{"id":871,"date":"2025-06-20T17:19:11","date_gmt":"2025-06-20T17:19:11","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=871"},"modified":"2025-09-11T16:03:00","modified_gmt":"2025-09-11T16:03:00","slug":"direction-fields-and-eulers-method-learn-it-1-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-learn-it-1-2\/","title":{"raw":"Introduction to Series: Learn It 1","rendered":"Introduction to Series: Learn It 1"},"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<p class=\"whitespace-normal break-words\">You've already worked with sequences\u2014ordered sets of terms like [latex]a_1, a_2, a_3, \\ldots[\/latex]. When you add the terms of a sequence together, you get a <strong>series<\/strong>.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">An infinite series is the sum of infinitely many terms, written as:<\/p>\r\n\r\n<div id=\"fs-id1169737839089\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+ \\cdots [\/latex].<\/div>\r\n<p id=\"fs-id1169738058915\">But what does it mean to add infinitely many terms? We can't simply add them the way we add a finite number of terms. Instead, we define the value of an infinite series using <strong>partial sums<\/strong>.<\/p>\r\nA partial sum is a finite sum of the first [latex]k[\/latex] terms:\r\n<div id=\"fs-id1169737906700\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+ \\cdots +{a}_{k}[\/latex].<\/div>\r\n<div data-type=\"equation\" data-label=\"\">The value of the infinite series is then defined as the limit of these partial sums as [latex]k \\to \\infty[\/latex].<\/div>\r\nLet's see how this works with a real scenario.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">Suppose oil seeps into a lake following this pattern:<\/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\">Week 1: [latex]1000 [\/latex] gallons enter the lake<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Week 2: [latex]500 [\/latex] gallons enter (half of previous week)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Week 3: [latex]250 [\/latex] gallons enter (half of previous week)<\/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\">Each week, half as much oil enters compared to the previous week. The question is: if this continues forever, will the total amount of oil grow without bound, or will it approach some finite limit?<\/p>\r\nLet [latex]S_k[\/latex] represent the total oil (in thousands of gallons) after [latex]k[\/latex] weeks:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{1}=1\\hfill \\\\ {S}_{2}=1+0.5=1+\\frac{1}{2}\\hfill \\\\ {S}_{3}=1+0.5+0.25=1+\\frac{1}{2}+\\frac{1}{4}\\hfill \\\\ {S}_{4}=1+0.5+0.25+0.125=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}\\hfill \\\\ {S}_{5}=1+0.5+0.25+0.125+0.0625=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}.\\hfill \\end{array}[\/latex]<\/p>\r\nLooking at this pattern, the total oil after [latex]k[\/latex] weeks is:\r\n<p style=\"text-align: center;\">[latex]{S}_{k}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+ \\cdots +\\frac{1}{{2}^{k - 1}}=\\displaystyle\\sum _{n=1}^{k}{\\left(\\frac{1}{2}\\right)}^{n - 1}[\/latex].<\/p>\r\nAs [latex]k \\to \\infty[\/latex], the total amount of oil is represented by the infinite series:\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\cdots [\/latex].<\/p>\r\n\r\n<\/section>The behavior of an infinite series depends entirely on what happens to its sequence of partial sums [latex]{S_k}[\/latex] as [latex]k \\to \\infty[\/latex].\r\n<p class=\"whitespace-normal break-words\">To find [latex]\\lim_{k \\to \\infty} S_k[\/latex], let's calculate some partial sums:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">[latex]S_1 = 1[\/latex] [latex]S_2 = 1 + \\frac{1}{2} = \\frac{3}{2} = 1.5[\/latex] [latex]S_3 = 1 + \\frac{1}{2} + \\frac{1}{4} = \\frac{7}{4} = 1.75[\/latex] [latex]S_4 = 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} = \\frac{15}{8} = 1.875[\/latex] [latex]S_5 = 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} = \\frac{31}{16} = 1.9375[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Plotting some of these values in Figure 1, it appears the sequence [latex]{S_k}[\/latex] might be approaching [latex]2[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_02_001\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234316\/CNX_Calc_Figure_09_02_001.jpg\" alt=\"This is a graph in quadrant 1with the x and y axes labeled n and S_n, respectively. From 1 to 5, points are plotted. They increase and seem to converge to 2 and n goes to infinity.\" width=\"325\" height=\"239\" data-media-type=\"image\/jpeg\" \/> Figure 1. The graph shows the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex]. It appears that the sequence is approaching the value [latex]2[\/latex].[\/caption]<\/figure>\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p id=\"fs-id1169737997707\"><strong>More Evidence:<\/strong> In the following table, we list the values of [latex]{S}_{k}[\/latex] for several values of [latex]k[\/latex].