{"id":1708,"date":"2021-07-23T19:58:02","date_gmt":"2021-07-23T19:58:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1708"},"modified":"2022-03-21T23:02:25","modified_gmt":"2022-03-21T23:02:25","slug":"sums-and-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/sums-and-series\/","title":{"raw":"Sums and Series","rendered":"Sums and Series"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Explain the meaning of the sum of an infinite series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1169738211913\" data-depth=\"1\">\r\n<p id=\"fs-id1169737911581\">An infinite series is a sum of infinitely many terms and is written in the form<\/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&nbsp;\r\n<p id=\"fs-id1169738058915\">But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the <em data-effect=\"italics\">limit<\/em> of partial sums. A partial sum of an infinite series is a finite sum of the form<\/p>\r\n\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&nbsp;\r\n<p id=\"fs-id1169737953074\">To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that [latex]1000[\/latex] gallons enters the lake the first week. During the second week, an additional [latex]500[\/latex] gallons of oil enters the lake. The third week, [latex]250[\/latex] more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after [latex]k[\/latex] weeks. Letting [latex]{S}_{k}[\/latex] denote the amount of oil in the lake (measured in thousands of gallons) after [latex]k[\/latex] weeks, we see that<\/p>\r\n\r\n<div id=\"fs-id1169738226744\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[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]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737848219\">Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after [latex]k[\/latex] weeks is<\/p>\r\n\r\n<div id=\"fs-id1169738113913\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[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].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738107779\">We are interested in what happens as [latex]k\\to \\infty [\/latex]. Symbolically, the amount of oil in the lake as [latex]k\\to \\infty [\/latex] is given by the infinite series<\/p>\r\n\r\n<div id=\"fs-id1169738044938\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[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].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737946990\">At the same time, as [latex]k\\to \\infty [\/latex], the amount of oil in the lake can be calculated by evaluating [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}[\/latex]. Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex]. If the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges, we say that the infinite series converges, and its sum is given by [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}[\/latex]. If the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex].<\/p>\r\n<p id=\"fs-id1169737844367\">First, simplifying some of these partial sums, we see that<\/p>\r\n\r\n<div id=\"fs-id1169737757878\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=1\\hfill \\\\ {S}_{2}=1+\\frac{1}{2}=\\frac{3}{2}\\hfill \\\\ {S}_{3}=1+\\frac{1}{2}+\\frac{1}{4}=\\frac{7}{4}\\hfill \\\\ {S}_{4}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}=\\frac{15}{8}\\hfill \\\\ {S}_{5}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}=\\frac{31}{16}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738067359\">Plotting some of these values in Figure 1, it appears that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] could be approaching 2.<\/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<p id=\"fs-id1169737997707\">Let\u2019s look for more convincing evidence. 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]. Later we will provide an analytic argument that can be used to prove that [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}=2[\/latex]. For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to [latex]2[\/latex]. Since this sequence of partial sums converges to [latex]2[\/latex], we say the infinite series converges to [latex]2[\/latex] and write<\/p>\r\n\r\n<div id=\"fs-id1169737845005\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}=2[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737713942\">Returning to the question about the oil in the lake, since this infinite series converges to [latex]2[\/latex], we conclude that the amount of oil in the lake will get arbitrarily close to [latex]2000[\/latex] gallons as the amount of time gets sufficiently large.<\/p>\r\n<p id=\"fs-id1169737906827\">This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.<\/p>\r\n\r\n<div id=\"fs-id1169737711861\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\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]. If the sequence of partial sums converges to a real number [latex]S[\/latex], the infinite series converges. If we can describe the <span data-type=\"term\">convergence of a series<\/span> to [latex]S[\/latex], we call [latex]S[\/latex] the sum of the series, and we write<\/p>\r\n\r\n<div id=\"fs-id1169738111008\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=S[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737935268\">If the sequence of partial sums diverges, we have the <strong>divergence of a series<\/strong>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737160608\" class=\"media-2\" data-type=\"note\">\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1169737948505\">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>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738183537\">Note that the index for a series need not begin with [latex]n=1[\/latex] but can begin with any value. For example, the series<\/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&nbsp;\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&nbsp;\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&nbsp;\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&nbsp;\r\n<div id=\"fs-id1169738068351\" data-type=\"example\">\r\n<div id=\"fs-id1169738148826\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738005574\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating Limits of Sequences of Partial Sums<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169737263906\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169737822602\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737822604\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<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\"><\/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<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169362[\/ohm_question]\r\n\r\n<\/div>\r\n<section id=\"fs-id1169737192263\" data-depth=\"2\"><\/section><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Explain the meaning of the sum of an infinite series<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1169738211913\" data-depth=\"1\">\n<p id=\"fs-id1169737911581\">An infinite series is a sum of infinitely many terms and is written in the form<\/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>&nbsp;<\/p>\n<p id=\"fs-id1169738058915\">But what does this mean? We cannot add an infinite number of terms in the same way we can add a finite number of terms. Instead, the value of an infinite series is defined in terms of the <em data-effect=\"italics\">limit<\/em> of partial sums. A partial sum of an infinite series is a finite sum of the form<\/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<p>&nbsp;<\/p>\n<p id=\"fs-id1169737953074\">To see how we use partial sums to evaluate infinite