{"id":876,"date":"2025-06-20T17:19:25","date_gmt":"2025-06-20T17:19:25","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=876"},"modified":"2025-09-09T19:57:20","modified_gmt":"2025-09-09T19:57:20","slug":"direction-fields-and-eulers-method-fresh-take-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-fresh-take-2\/","title":{"raw":"Introduction to Series: Fresh Take","rendered":"Introduction to Series: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Understand what we mean by the sum of an infinite series<\/li>\r\n \t<li>Find the sum of a geometric series<\/li>\r\n \t<li>Calculate the sum of a telescoping series<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Sums and 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\">An infinite series is what you get when you try to add infinitely many terms together. Since you can't actually perform infinite addition, we define a series through the clever concept of partial sums\u2014finite sums that we can calculate.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The core concept:<\/strong> For series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex], create the sequence of partial sums:<\/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\">[latex]S_1 = a_1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_2 = a_1 + a_2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_3 = a_1 + a_2 + a_3[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">And so on...<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] converges to [latex]S[\/latex] if and only if the sequence of partial sums [latex]{S_k}[\/latex] converges to [latex]S[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">To analyze any series, compute several partial sums and look for patterns. Does [latex]S_k[\/latex] seem to approach a finite limit? Does it grow without bound? Does it oscillate?<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"http:\/\/web.archive.org\/web\/20200216163744\/http:\/\/mathdemos.org\/mathdemos\/donut-demo\/\" target=\"_blank\" rel=\"noopener\">this\u00a0website for a whimsical demonstration of series using donuts<\/a>.<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169737822604\" data-type=\"problem\">\r\n<p id=\"fs-id1169737297457\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\left(n+1\\right)}{n}[\/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-id1169737294981\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169738073257\">Look at the sequence of partial sums.<\/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-id1169737743047\" data-type=\"solution\">\r\n<p id=\"fs-id1169738155184\">The series diverges because the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}&gt;k[\/latex].<\/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\/VZkRi47848U?controls=0&amp;start=483&amp;end=620&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\/5.2.1_483to620_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.2.1\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">The Harmonic 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 harmonic series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n} = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\cdots[\/latex] is one of mathematics' most deceptive examples\u2014it looks like it should converge, but it actually diverges.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The proof strategy:<\/strong> Group terms cleverly to show the partial sums are unbounded:<\/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\">[latex]S_2 = 1 + \\frac{1}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_4 &gt; 1 + \\frac{1}{2} + \\left(\\frac{1}{4} + \\frac{1}{4}\\right) = 1 + 2 \\cdot \\frac{1}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_8 &gt; 1 + 3 \\cdot \\frac{1}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">In general: [latex]S_{2^j} &gt; 1 + j \\cdot \\frac{1}{2}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>The key insight:<\/strong> Since [latex]1 + j \\cdot \\frac{1}{2} \\to \\infty[\/latex] as [latex]j \\to \\infty[\/latex], the partial sums are unbounded, so the series diverges.<\/p>\r\nThe harmonic series demonstrates that having [latex]\\lim_{n \\to \\infty} a_n = 0[\/latex] is necessary but not sufficient for convergence. The terms must approach zero \"fast enough\"\u2014the harmonic series shows the borderline between convergence and divergence.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738201811\" data-type=\"problem\">\r\n<p id=\"fs-id1169738044543\">Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{5}{{2}^{n - 1}}[\/latex].<\/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-id1169737910071\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737910078\">Rewrite as [latex]\\displaystyle\\sum _{n=1}^{\\infty }5{\\left(\\frac{1}{2}\\right)}^{n - 1}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1169737394304\" data-type=\"solution\">\r\n<p id=\"fs-id1169737394306\">[latex]10[\/latex].<\/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\/VZkRi47848U?controls=0&amp;start=1056&amp;end=1095&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/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\/5.2.1_1056to1095_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.2.1\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169737168534\" data-type=\"problem\">\r\n<p id=\"fs-id1169737168537\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{-2}{5}\\right)}^{n - 1}[\/latex] converges or diverges. If it converges, find its sum.