{"id":1703,"date":"2021-07-23T19:43:51","date_gmt":"2021-07-23T19:43:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1703"},"modified":"2022-03-21T23:03:05","modified_gmt":"2022-03-21T23:03:05","slug":"different-types-of-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/different-types-of-series\/","title":{"raw":"Different Types of Series","rendered":"Different Types of Series"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the sum of a geometric series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">The Harmonic Series<\/h2>\r\n<p id=\"fs-id1169737935177\">A useful series to know about is the <span data-type=\"term\">harmonic series<\/span>. The harmonic series is defined as<\/p>\r\n\r\n<div id=\"fs-id1169738057265\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737393217\">This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] in the following table.<\/p>\r\n\r\n<table id=\"fs-id1169738150006\" class=\"unnumbered\" summary=\"This is a table with two rows and seven 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]10[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]100[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]1000[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]10,000[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]100,000[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]1,000,000[\/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]2.92897[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]5.18738[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]7.48547[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]9.78761[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]12.09015[\/latex]<\/td>\r\n<td data-valign=\"top\" data-align=\"left\">[latex]14.39273[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1169737910299\">Even after [latex]1,000,000[\/latex] terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.<\/p>\r\n<p id=\"fs-id1169737168935\">To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:<\/p>\r\n\r\n<div id=\"fs-id1169737292630\" 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}\\hfill \\\\ {S}_{3}=1+\\frac{1}{2}+\\frac{1}{3}\\hfill \\\\ {S}_{4}=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738078962\">Notice that for the last two terms in [latex]{S}_{4}[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169738078297\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{3}+\\frac{1}{4}&gt;\\frac{1}{4}+\\frac{1}{4}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737818894\">Therefore, we conclude that<\/p>\r\n\r\n<div id=\"fs-id1169738250171\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{4}&gt;1+\\frac{1}{2}+\\left(\\frac{1}{4}+\\frac{1}{4}\\right)=1+\\frac{1}{2}+\\frac{1}{2}=1+2\\left(\\frac{1}{2}\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737265736\">Using the same idea for [latex]{S}_{8}[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1169737809241\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {S}_{8}&amp; =1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}&gt;1+\\frac{1}{2}+\\left(\\frac{1}{4}+\\frac{1}{4}\\right)+\\left(\\frac{1}{8}+\\frac{1}{8}+\\frac{1}{8}+\\frac{1}{8}\\right)\\hfill \\\\ &amp; =1+\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}=1+3\\left(\\frac{1}{2}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738165396\">From this pattern, we see that [latex]{S}_{1}=1[\/latex], [latex]{S}_{2}=1+\\frac{1}{2}[\/latex], [latex]{S}_{4}&gt;1+2\\left(\\frac{1}{2}\\right)[\/latex], and [latex]{S}_{8}&gt;1+3\\left(\\frac{1}{2}\\right)[\/latex]. More generally, it can be shown that [latex]{S}_{{2}^{j}}&gt;1+j\\left(\\frac{1}{2}\\right)[\/latex] for all [latex]j&gt;1[\/latex]. Since [latex]1+j\\left(\\frac{1}{2}\\right)\\to \\infty [\/latex], we conclude that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded, it diverges. Thus, the harmonic series diverges.<\/p>\r\n\r\n<section id=\"fs-id1169738056466\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Algebraic Properties of Convergent Series<\/h2>\r\n<p id=\"fs-id1169738188266\">Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.<\/p>\r\n\r\n<div id=\"fs-id1169738189943\" class=\"theorem\" 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\">Theorem: Algebraic Properties of Convergent Series<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169737179327\">Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] be convergent series. Then the following algebraic properties hold.