{"id":888,"date":"2025-06-20T17:20:05","date_gmt":"2025-06-20T17:20:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=888"},"modified":"2025-09-09T20:02:25","modified_gmt":"2025-09-09T20:02:25","slug":"separation-of-variables-fresh-take-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/separation-of-variables-fresh-take-2\/","title":{"raw":"The Divergence and Integral Tests: Fresh Take","rendered":"The Divergence and Integral Tests: Fresh Take"},"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<h2>Divergence Test<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The divergence test is the simplest and most fundamental test for series convergence. It's based on a logical necessity: if you're adding infinitely many terms and want a finite sum, those terms must be getting smaller and smaller\u2014specifically, they must approach zero.<\/p>\r\n<p class=\"whitespace-normal break-words\">If [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] converges to [latex]S[\/latex], then consecutive partial sums [latex]S_k[\/latex] and [latex]S_{k-1}[\/latex] both approach [latex]S[\/latex]. Since [latex]a_k = S_k - S_{k-1}[\/latex], we must have [latex]a_k \\to S - S = 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The test:<\/strong> If [latex]\\lim_{n \\to \\infty} a_n \\neq 0[\/latex] (or the limit doesn't exist), then [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] diverges.<\/p>\r\n<p class=\"whitespace-normal break-words\">The divergence test can only prove divergence, never convergence. When [latex]\\lim_{n \\to \\infty} a_n = 0[\/latex], the test is inconclusive\u2014the series might converge or diverge.<\/p>\r\nAlways apply the divergence test first! It's quick and can immediately rule out many series. Look for terms that don't approach zero, like [latex]\\frac{n}{3n-1} \\to \\frac{1}{3}[\/latex] or [latex]e^{\\frac{1}{n^2}} \\to e^0 = 1[\/latex].\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169737949827\" data-type=\"problem\">\r\n<p id=\"fs-id1169738149942\">What does the divergence test tell us about the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\cos\\left(\\frac{1}{{n}^{2}}\\right)\\text{?}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1169738226267\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737946990\">Look at [latex]\\underset{n\\to \\infty }{\\text{lim}}\\cos\\left(\\frac{1}{{n}^{2}}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1169738186762\" data-type=\"solution\">\r\n<p id=\"fs-id1169738093054\">The series diverges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/-y848AvISyY?controls=0&amp;start=150&amp;end=186&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.3.1_150to186_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3.1\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Integral Test<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The integral test provides a powerful way to determine series convergence by comparing an infinite sum to an improper integral. Since both represent \"adding up\" infinitely many values, their convergence behaviors are linked.<\/p>\r\n<p class=\"whitespace-normal break-words\">For a series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] with positive terms, find a continuous, decreasing function [latex]f(x)[\/latex] where [latex]f(n) = a_n[\/latex]. Then [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] and [latex]\\int_1^{\\infty} f(x)dx[\/latex] either both converge or both diverge.<\/p>\r\n<p class=\"whitespace-normal break-words\">Think of rectangles representing series terms compared to the area under a curve. When the rectangles sit above the curve (like with [latex]\\frac{1}{n}[\/latex] and [latex]\\frac{1}{x}[\/latex]), if the integral diverges, the series must also diverge. When rectangles sit below the curve (like with [latex]\\frac{1}{n^2}[\/latex] and [latex]\\frac{1}{x^2}[\/latex]), if the integral converges, the series is bounded and must converge.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key requirements:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Series must have positive terms<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Function [latex]f(x)[\/latex] must be continuous, positive, and decreasing<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f(n) = a_n[\/latex] for all sufficiently large [latex]n[\/latex]<\/li>\r\n<\/ul>\r\nThe integral test only tells you about convergence\/divergence, not the actual sum. The series sum and integral value are generally different (like [latex]\\sum(\\frac{1}{e})^n = \\frac{1}{e-1}[\/latex] versus [latex]\\int_1^{\\infty}(\\frac{1}{e})^x dx = \\frac{1}{e}[\/latex]).