series, consider the following example. Suppose oil is seeping into a lake such that [latex]1000[\/latex] gallons enters the lake the first week. During the second week, an additional [latex]500[\/latex] gallons of oil enters the lake. The third week, [latex]250[\/latex] more gallons enters the lake. Assume this pattern continues such that each week half as much oil enters the lake as did the previous week. If this continues forever, what can we say about the amount of oil in the lake? Will the amount of oil continue to get arbitrarily large, or is it possible that it approaches some finite amount? To answer this question, we look at the amount of oil in the lake after [latex]k[\/latex] weeks. Letting [latex]{S}_{k}[\/latex] denote the amount of oil in the lake (measured in thousands of gallons) after [latex]k[\/latex] weeks, we see that<\/p>\n<div id=\"fs-id1169738226744\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[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]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737848219\">Looking at this pattern, we see that the amount of oil in the lake (in thousands of gallons) after [latex]k[\/latex] weeks is<\/p>\n<div id=\"fs-id1169738113913\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[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].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738107779\">We are interested in what happens as [latex]k\\to \\infty[\/latex]. Symbolically, the amount of oil in the lake as [latex]k\\to \\infty[\/latex] is given by the infinite series<\/p>\n<div id=\"fs-id1169738044938\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[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].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737946990\">At the same time, as [latex]k\\to \\infty[\/latex], the amount of oil in the lake can be calculated by evaluating [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}[\/latex]. Therefore, the behavior of the infinite series can be determined by looking at the behavior of the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex]. If the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] converges, we say that the infinite series converges, and its sum is given by [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}[\/latex]. If the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] diverges, we say the infinite series diverges. We now turn our attention to determining the limit of this sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex].<\/p>\n<p id=\"fs-id1169737844367\">First, simplifying some of these partial sums, we see that<\/p>\n<div id=\"fs-id1169737757878\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=1\\hfill \\\\ {S}_{2}=1+\\frac{1}{2}=\\frac{3}{2}\\hfill \\\\ {S}_{3}=1+\\frac{1}{2}+\\frac{1}{4}=\\frac{7}{4}\\hfill \\\\ {S}_{4}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}=\\frac{15}{8}\\hfill \\\\ {S}_{5}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}=\\frac{31}{16}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738067359\">Plotting some of these values in Figure 1, it appears that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] could be approaching 2.<\/p>\n<figure id=\"CNX_Calc_Figure_09_02_001\"><figcaption><\/figcaption><div style=\"width: 335px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1169737997707\">Let\u2019s look for more convincing evidence. 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]. Later we will provide an analytic argument that can be used to prove that [latex]\\underset{k\\to \\infty }{\\text{lim}}{S}_{k}=2[\/latex]. For now, we rely on the numerical and graphical data to convince ourselves that the sequence of partial sums does actually converge to [latex]2[\/latex]. Since this sequence of partial sums converges to [latex]2[\/latex], we say the infinite series converges to [latex]2[\/latex] and write<\/p>\n<div id=\"fs-id1169737845005\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}=2[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737713942\">Returning to the question about the oil in the lake, since this infinite series converges to [latex]2[\/latex], we conclude that the amount of oil in the lake will get arbitrarily close to [latex]2000[\/latex] gallons as the amount of time gets sufficiently large.<\/p>\n<p id=\"fs-id1169737906827\">This series is an example of a geometric series. We discuss geometric series in more detail later in this section. First, we summarize what it means for an infinite series to converge.<\/p>\n<div id=\"fs-id1169737711861\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\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]. If the sequence of partial sums converges to a real number [latex]S[\/latex], the infinite series converges. If we can describe the <span data-type=\"term\">convergence of a series<\/span> to [latex]S[\/latex], we call [latex]S[\/latex] the sum of the series, and we write<\/p>\n<div id=\"fs-id1169738111008\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=S[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737935268\">If the sequence of partial sums diverges, we have the <strong>divergence of a series<\/strong>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737160608\" class=\"media-2\" data-type=\"note\">\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1169737948505\">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>.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738183537\">Note that the index for a series need not begin with [latex]n=1[\/latex] but can begin with any value. For example, the series<\/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>&nbsp;<\/p>\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>&nbsp;<\/p>\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>&nbsp;<\/p>\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<p>&nbsp;<\/p>\n<div id=\"fs-id1169738068351\" data-type=\"example\">\n<div id=\"fs-id1169738148826\" data-type=\"exercise\">\n<div id=\"fs-id1169738005574\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating Limits of Sequences of Partial Sums<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737263906\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169737822602\" data-type=\"exercise\">\n<div id=\"fs-id1169737822604\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/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\"><\/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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169362\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169362&theme=oea&iframe_resize_id=ohm169362&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<section id=\"fs-id1169737192263\" data-depth=\"2\"><\/section>\n<\/section>\n<div data-type=\"glossary\"><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1708\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>5.2.1. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.2.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1708","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1708","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":5,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1708\/revisions"}],"predecessor-version":[{"id":2212,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1708\/revisions\/2212"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1708\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1708"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1708"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1708"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}