<\/p>\r\n\r\n<ol id=\"fs-id1169737947446\" type=\"i\">\r\n \t<li>The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. (Sum Rule)<\/li>\r\n \t<li>The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. (Difference Rule)<\/li>\r\n \t<li>For any real number [latex]c[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}=c\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. (Constant Multiple Rule)<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169738055547\" data-type=\"example\">\r\n<div id=\"fs-id1169738055550\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738148599\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Using Algebraic Properties of Convergent Series<\/h3>\r\n<div id=\"fs-id1169738148599\" data-type=\"problem\">\r\n<p id=\"fs-id1169738058908\">Evaluate<\/p>\r\n\r\n<div id=\"fs-id1169738058911\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right][\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169738042181\" data-type=\"solution\">\r\n<p id=\"fs-id1169738042184\">We showed earlier that<\/p>\r\n\r\n<div id=\"fs-id1169737949111\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737297603\">and<\/p>\r\n\r\n<div id=\"fs-id1169738189920\" 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-id1169737209441\">Since both of those series converge, we can apply the Alegraic Properties of Convergent Series to evaluate<\/p>\r\n\r\n<div id=\"fs-id1169737355719\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right][\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738146272\">Using the sum rule, write<\/p>\r\n\r\n<div id=\"fs-id1169738146275\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right]=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3}{n\\left(n+1\\right)}\\underset{n=1}{\\overset{\\infty }{+\\displaystyle\\sum }}{\\left(\\frac{1}{2}\\right)}^{n - 2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738248902\">Then, using the constant multiple rule and the sums above, we can conclude that<\/p>\r\n\r\n<div id=\"fs-id1169738086671\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ll}{\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3}{n\\left(n+1\\right)}+\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 2}}\\hfill &amp; =3{\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{-1}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}}\\hfill \\\\ &amp; =3\\left(1\\right)+{\\left(\\frac{1}{2}\\right)}^{-1}\\left(2\\right)=3+2\\left(2\\right)=7.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n\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-id1169738201806\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738201809\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738201811\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/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=1056&amp;end=1095&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><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 id=\"fs-id1169737395366\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Geometric Series<\/h2>\r\n<p id=\"fs-id1169737395484\">A <strong>geometric series<\/strong> is any series that we can write in the form<\/p>\r\n\r\n<div id=\"fs-id1169738080256\" style=\"text-align: center;\" data-type=\"equation\">[latex]a+ar+a{r}^{2}+a{r}^{3}+\\cdots =\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737796160\">Because the ratio of each term in this series to the previous term is <em data-effect=\"italics\">r<\/em>, the number <em data-effect=\"italics\">r<\/em> is called the ratio. We refer to <em data-effect=\"italics\">a<\/em> as the initial term because it is the first term in the series. For example, the series<\/p>\r\n\r\n<div id=\"fs-id1169737179889\" 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}+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737227806\">is a geometric series with initial term [latex]a=1[\/latex] and ratio [latex]r=\\frac{1}{2}[\/latex].<\/p>\r\n<p id=\"fs-id1169738234421\">In general, when does a geometric series converge? Consider the geometric series<\/p>\r\n\r\n<div id=\"fs-id1169738234424\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738046261\">when [latex]a&gt;0[\/latex]. Its sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1169737934871\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}a{r}^{n - 1}=a+ar+a{r}^{2}+\\cdots +a{r}^{k - 1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737394756\">Consider the case when [latex]r=1[\/latex]. In that case,<\/p>\r\n\r\n<div id=\"fs-id1169738080270\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=a+a\\left(1\\right)+a{\\left(1\\right)}^{2}+\\cdots +a{\\left(1\\right)}^{k - 1}=ak[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737934742\">Since [latex]a&gt;0[\/latex], we know [latex]ak\\to \\infty [\/latex] as [latex]k\\to \\infty [\/latex]. Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for [latex]r=1[\/latex]. For [latex]r\\ne 1[\/latex], to find the limit of [latex]\\left\\{{S}_{k}\\right\\}[\/latex], multiply the geometric series general equation by [latex]1-r[\/latex]. Doing so, we see that<\/p>\r\n\r\n<div id=\"fs-id1169738171219\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\left(1-r\\right){S}_{k}&amp; =a\\left(1-r\\right)\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)\\hfill \\\\ &amp; =a\\left[\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)-\\left(r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k}\\right)\\right]\\hfill \\\\ &amp; =a\\left(1-{r}^{k}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737168821\">All the other terms cancel out.