\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738164454\" data-type=\"problem\">\r\n<p id=\"fs-id1169738164456\">Use the integral test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3{n}^{2}+1}[\/latex] converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1169738249441\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169738250182\">Compare to the integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{x}{3{x}^{2}+1}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169738068120\" data-type=\"solution\">\r\n<p id=\"fs-id1169738068122\">The series diverges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_zV7bzn3jUE?controls=0&amp;start=235&amp;end=342&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.3.2_235to342_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3.2\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2>The p-Series<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The p-series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] represents one of the most fundamental families of series in mathematics. Understanding p-series gives you a benchmark for testing many other series through comparison techniques.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The simple convergence rule:<\/strong> The p-series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] converges if and only if [latex]p &gt; 1[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why this cutoff at [latex]p = 1[\/latex]:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">When [latex]p \\leq 0[\/latex]: Terms don't approach zero, so divergence test kills it immediately<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]0 &lt; p &lt; 1[\/latex]: Terms approach zero too slowly\u2014integral test with [latex]\\int_1^{\\infty} \\frac{1}{x^p}dx = \\infty[\/latex] shows divergence<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]p = 1[\/latex]: This is the harmonic series, which diverges (borderline case)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]p &gt; 1[\/latex]: Terms approach zero fast enough\u2014integral test with [latex]\\int_1^{\\infty} \\frac{1}{x^p}dx = \\frac{1}{p-1}[\/latex] shows convergence<\/li>\r\n<\/ul>\r\nThe value [latex]p = 1[\/latex] represents the boundary between convergence and divergence. The harmonic series shows that even when terms approach zero, they must do so \"fast enough\"\u2014faster than [latex]\\frac{1}{n}[\/latex].\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738055556\" data-type=\"problem\">\r\n<p id=\"fs-id1169738055558\">Does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{5}{4}}}[\/latex] converge or diverge?<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1169737209098\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169738080287\">[latex]p=\\frac{5}{4}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1169737209091\" data-type=\"solution\">\r\n<p id=\"fs-id1169737209093\">The series converges.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/1HTIl9x92UU?controls=0&amp;start=32&amp;end=44&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.3.4_32to44_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3.4\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2>Estimating Series Value<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">When a series converges, you often want to know its actual sum, not just that it converges. The integral test provides a powerful way to estimate how close your partial sums are to the true infinite sum.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The remainder concept:<\/strong> For a convergent series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex], the remainder [latex]R_N = \\sum_{n=1}^{\\infty} a_n - S_N[\/latex] represents the \"error\" when you approximate the infinite sum with the [latex]N[\/latex]th partial sum [latex]S_N[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The key insight:<\/strong> If your series satisfies the integral test conditions, then: [latex]\\int_{N+1}^{\\infty} f(x)dx &lt; R_N &lt; \\int_N^{\\infty} f(x)dx[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This gives you concrete upper and lower bounds on your approximation error.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Calculate [latex]S_N[\/latex] for some reasonable [latex]N[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate [latex]\\int_N^{\\infty} f(x)dx[\/latex] to get an upper bound on the error<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Add this bound to [latex]S_N[\/latex] to get an overestimate of the series sum<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If you need the error below a certain threshold, solve [latex]\\int_N^{\\infty} f(x)dx &lt; \\text{desired error}[\/latex] for [latex]N[\/latex]<\/li>\r\n<\/ol>\r\nThe integral test doesn't just tell you about convergence\u2014it quantifies how quickly the series approaches its limit, letting you control approximation accuracy.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738249586\" data-type=\"problem\">\r\n<p id=\"fs-id1169738249588\">For [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex], calculate [latex]{S}_{5}[\/latex] and estimate the error [latex]{R}_{5}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1169737265746\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737265752\">Use the remainder estimate [latex]{R}_{N}&lt;{\\displaystyle\\int }_{N}^{\\infty }\\frac{1}{{x}^{4}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1169737394110\" data-type=\"solution\">\r\n<p id=\"fs-id1169737394112\">[latex]{S}_{5}\\approx 1.09035[\/latex], [latex]{R}_{5}&lt;0.00267[\/latex]<\/p>\r\n\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<h2>Divergence Test<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The divergence test is the simplest and most fundamental test for series convergence. It&#8217;s based on a logical necessity: if you&#8217;re adding infinitely many terms and want a finite sum, those terms must be getting smaller and smaller\u2014specifically, they must approach zero.