{"id":883,"date":"2025-06-20T17:19:51","date_gmt":"2025-06-20T17:19:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=883"},"modified":"2025-09-11T16:09:48","modified_gmt":"2025-09-11T16:09:48","slug":"separation-of-variables-learn-it-1-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/separation-of-variables-learn-it-1-2\/","title":{"raw":"The Divergence and Integral Tests: Learn It 1","rendered":"The Divergence and Integral Tests: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the divergence test to check if a series might converge<\/li>\r\n \t<li>Apply the integral test to determine if a series converges<\/li>\r\n \t<li>Estimate how close a partial sum is to the actual sum of a series<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">In the previous section, we determined whether series converged by calculating limits of partial sums [latex]{S_k}[\/latex]. While this direct approach works, it's often difficult or impossible to find these limits explicitly.<\/p>\r\n<p class=\"whitespace-normal break-words\">Fortunately, mathematicians have developed several tests that let us determine convergence without finding explicit limits. We'll start with two fundamental tests: the divergence test and the integral test.<\/p>\r\n\r\n<h2>Divergence Test<\/h2>\r\n<p class=\"whitespace-normal break-words\">The divergence test comes from a basic requirement for convergence: <strong>if a series converges, its terms must approach zero<\/strong>.<\/p>\r\n<p class=\"whitespace-normal break-words\">Here's the reasoning: for a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] to converge to sum [latex]S[\/latex], we need [latex]\\underset{k\\to \\infty}{\\text{lim}} S_k = S[\/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 class=\"whitespace-normal break-words\">This gives us a powerful test for divergence.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>divergence test<\/h3>\r\nIf [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.\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"62\" height=\"55\" \/>The divergence test can only prove divergence\u2014never convergence. Even if [latex]\\underset{n\\to \\infty}{\\text{lim}} a_n = 0[\/latex], the series might still diverge.\r\n<p class=\"whitespace-normal break-words\">The harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n}[\/latex] has [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{1}{n} = 0[\/latex], but it diverges. When [latex]a_n \\to 0[\/latex], the divergence test is <strong>inconclusive<\/strong>\u2014we need other methods.<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the divergence test to check if a series might converge<\/li>\n<li>Apply the integral test to determine if a series converges<\/li>\n<li>Estimate how close a partial sum is to the actual sum of a series<\/li>\n<\/ul>\n<\/section>\n<p class=\"whitespace-normal break-words\">In the previous section, we determined whether series converged by calculating limits of partial sums [latex]{S_k}[\/latex]. While this direct approach works, it&#8217;s often difficult or impossible to find these limits explicitly.<\/p>\n<p class=\"whitespace-normal break-words\">Fortunately, mathematicians have developed several tests that let us determine convergence without finding explicit limits. We&#8217;ll start with two fundamental tests: the divergence test and the integral test.<\/p>\n<h2>Divergence Test<\/h2>\n<p class=\"whitespace-normal break-words\">The divergence test comes from a basic requirement for convergence: <strong>if a series converges, its terms must approach zero<\/strong>.<\/p>\n<p class=\"whitespace-normal break-words\">Here&#8217;s the reasoning: for a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] to converge to sum [latex]S[\/latex], we need [latex]\\underset{k\\to \\infty}{\\text{lim}} S_k = S[\/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 class=\"whitespace-normal break-words\">This gives us a powerful test for divergence.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>divergence test<\/h3>\n<p>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<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"62\" height=\"55\" \/>The divergence test can only prove divergence\u2014never convergence. Even if [latex]\\underset{n\\to \\infty}{\\text{lim}} a_n = 0[\/latex], the series might still diverge.<\/p>\n<p class=\"whitespace-normal break-words\">The harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n}[\/latex] has [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{1}{n} = 0[\/latex], but it diverges. When [latex]a_n \\to 0[\/latex], the divergence test is <strong>inconclusive<\/strong>\u2014we need other methods.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><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-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<\/section>\n","protected":false},"author":15,"menu_order":18,"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\/883"}],"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":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/883\/revisions"}],"predecessor-version":[{"id":2319,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/883\/revisions\/2319"}],"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\/883\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=883"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=883"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=883"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=883"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}