<\/p>\r\n\r\n<table id=\"fs-id1169737896803\" class=\"unnumbered\" summary=\"This is a table with two rows and five columns. The first row is labeled \" data-label=\"\">\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"left\">[latex]k[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]10[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]15[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]20[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td data-valign=\"top\" data-align=\"left\">[latex]{S}_{k}[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]1.9375[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]1.998[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]1.999939[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]1.999998[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1169738226260\">These data supply more evidence suggesting that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges to [latex]2[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">This numerical evidence strongly suggests that [latex]\\lim_{k \\to \\infty} S_k = 2[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Since the sequence of partial sums converges to [latex]2[\/latex], we say the infinite series <strong>converges to [latex]2[\/latex]<\/strong> and write:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\sum_{n=1}^{\\infty} \\left(\\frac{1}{2}\\right)^{n-1} = 2[\/latex]<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Returning to our question about the oil in the lake, since this series converges to [latex]2[\/latex], the total amount of oil in the lake will approach [latex]2000 [\/latex] gallons (remember our units were thousands of gallons), no matter how long the spill continues.<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>infinite series<\/h3>\r\n<p id=\"fs-id1169738143828\">An <strong>infinite series<\/strong> is an expression of the form<\/p>\r\n\r\n<div id=\"fs-id1169738150202\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737256022\">For each positive integer [latex]k[\/latex], the sum<\/p>\r\n\r\n<div id=\"fs-id1169737841847\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots +{a}_{k}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737712136\">is called the [latex]k\\text{th}[\/latex] <strong>partial sum<\/strong> of the infinite series. The partial sums form a sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex].<\/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]\\lim_{k \\to \\infty} S_k = S[\/latex] (a finite number), the series <strong>converges<\/strong> to [latex]S[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]\\lim_{k \\to \\infty} S_k[\/latex] does not exist or is infinite, the series <strong>diverges<\/strong><\/li>\r\n<\/ul>\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">The starting index of a series can be adjusted for convenience. For example:<\/p>\r\n\r\n<div id=\"fs-id1169737850303\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{2}\\right)}^{n - 1}[\/latex]<\/div>\r\n<p id=\"fs-id1169738053534\">can also be written as<\/p>\r\n\r\n<div id=\"fs-id1169737142209\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\frac{1}{2}\\right)}^{n}\\text{or}{\\displaystyle\\sum _{n=5}^{\\infty }\\left(\\frac{1}{2}\\right)}^{n - 5}[\/latex].<\/div>\r\n<p id=\"fs-id1169737954070\">Often it is convenient for the index to begin at [latex]1[\/latex], so if for some reason it begins at a different value, we can reindex by making a change of variables. For example, consider the series<\/p>\r\n\r\n<div id=\"fs-id1169737806291\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\r\n<p id=\"fs-id1169737931594\">By introducing the variable [latex]m=n - 1[\/latex], so that [latex]n=m+1[\/latex], we can rewrite the series as<\/p>\r\n\r\n<div id=\"fs-id1169737846917\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{m=1}^{\\infty }\\frac{1}{{\\left(m+1\\right)}^{2}}[\/latex].<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738005574\" data-type=\"problem\">\r\n<p id=\"fs-id1169737796336\">For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169737856979\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{n+1}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\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-id1169738005710\" data-type=\"solution\">\r\n<ol id=\"fs-id1169737792322\" type=\"a\">\r\n \t<li>The sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] satisfies<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737796047\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\frac{1}{2}+\\frac{2}{3}\\hfill \\\\ {S}_{3}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}\\hfill \\\\ {S}_{4}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}+\\frac{4}{5}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNotice that each term added is greater than [latex]\\frac{1}{2}[\/latex]. As a result, we see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737700342\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\frac{1}{2}+\\frac{2}{3}&gt;\\frac{1}{2}+\\frac{1}{2}=2\\left(\\frac{1}{2}\\right)\\hfill \\\\ {S}_{3}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}&gt;\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}=3\\left(\\frac{1}{2}\\right)\\hfill \\\\ {S}_{4}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}+\\frac{4}{5}&gt;\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}=4\\left(\\frac{1}{2}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFrom this pattern we can see that [latex]{S}_{k}&gt;k\\left(\\frac{1}{2}\\right)[\/latex] for every integer [latex]k[\/latex]. Therefore, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded and consequently, diverges. Therefore, the infinite series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{\\left(n+1\\right)}[\/latex] diverges.