li>\n<\/ul>\n<\/section>\n<p class=\"whitespace-normal break-words\">We&#8217;ve seen that the integral test determines convergence by comparing a series to a related improper integral. Now we&#8217;ll learn how to determine convergence by comparing a series to another series whose behavior we already know. Comparison tests are particularly useful for series that resemble geometric series or <em>p<\/em>-series\u2014two types whose convergence patterns we understand completely.<\/p>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Comparison Test<\/h2>\n<p>In previous sections, we studied geometric series and <em>p<\/em>-series and learned exactly when they converge or diverge. The <strong>comparison test<\/strong> lets us use the known behavior of these series to determine convergence for other, similar series.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\">Before diving into the test, recall this important inequality behavior:<\/p>\n<p style=\"text-align: center;\">If [latex]0 < a_n \\le b_n[\/latex], then [latex]\\frac{1}{a_n} \\ge \\frac{1}{b_n}[\/latex]<\/p>\n<p>In other words, a larger denominator corresponds to a smaller fraction.<\/p>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Comparing Similar Series<\/h3>\n<p class=\"whitespace-normal break-words\">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 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 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 class=\"whitespace-normal break-words\">Notice that for all positive integers [latex]n[\/latex]:<\/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<div data-type=\"equation\" data-label=\"\">This inequality holds because [latex]n^2 + 1 > n^2[\/latex], making [latex]\\frac{1}{n^2 + 1}[\/latex] smaller than [latex]\\frac{1}{n^2}[\/latex].<\/div>\n<p>Therefore, the [latex]k[\/latex]th 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 id=\"fs-id1169739189474\">(See Figure 1 (a) and the following table.)<\/p>\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<p class=\"whitespace-normal break-words\">Since the series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex] converges, the sequence [latex]{S_k}[\/latex] is bounded above. We now have a sequence that is both monotone increasing and bounded above.<\/p>\n<p class=\"whitespace-normal break-words\">By the Monotone Convergence Theorem, [latex]{S_k}[\/latex] must converge. Therefore, [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2 + 1}[\/latex] <strong>converges<\/strong>.<\/p>\n<p>Now 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 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 id=\"fs-id1169739301476\">The sequence of partial sums for each series is monotone increasing, and we can compare the terms:<\/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 class=\"whitespace-normal break-words\">This inequality holds because [latex]n - \\frac{1}{2} < n[\/latex], making [latex]\\frac{1}{n - \\frac{1}{2}}[\/latex] larger than [latex]\\frac{1}{n}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, the [latex]k[\/latex]th 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 id=\"fs-id1169739004447\">(See Figure 1 (a) and the following table.)<\/p>\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<p class=\"whitespace-normal break-words\">Since the harmonic 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.<\/p>\n<p class=\"whitespace-normal break-words\">Consequently, [latex]{S_k}[\/latex] is also unbounded and therefore diverges. We conclude that [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n - \\frac{1}{2}}[\/latex] <strong>diverges<\/strong>.<\/p>\n<figure id=\"CNX_Calc_Figure_09_04_001\"><figcaption><\/figcaption><figure style=\"width: 731px\" 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\" \/><figcaption 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.<\/figcaption><\/figure>\n<\/figure>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>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].<br \/>\nIf [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] <strong>converges<\/strong>, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] <strong>converges<\/strong>.<\/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].<br \/>\nIf [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] <strong>diverges<\/strong>, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] <strong>diverges<\/strong>.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\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<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\"><strong>How to: Apply the Comparison Test<\/strong><\/p>\n<p>To use the comparison test for a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex], you need to find a suitable comparison series. Since we know the convergence properties of geometric series and [latex]p[\/latex]-series, these are the most commonly used comparison series.<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Strategy for convergence<\/strong>: If you can show that [latex]a_n \\leq b_n[\/latex] for all [latex]n \\geq N[\/latex] (for some integer [latex]N[\/latex]), where [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] is a known convergent series, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] converges.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Strategy for divergence<\/strong>: If you can show that [latex]a_n \\geq b_n[\/latex] for all [latex]n \\geq N[\/latex], where [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] is a known divergent series, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] diverges.<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-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<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311398\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311398&theme=lumen&iframe_resize_id=ohm311398&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":24,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/895"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/895\/revisions"}],"predecessor-version":[{"id":2321,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/895\/revisions\/2321"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/895\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=895"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=895"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=895"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=895"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}