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1169737233901\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737233907\">[latex]r=-\\frac{2}{5}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1169737233887\" data-type=\"solution\">\r\n<p id=\"fs-id1169737233889\">[latex]\\frac{5}{7}[\/latex]<\/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\/05JxlmgOCkM?controls=0&amp;start=0&amp;end=29&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\/5.2.2_0to29_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.2.2\" here (opens in new window)<\/a>.<\/section>\r\n<h2 data-type=\"title\">Algebraic Properties of Convergent 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\">Once you know that individual series converge, you can combine them using familiar algebraic operations. These rules work exactly like you'd expect from your experience with finite sums and limits.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The three fundamental rules:<\/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\"><strong>Sum Rule:<\/strong> [latex]\\sum_{n=1}^{\\infty} (a_n + b_n) = \\sum_{n=1}^{\\infty} a_n + \\sum_{n=1}^{\\infty} b_n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Difference Rule:<\/strong> [latex]\\sum_{n=1}^{\\infty} (a_n - b_n) = \\sum_{n=1}^{\\infty} a_n - \\sum_{n=1}^{\\infty} b_n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Constant Multiple Rule:<\/strong> [latex]\\sum_{n=1}^{\\infty} ca_n = c\\sum_{n=1}^{\\infty} a_n[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Important requirement:<\/strong> Both series must converge individually before you can apply these rules. These rules only apply to convergent series. If either series diverges, you cannot use these algebraic properties.<\/p>\r\nWhen facing a complex series, try to break it apart into simpler series that you can evaluate separately, then recombine using these rules.\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Geometric 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\">Geometric series are the foundation of series theory because they have a clear pattern and a simple convergence test. Every geometric series has the form [latex]\\sum_{n=1}^{\\infty} ar^{n-1} = a + ar + ar^2 + ar^3 + \\cdots[\/latex] where each term is the previous term multiplied by the common ratio [latex]r[\/latex].<\/p>\r\nGeometric series give you an exact formula for the sum when they converge, making them the most useful and predictable type of series you'll encounter.\r\n<p class=\"whitespace-normal break-words\"><strong>The convergence rule is simple:<\/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\">If [latex]|r| &lt; 1[\/latex]: the series converges to [latex]\\frac{a}{1-r}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]|r| \\geq 1[\/latex]: the series diverges<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Why does this work? The partial sum formula [latex]S_k = \\frac{a(1-r^k)}{1-r}[\/latex] (for [latex]r \\neq 1[\/latex]) depends entirely on what happens to [latex]r^k[\/latex] as [latex]k \\to \\infty[\/latex]. When [latex]|r| &lt; 1[\/latex], we have [latex]r^k \\to 0[\/latex], so [latex]S_k \\to \\frac{a}{1-r}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Look for a constant ratio between consecutive terms. The series doesn't have to start with [latex]n=1[\/latex] or have exponent [latex]n-1[\/latex]\u2014you can always rewrite it in standard form by factoring out appropriate terms.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key technique:<\/strong> For series like [latex]\\sum_{n=0}^{\\infty} \\left(\\frac{2}{3}\\right)^{n+2}[\/latex], write out the first few terms to identify [latex]a[\/latex] and [latex]r[\/latex], then apply the convergence test.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738115130\" data-type=\"problem\">\r\n<p id=\"fs-id1169738115132\">Write [latex]5.2\\overline{7}[\/latex] as a fraction of integers.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558879\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1169737394617\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<div data-type=\"title\"><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">By expressing this number as a series, find a geometric series with initial term [latex]a=\\frac{7}{100}[\/latex] and ratio [latex]r=\\frac{1}{10}[\/latex].<\/span><\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1169738115149\" data-type=\"solution\">\r\n<p id=\"fs-id1169738115151\">[latex]\\frac{475}{90}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2>Telescoping 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\">Telescoping series are special because most terms cancel out when you compute partial sums, leaving only a few terms at the beginning and end. This makes them surprisingly easy to analyze despite looking complicated at first glance.<\/p>\r\n<p class=\"whitespace-normal break-words\">Telescoping series have the form [latex]\\sum_{n=1}^{\\infty}[b_n - b_{n+1}][\/latex], where consecutive terms are designed to cancel each other out.