<\/li>\r\n \t<li>The sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] satisfies<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738144263\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=-1\\hfill \\\\ {S}_{2}=-1+1=0\\hfill \\\\ {S}_{3}=-1+1 - 1=-1\\hfill \\\\ {S}_{4}=-1+1 - 1+1=0.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFrom this pattern we can see the sequence of partial sums is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737790076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{{S}_{k}\\right\\}=\\left\\{-1,0,-1,0\\text{,\\ldots }\\right\\}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince this sequence diverges, the infinite series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}[\/latex] diverges.<\/li>\r\n \t<li>The sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] satisfies<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738064000\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=\\frac{1}{1\\cdot 2}=\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}=\\frac{1}{2}+\\frac{1}{6}=\\frac{2}{3}\\hfill \\\\ {S}_{3}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}+\\frac{1}{3\\cdot 4}=\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}=\\frac{3}{4}\\hfill \\\\ {S}_{4}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}+\\frac{1}{3\\cdot 4}+\\frac{1}{4\\cdot 5}=\\frac{4}{5}\\hfill \\\\ {S}_{5}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}+\\frac{1}{3\\cdot 4}+\\frac{1}{4\\cdot 5}+\\frac{1}{5\\cdot 6}=\\frac{5}{6}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFrom this pattern, we can see that the [latex]k\\text{th}[\/latex] partial sum is given by the explicit formula<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737806522\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\frac{k}{k+1}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\frac{k}{\\left(k+1\\right)}\\to 1[\/latex], we conclude that the sequence of partial sums converges, and therefore the infinite series converges to [latex]1[\/latex]. We have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737933604\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}=1[\/latex].<\/div>\r\n&nbsp;<\/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>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<p class=\"whitespace-normal break-words\">You&#8217;ve already worked with sequences\u2014ordered sets of terms like [latex]a_1, a_2, a_3, \\ldots[\/latex]. When you add the terms of a sequence together, you get a <strong>series<\/strong>.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">An infinite series is the sum of infinitely many terms, written as:<\/p>\n<div id=\"fs-id1169737839089\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+ \\cdots[\/latex].<\/div>\n<p id=\"fs-id1169738058915\">But what does it mean to add infinitely many terms? We can&#8217;t simply add them the way we add a finite number of terms. Instead, we define the value of an infinite series using <strong>partial sums<\/strong>.<\/p>\n<p>A partial sum is a finite sum of the first [latex]k[\/latex] terms:<\/p>\n<div id=\"fs-id1169737906700\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+ \\cdots +{a}_{k}[\/latex].<\/div>\n<div data-type=\"equation\" data-label=\"\">The value of the infinite series is then defined as the limit of these partial sums as [latex]k \\to \\infty[\/latex].<\/div>\n<p>Let&#8217;s see how this works with a real scenario.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">Suppose oil seeps into a lake following this pattern:<\/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\">Week 1: [latex]1000[\/latex] gallons enter the lake<\/li>\n<li class=\"whitespace-normal break-words\">Week 2: [latex]500[\/latex] gallons enter (half of previous week)<\/li>\n<li class=\"whitespace-normal break-words\">Week 3: [latex]250[\/latex] gallons enter (half of previous week)<\/li>\n<li class=\"whitespace-normal break-words\">And so on&#8230;<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Each week, half as much oil enters compared to the previous week. The question is: if this continues forever, will the total amount of oil grow without bound, or will it approach some finite limit?<\/p>\n<p>Let [latex]S_k[\/latex] represent the total oil (in thousands of gallons) after [latex]k[\/latex] weeks:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{1}=1\\hfill \\\\ {S}_{2}=1+0.5=1+\\frac{1}{2}\\hfill \\\\ {S}_{3}=1+0.5+0.25=1+\\frac{1}{2}+\\frac{1}{4}\\hfill \\\\ {S}_{4}=1+0.5+0.25+0.125=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}\\hfill \\\\ {S}_{5}=1+0.5+0.25+0.125+0.0625=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Looking at this pattern, the total oil after [latex]k[\/latex] weeks is:<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{k}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+ \\cdots +\\frac{1}{{2}^{k - 1}}=\\displaystyle\\sum _{n=1}^{k}{\\left(\\frac{1}{2}\\right)}^{n - 1}[\/latex].<\/p>\n<p>As [latex]k \\to \\infty[\/latex], the total amount of oil is represented by the infinite series:<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\cdots[\/latex].