<\/p>\n<p class=\"whitespace-normal break-words\">If [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] converges to [latex]S[\/latex], then consecutive partial sums [latex]S_k[\/latex] and [latex]S_{k-1}[\/latex] both approach [latex]S[\/latex]. Since [latex]a_k = S_k - S_{k-1}[\/latex], we must have [latex]a_k \\to S - S = 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The test:<\/strong> If [latex]\\lim_{n \\to \\infty} a_n \\neq 0[\/latex] (or the limit doesn&#8217;t exist), then [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] diverges.<\/p>\n<p class=\"whitespace-normal break-words\">The divergence test can only prove divergence, never convergence. When [latex]\\lim_{n \\to \\infty} a_n = 0[\/latex], the test is inconclusive\u2014the series might converge or diverge.<\/p>\n<p>Always apply the divergence test first! It&#8217;s quick and can immediately rule out many series. Look for terms that don&#8217;t approach zero, like [latex]\\frac{n}{3n-1} \\to \\frac{1}{3}[\/latex] or [latex]e^{\\frac{1}{n^2}} \\to e^0 = 1[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169737949827\" data-type=\"problem\">\n<p id=\"fs-id1169738149942\">What does the divergence test tell us about the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\cos\\left(\\frac{1}{{n}^{2}}\\right)\\text{?}[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738226267\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737946990\">Look at [latex]\\underset{n\\to \\infty }{\\text{lim}}\\cos\\left(\\frac{1}{{n}^{2}}\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738186762\" data-type=\"solution\">\n<p id=\"fs-id1169738093054\">The series diverges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/-y848AvISyY?controls=0&amp;start=150&amp;end=186&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.3.1_150to186_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3.1&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2 data-type=\"title\">Integral Test<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The integral test provides a powerful way to determine series convergence by comparing an infinite sum to an improper integral. Since both represent &#8220;adding up&#8221; infinitely many values, their convergence behaviors are linked.<\/p>\n<p class=\"whitespace-normal break-words\">For a series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] with positive terms, find a continuous, decreasing function [latex]f(x)[\/latex] where [latex]f(n) = a_n[\/latex]. Then [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] and [latex]\\int_1^{\\infty} f(x)dx[\/latex] either both converge or both diverge.<\/p>\n<p class=\"whitespace-normal break-words\">Think of rectangles representing series terms compared to the area under a curve. When the rectangles sit above the curve (like with [latex]\\frac{1}{n}[\/latex] and [latex]\\frac{1}{x}[\/latex]), if the integral diverges, the series must also diverge. When rectangles sit below the curve (like with [latex]\\frac{1}{n^2}[\/latex] and [latex]\\frac{1}{x^2}[\/latex]), if the integral converges, the series is bounded and must converge.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key requirements:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Series must have positive terms<\/li>\n<li class=\"whitespace-normal break-words\">Function [latex]f(x)[\/latex] must be continuous, positive, and decreasing<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f(n) = a_n[\/latex] for all sufficiently large [latex]n[\/latex]<\/li>\n<\/ul>\n<p>The integral test only tells you about convergence\/divergence, not the actual sum. The series sum and integral value are generally different (like [latex]\\sum(\\frac{1}{e})^n = \\frac{1}{e-1}[\/latex] versus [latex]\\int_1^{\\infty}(\\frac{1}{e})^x dx = \\frac{1}{e}[\/latex]).<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738164454\" data-type=\"problem\">\n<p id=\"fs-id1169738164456\">Use the integral test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{3{n}^{2}+1}[\/latex] converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Hint<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738249441\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169738250182\">Compare to the integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{x}{3{x}^{2}+1}dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Show Solution<\/button><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738068120\" data-type=\"solution\">\n<p id=\"fs-id1169738068122\">The series diverges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/_zV7bzn3jUE?controls=0&amp;start=235&amp;end=342&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.3.2_235to342_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3.2&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2>The p-Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The p-series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] represents one of the most fundamental families of series in mathematics. Understanding p-series gives you a benchmark for testing many other series through comparison techniques.