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>How the cancellation works:<\/strong> When you write out the partial sums:<\/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\">[latex]S_1 = b_1 - b_2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_2 = (b_1 - b_2) + (b_2 - b_3) = b_1 - b_3[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]S_3 = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) = b_1 - b_4[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">The middle terms vanish, leaving [latex]S_k = b_1 - b_{k+1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The series [latex]\\sum_{n=1}^{\\infty}[b_n - b_{n+1}][\/latex] converges if and only if [latex]\\lim_{k \\to \\infty} b_{k+1}[\/latex] exists. If this limit is [latex]B[\/latex], then the series converges to [latex]b_1 - B[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Look for series where you can use partial fractions or other algebraic techniques to rewrite terms as differences. For example, [latex]\\frac{1}{n(n+1)} = \\frac{1}{n} - \\frac{1}{n+1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Common telescoping forms:<\/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\">[latex]\\frac{1}{n(n+1)} = \\frac{1}{n} - \\frac{1}{n+1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{n(n+k)} = \\frac{1}{k}\\left(\\frac{1}{n} - \\frac{1}{n+k}\\right)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Functions like [latex]\\cos(\\frac{1}{n}) - \\cos(\\frac{1}{n+1})[\/latex]<\/li>\r\n<\/ul>\r\nTelescoping series give you exact answers without requiring complex partial sum formulas. They're one of the few types of series where you can find the exact sum rather than just determining convergence.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738185126\" data-type=\"problem\">\r\n<p id=\"fs-id1169738185128\">Determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{e}^{\\frac{1}{n}}-{e}^{\\frac{1}{\\left(n+1\\right)}}\\right][\/latex] converges or diverges. If it converges, find its sum.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558849\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1169737160551\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737160558\">Write out the sequence of partial sums to see which terms cancel.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1169737160537\" data-type=\"solution\">\r\n<p id=\"fs-id1169737160539\">[latex]e - 1[\/latex]<\/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\/Catp5TwohC8?controls=0&amp;start=192&amp;end=255&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\/5.2.3_192to255_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.2.3\" here (opens in new window)<\/a>.\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Understand what we mean by the sum of an infinite series<\/li>\n<li>Find the sum of a geometric series<\/li>\n<li>Calculate the sum of a telescoping series<\/li>\n<\/ul>\n<\/section>\n<h2>Sums and Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">An infinite series is what you get when you try to add infinitely many terms together. Since you can&#8217;t actually perform infinite addition, we define a series through the clever concept of partial sums\u2014finite sums that we can calculate.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The core concept:<\/strong> For series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex], create the sequence of partial sums:<\/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\">[latex]S_1 = a_1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_2 = a_1 + a_2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_3 = a_1 + a_2 + a_3[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">And so on&#8230;<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The series [latex]\\sum_{n=1}^{\\infty} a_n[\/latex] converges to [latex]S[\/latex] if and only if the sequence of partial sums [latex]{S_k}[\/latex] converges to [latex]S[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">To analyze any series, compute several partial sums and look for patterns. Does [latex]S_k[\/latex] seem to approach a finite limit? Does it grow without bound? Does it oscillate?<\/p>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"http:\/\/web.archive.org\/web\/20200216163744\/http:\/\/mathdemos.org\/mathdemos\/donut-demo\/\" target=\"_blank\" rel=\"noopener\">this\u00a0website for a whimsical demonstration of series using donuts<\/a>.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169737822604\" data-type=\"problem\">\n<p id=\"fs-id1169737297457\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\left(n+1\\right)}{n}[\/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=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737294981\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169738073257\">Look at the sequence of partial sums.<\/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-id1169737743047\" data-type=\"solution\">\n<p id=\"fs-id1169738155184\">The series diverges because the [latex]k\\text{th}[\/latex] partial sum [latex]{S}_{k}>k[\/latex].<\/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\/VZkRi47848U?controls=0&amp;start=483&amp;end=620&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\/5.2.1_483to620_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.2.1&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2 data-type=\"title\">The Harmonic Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The harmonic series [latex]\\sum_{n=1}^{\\infty} \\frac{1}{n} = 1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\cdots[\/latex] is one of mathematics&#8217; most deceptive examples\u2014it looks like it should converge, but it actually diverges.