<\/p>\n<\/section>\n<p>The behavior of an infinite series depends entirely on what happens to its sequence of partial sums [latex]{S_k}[\/latex] as [latex]k \\to \\infty[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">To find [latex]\\lim_{k \\to \\infty} S_k[\/latex], let&#8217;s calculate some partial sums:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]S_1 = 1[\/latex] [latex]S_2 = 1 + \\frac{1}{2} = \\frac{3}{2} = 1.5[\/latex] [latex]S_3 = 1 + \\frac{1}{2} + \\frac{1}{4} = \\frac{7}{4} = 1.75[\/latex] [latex]S_4 = 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} = \\frac{15}{8} = 1.875[\/latex] [latex]S_5 = 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\frac{1}{16} = \\frac{31}{16} = 1.9375[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Plotting some of these values in Figure 1, it appears the sequence [latex]{S_k}[\/latex] might be approaching [latex]2[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_09_02_001\"><figcaption><\/figcaption><figure style=\"width: 325px\" 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\/11234316\/CNX_Calc_Figure_09_02_001.jpg\" alt=\"This is a graph in quadrant 1with the x and y axes labeled n and S_n, respectively. From 1 to 5, points are plotted. They increase and seem to converge to 2 and n goes to infinity.\" width=\"325\" height=\"239\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. The graph shows the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex]. It appears that the sequence is approaching the value [latex]2[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p id=\"fs-id1169737997707\"><strong>More Evidence:<\/strong> In the following table, we list the values of [latex]{S}_{k}[\/latex] for several values of [latex]k[\/latex].<\/p>\n<table id=\"fs-id1169737896803\" class=\"unnumbered\" summary=\"This is a table with two rows and five columns. The first row is labeled\" data-label=\"\">\n<tbody>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"left\">[latex]k[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]5[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]10[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]15[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]20[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td data-valign=\"top\" data-align=\"left\">[latex]{S}_{k}[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]1.9375[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]1.998[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]1.999939[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]1.999998[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1169738226260\">These data supply more evidence suggesting that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges to [latex]2[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">This numerical evidence strongly suggests that [latex]\\lim_{k \\to \\infty} S_k = 2[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Since the sequence of partial sums converges to [latex]2[\/latex], we say the infinite series <strong>converges to [latex]2[\/latex]<\/strong> and write:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\sum_{n=1}^{\\infty} \\left(\\frac{1}{2}\\right)^{n-1} = 2[\/latex]<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Returning to our question about the oil in the lake, since this series converges to [latex]2[\/latex], the total amount of oil in the lake will approach [latex]2000[\/latex] gallons (remember our units were thousands of gallons), no matter how long the spill continues.<\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>infinite series<\/h3>\n<p id=\"fs-id1169738143828\">An <strong>infinite series<\/strong> is an expression of the form<\/p>\n<div id=\"fs-id1169738150202\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737256022\">For each positive integer [latex]k[\/latex], the sum<\/p>\n<div id=\"fs-id1169737841847\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}{a}_{n}={a}_{1}+{a}_{2}+{a}_{3}+\\cdots +{a}_{k}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737712136\">is called the [latex]k\\text{th}[\/latex] <strong>partial sum<\/strong> of the infinite series. The partial sums form a sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex].<\/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]\\lim_{k \\to \\infty} S_k = S[\/latex] (a finite number), the series <strong>converges<\/strong> to [latex]S[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]\\lim_{k \\to \\infty} S_k[\/latex] does not exist or is infinite, the series <strong>diverges<\/strong><\/li>\n<\/ul>\n<\/section>\n<p class=\"whitespace-normal break-words\">The starting index of a series can be adjusted for convenience. For example:<\/p>\n<div id=\"fs-id1169737850303\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\sum _{n=1}^{\\infty }\\left(\\frac{1}{2}\\right)}^{n - 1}[\/latex]<\/div>\n<p id=\"fs-id1169738053534\">can also be written as<\/p>\n<div id=\"fs-id1169737142209\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\frac{1}{2}\\right)}^{n}\\text{or}{\\displaystyle\\sum _{n=5}^{\\infty }\\left(\\frac{1}{2}\\right)}^{n - 5}[\/latex].