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The simple convergence rule:<\/strong> The p-series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] converges if and only if [latex]p > 1[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why this cutoff at [latex]p = 1[\/latex]:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">When [latex]p \\leq 0[\/latex]: Terms don&#8217;t approach zero, so divergence test kills it immediately<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]0 < p < 1[\/latex]: Terms approach zero too slowly\u2014integral test with [latex]\\int_1^{\\infty} \\frac{1}{x^p}dx = \\infty[\/latex] shows divergence<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]p = 1[\/latex]: This is the harmonic series, which diverges (borderline case)<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]p > 1[\/latex]: Terms approach zero fast enough\u2014integral test with [latex]\\int_1^{\\infty} \\frac{1}{x^p}dx = \\frac{1}{p-1}[\/latex] shows convergence<\/li>\n<\/ul>\n<p>The value [latex]p = 1[\/latex] represents the boundary between convergence and divergence. The harmonic series shows that even when terms approach zero, they must do so &#8220;fast enough&#8221;\u2014faster than [latex]\\frac{1}{n}[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738055556\" data-type=\"problem\">\n<p id=\"fs-id1169738055558\">Does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{5}{4}}}[\/latex] converge or diverge?<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Hint<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737209098\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169738080287\">[latex]p=\\frac{5}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558893\">Show Solution<\/button><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737209091\" data-type=\"solution\">\n<p id=\"fs-id1169737209093\">The series converges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/1HTIl9x92UU?controls=0&amp;start=32&amp;end=44&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.3.4_32to44_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3.4&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2>Estimating Series Value<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When a series converges, you often want to know its actual sum, not just that it converges. The integral test provides a powerful way to estimate how close your partial sums are to the true infinite sum.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The remainder concept:<\/strong> For a convergent series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex], the remainder [latex]R_N = \\sum_{n=1}^{\\infty} a_n - S_N[\/latex] represents the &#8220;error&#8221; when you approximate the infinite sum with the [latex]N[\/latex]th partial sum [latex]S_N[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The key insight:<\/strong> If your series satisfies the integral test conditions, then: [latex]\\int_{N+1}^{\\infty} f(x)dx < R_N < \\int_N^{\\infty} f(x)dx[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This gives you concrete upper and lower bounds on your approximation error.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Calculate [latex]S_N[\/latex] for some reasonable [latex]N[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate [latex]\\int_N^{\\infty} f(x)dx[\/latex] to get an upper bound on the error<\/li>\n<li class=\"whitespace-normal break-words\">Add this bound to [latex]S_N[\/latex] to get an overestimate of the series sum<\/li>\n<li class=\"whitespace-normal break-words\">If you need the error below a certain threshold, solve [latex]\\int_N^{\\infty} f(x)dx < \\text{desired error}[\/latex] for [latex]N[\/latex]<\/li>\n<\/ol>\n<p>The integral test doesn&#8217;t just tell you about convergence\u2014it quantifies how quickly the series approaches its limit, letting you control approximation accuracy.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738249586\" data-type=\"problem\">\n<p id=\"fs-id1169738249588\">For [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex], calculate [latex]{S}_{5}[\/latex] and estimate the error [latex]{R}_{5}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558890\">Hint<\/button><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737265746\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737265752\">Use the remainder estimate [latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{1}{{x}^{4}}dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Show Solution<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737394110\" data-type=\"solution\">\n<p id=\"fs-id1169737394112\">[latex]{S}_{5}\\approx 1.09035[\/latex], [latex]{R}_{5}<0.00267[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":23,"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 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