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The proof strategy:<\/strong> Group terms cleverly to show the partial sums are unbounded:<\/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\">[latex]S_2 = 1 + \\frac{1}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_4 > 1 + \\frac{1}{2} + \\left(\\frac{1}{4} + \\frac{1}{4}\\right) = 1 + 2 \\cdot \\frac{1}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_8 > 1 + 3 \\cdot \\frac{1}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">In general: [latex]S_{2^j} > 1 + j \\cdot \\frac{1}{2}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>The key insight:<\/strong> Since [latex]1 + j \\cdot \\frac{1}{2} \\to \\infty[\/latex] as [latex]j \\to \\infty[\/latex], the partial sums are unbounded, so the series diverges.<\/p>\n<p>The harmonic series demonstrates that having [latex]\\lim_{n \\to \\infty} a_n = 0[\/latex] is necessary but not sufficient for convergence. The terms must approach zero &#8220;fast enough&#8221;\u2014the harmonic series shows the borderline between convergence and divergence.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738201811\" data-type=\"problem\">\n<p id=\"fs-id1169738044543\">Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{5}{{2}^{n - 1}}[\/latex].<\/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-id1169737910071\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737910078\">Rewrite as [latex]\\displaystyle\\sum _{n=1}^{\\infty }5{\\left(\\frac{1}{2}\\right)}^{n - 1}[\/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=\"q44558896\">Show Solution<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737394304\" data-type=\"solution\">\n<p id=\"fs-id1169737394306\">[latex]10[\/latex].<\/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\/VZkRi47848U?controls=0&amp;start=1056&amp;end=1095&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/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\/5.2.1_1056to1095_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.2.1&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169737168534\" data-type=\"problem\">\n<p id=\"fs-id1169737168537\">Determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{-2}{5}\\right)}^{n - 1}[\/latex] converges or diverges. If it converges, find its sum.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Hint<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737233901\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737233907\">[latex]r=-\\frac{2}{5}[\/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=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737233887\" data-type=\"solution\">\n<p id=\"fs-id1169737233889\">[latex]\\frac{5}{7}[\/latex]<\/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\/05JxlmgOCkM?controls=0&amp;start=0&amp;end=29&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\/5.2.2_0to29_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.2.2&#8221; here (opens in new window)<\/a>.<\/section>\n<h2 data-type=\"title\">Algebraic Properties of Convergent Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Once you know that individual series converge, you can combine them using familiar algebraic operations. These rules work exactly like you&#8217;d expect from your experience with finite sums and limits.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The three fundamental rules:<\/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\"><strong>Sum Rule:<\/strong> [latex]\\sum_{n=1}^{\\infty} (a_n + b_n) = \\sum_{n=1}^{\\infty} a_n + \\sum_{n=1}^{\\infty} b_n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Difference Rule:<\/strong> [latex]\\sum_{n=1}^{\\infty} (a_n - b_n) = \\sum_{n=1}^{\\infty} a_n - \\sum_{n=1}^{\\infty} b_n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Constant Multiple Rule:<\/strong> [latex]\\sum_{n=1}^{\\infty} ca_n = c\\sum_{n=1}^{\\infty} a_n[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Important requirement:<\/strong> Both series must converge individually before you can apply these rules. These rules only apply to convergent series. If either series diverges, you cannot use these algebraic properties.<\/p>\n<p>When facing a complex series, try to break it apart into simpler series that you can evaluate separately, then recombine using these rules.<\/p>\n<\/div>\n<h2 data-type=\"title\">Geometric Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Geometric series are the foundation of series theory because they have a clear pattern and a simple convergence test. Every geometric series has the form [latex]\\sum_{n=1}^{\\infty} ar^{n-1} = a + ar + ar^2 + ar^3 + \\cdots[\/latex] where each term is the previous term multiplied by the common ratio [latex]r[\/latex].<\/p>\n<p>Geometric series give you an exact formula for the sum when they converge, making them the most useful and predictable type of series you&#8217;ll encounter.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The convergence rule is simple:<\/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\">If [latex]|r| < 1[\/latex]: the series converges to [latex]\\frac{a}{1-r}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]|r| \\geq 1[\/latex]: the series diverges<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Why does this work? The partial sum formula [latex]S_k = \\frac{a(1-r^k)}{1-r}[\/latex] (for [latex]r \\neq 1[\/latex]) depends entirely on what happens to [latex]r^k[\/latex] as [latex]k \\to \\infty[\/latex]. When [latex]|r| < 1[\/latex], we have [latex]r^k \\to 0[\/latex], so [latex]S_k \\to \\frac{a}{1-r}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Look for a constant ratio between consecutive terms. The series doesn&#8217;t have to start with [latex]n=1[\/latex] or have exponent [latex]n-1[\/latex]\u2014you can always rewrite it in standard form by factoring out appropriate terms.