<\/div>\n<p id=\"fs-id1169737954070\">Often it is convenient for the index to begin at [latex]1[\/latex], so if for some reason it begins at a different value, we can reindex by making a change of variables. For example, consider the series<\/p>\n<div id=\"fs-id1169737806291\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\n<p id=\"fs-id1169737931594\">By introducing the variable [latex]m=n - 1[\/latex], so that [latex]n=m+1[\/latex], we can rewrite the series as<\/p>\n<div id=\"fs-id1169737846917\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{m=1}^{\\infty }\\frac{1}{{\\left(m+1\\right)}^{2}}[\/latex].<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738005574\" data-type=\"problem\">\n<p id=\"fs-id1169737796336\">For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.<\/p>\n<ol id=\"fs-id1169737856979\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{n+1}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\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-id1169738005710\" data-type=\"solution\">\n<ol id=\"fs-id1169737792322\" type=\"a\">\n<li>The sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] satisfies<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737796047\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\frac{1}{2}+\\frac{2}{3}\\hfill \\\\ {S}_{3}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}\\hfill \\\\ {S}_{4}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}+\\frac{4}{5}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNotice that each term added is greater than [latex]\\frac{1}{2}[\/latex]. As a result, we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737700342\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\frac{1}{2}+\\frac{2}{3}>\\frac{1}{2}+\\frac{1}{2}=2\\left(\\frac{1}{2}\\right)\\hfill \\\\ {S}_{3}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}>\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}=3\\left(\\frac{1}{2}\\right)\\hfill \\\\ {S}_{4}=\\frac{1}{2}+\\frac{2}{3}+\\frac{3}{4}+\\frac{4}{5}>\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}=4\\left(\\frac{1}{2}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFrom this pattern we can see that [latex]{S}_{k}>k\\left(\\frac{1}{2}\\right)[\/latex] for every integer [latex]k[\/latex]. Therefore, [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded and consequently, diverges. Therefore, the infinite series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{n}{\\left(n+1\\right)}[\/latex] diverges.<\/li>\n<li>The sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] satisfies<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738144263\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=-1\\hfill \\\\ {S}_{2}=-1+1=0\\hfill \\\\ {S}_{3}=-1+1 - 1=-1\\hfill \\\\ {S}_{4}=-1+1 - 1+1=0.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFrom this pattern we can see the sequence of partial sums is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737790076\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{{S}_{k}\\right\\}=\\left\\{-1,0,-1,0\\text{,\\ldots }\\right\\}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince this sequence diverges, the infinite series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}[\/latex] diverges.<\/li>\n<li>The sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] satisfies<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738064000\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=\\frac{1}{1\\cdot 2}=\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}=\\frac{1}{2}+\\frac{1}{6}=\\frac{2}{3}\\hfill \\\\ {S}_{3}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}+\\frac{1}{3\\cdot 4}=\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{12}=\\frac{3}{4}\\hfill \\\\ {S}_{4}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}+\\frac{1}{3\\cdot 4}+\\frac{1}{4\\cdot 5}=\\frac{4}{5}\\hfill \\\\ {S}_{5}=\\frac{1}{1\\cdot 2}+\\frac{1}{2\\cdot 3}+\\frac{1}{3\\cdot 4}+\\frac{1}{4\\cdot 5}+\\frac{1}{5\\cdot 6}=\\frac{5}{6}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFrom this pattern, we can see that the [latex]k\\text{th}[\/latex] partial sum is given by the explicit formula<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737806522\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\frac{k}{k+1}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\frac{k}{\\left(k+1\\right)}\\to 1[\/latex], we conclude that the sequence of partial sums converges, and therefore the infinite series converges to [latex]1[\/latex]. We have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737933604\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}=1[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":12,"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\/871"}],"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":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/871\/revisions"}],"predecessor-version":[{"id":2317,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/871\/revisions\/2317"}],"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\/871\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=871"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=871"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=871"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=871"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}