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key technique:<\/strong> For series like [latex]\\sum_{n=0}^{\\infty} \\left(\\frac{2}{3}\\right)^{n+2}[\/latex], write out the first few terms to identify [latex]a[\/latex] and [latex]r[\/latex], then apply the convergence test.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738115130\" data-type=\"problem\">\n<p id=\"fs-id1169738115132\">Write [latex]5.2\\overline{7}[\/latex] as a fraction of integers.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Hint<\/button><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737394617\" data-type=\"commentary\" data-element-type=\"hint\">\n<div data-type=\"title\"><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">By expressing this number as a series, find a geometric series with initial term [latex]a=\\frac{7}{100}[\/latex] and ratio [latex]r=\\frac{1}{10}[\/latex].<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558889\">Show Solution<\/button><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738115149\" data-type=\"solution\">\n<p id=\"fs-id1169738115151\">[latex]\\frac{475}{90}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2>Telescoping Series<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Telescoping series are special because most terms cancel out when you compute partial sums, leaving only a few terms at the beginning and end. This makes them surprisingly easy to analyze despite looking complicated at first glance.<\/p>\n<p class=\"whitespace-normal break-words\">Telescoping series have the form [latex]\\sum_{n=1}^{\\infty}[b_n - b_{n+1}][\/latex], where consecutive terms are designed to cancel each other out.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>How the cancellation works:<\/strong> When you write out the partial sums:<\/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\">[latex]S_1 = b_1 - b_2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_2 = (b_1 - b_2) + (b_2 - b_3) = b_1 - b_3[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]S_3 = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) = b_1 - b_4[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">The middle terms vanish, leaving [latex]S_k = b_1 - b_{k+1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The series [latex]\\sum_{n=1}^{\\infty}[b_n - b_{n+1}][\/latex] converges if and only if [latex]\\lim_{k \\to \\infty} b_{k+1}[\/latex] exists. If this limit is [latex]B[\/latex], then the series converges to [latex]b_1 - B[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Look for series where you can use partial fractions or other algebraic techniques to rewrite terms as differences. For example, [latex]\\frac{1}{n(n+1)} = \\frac{1}{n} - \\frac{1}{n+1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Common telescoping forms:<\/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\">[latex]\\frac{1}{n(n+1)} = \\frac{1}{n} - \\frac{1}{n+1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{1}{n(n+k)} = \\frac{1}{k}\\left(\\frac{1}{n} - \\frac{1}{n+k}\\right)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Functions like [latex]\\cos(\\frac{1}{n}) - \\cos(\\frac{1}{n+1})[\/latex]<\/li>\n<\/ul>\n<p>Telescoping series give you exact answers without requiring complex partial sum formulas. They&#8217;re one of the few types of series where you can find the exact sum rather than just determining convergence.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738185126\" data-type=\"problem\">\n<p id=\"fs-id1169738185128\">Determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{e}^{\\frac{1}{n}}-{e}^{\\frac{1}{\\left(n+1\\right)}}\\right][\/latex] converges or diverges. If it converges, find its sum.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558849\">Hint<\/button><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737160551\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737160558\">Write out the sequence of partial sums to see which terms cancel.<\/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=\"q44558859\">Show Solution<\/button><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737160537\" data-type=\"solution\">\n<p id=\"fs-id1169737160539\">[latex]e - 1[\/latex]<\/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\/Catp5TwohC8?controls=0&amp;start=192&amp;end=255&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\/5.2.3_192to255_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.2.3&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":17,"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\/876"}],"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":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/876\/revisions"}],"predecessor-version":[{"id":2264,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/876\/revisions\/2264"}],"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\/876\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=876"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=876"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=876"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=876"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}