<\/p>\r\n<p id=\"fs-id1169737168824\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1169738228450\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\frac{a\\left(1-{r}^{k}\\right)}{1-r}\\text{for }r\\ne 1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737178277\">From our discussion in the previous section, we know that the geometric sequence [latex]{r}^{k}\\to 0[\/latex] if [latex]|r|&lt;1[\/latex] and that [latex]{r}^{k}[\/latex] diverges if [latex]|r|&gt;1[\/latex] or [latex]r= \\pm{1}[\/latex]. Therefore, for [latex]|r|&lt;1[\/latex], [latex]{S}_{k}\\to \\frac{a}{\\left(1-r\\right)}[\/latex] and we have<\/p>\r\n\r\n<div id=\"fs-id1169737234419\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}\\text{ if }|r|&lt;1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737178260\">If [latex]|r|\\ge 1[\/latex], [latex]{S}_{k}[\/latex] diverges, and therefore<\/p>\r\n\r\n<div id=\"fs-id1169738168112\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}\\text{ diverges if }|r|\\ge 1[\/latex].<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1169738166344\" 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-id1169738166347\">A geometric series is a series of the form<\/p>\r\n\r\n<div id=\"fs-id1169738166350\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=a+ar+a{r}^{2}+a{r}^{3}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737168122\">If [latex]|r|&lt;1[\/latex], the series converges, and<\/p>\r\n\r\n<div id=\"fs-id1169737934790\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}\\text{ for }|r|&lt;1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738127497\">If [latex]|r|\\ge 1[\/latex], the series diverges.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738171242\">Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than [latex]n=1[\/latex] or the exponent involves a linear expression for [latex]n[\/latex] other than [latex]n - 1[\/latex]. As long as we can rewrite the series in the form given by the harmonic series general equation, it is a geometric series. For example, consider the series<\/p>\r\n\r\n<div id=\"fs-id1169737360116\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737392660\">To see that this is a geometric series, we write out the first several terms:<\/p>\r\n\r\n<div id=\"fs-id1169737392664\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}&amp; ={\\left(\\frac{2}{3}\\right)}^{2}+{\\left(\\frac{2}{3}\\right)}^{3}+{\\left(\\frac{2}{3}\\right)}^{4}+\\cdots \\hfill \\\\ &amp; =\\frac{4}{9}+\\frac{4}{9}\\cdot \\left(\\frac{2}{3}\\right)+\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738153470\">We see that the initial term is [latex]a=\\frac{4}{9}[\/latex] and the ratio is [latex]r=\\frac{2}{3}[\/latex]. Therefore, the series can be written as<\/p>\r\n\r\n<div id=\"fs-id1169737392509\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738216526\">Since [latex]r=\\frac{2}{3}&lt;1[\/latex], this series converges, and its sum is given by<\/p>\r\n\r\n<div id=\"fs-id1169738191148\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}=\\frac{\\frac{4}{9}}{1 - \\frac{2}{3}}=\\frac{4}{3}[\/latex].<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1169738194429\" data-type=\"example\">\r\n<div id=\"fs-id1169738194431\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738194433\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Determining Convergence or Divergence of a Geometric Series<\/h3>\r\n<div id=\"fs-id1169738194433\" data-type=\"problem\">\r\n<p id=\"fs-id1169738194438\">Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.<\/p>\r\n\r\n<ol id=\"fs-id1169738194443\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-3\\right)}^{n+1}}{{4}^{n - 1}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{2n}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1169737162785\" data-type=\"solution\">\r\n<ol id=\"fs-id1169737162787\" type=\"a\">\r\n \t<li>Writing out the first several terms in the series, we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737162796\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-3\\right)}^{n+1}}{{4}^{n - 1}}&amp; =\\frac{{\\left(-3\\right)}^{2}}{{4}^{0}}+\\frac{{\\left(-3\\right)}^{3}}{4}+\\frac{{\\left(-3\\right)}^{4}}{{4}^{2}}+\\cdots \\hfill \\\\ &amp; ={\\left(-3\\right)}^{2}+{\\left(-3\\right)}^{2}\\cdot \\left(\\frac{-3}{4}\\right)+{\\left(-3\\right)}^{2}\\cdot {\\left(\\frac{-3}{4}\\right)}^{2}+\\cdots \\hfill \\\\ &amp; =9+9\\cdot \\left(\\frac{-3}{4}\\right)+9\\cdot {\\left(\\frac{-3}{4}\\right)}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe initial term [latex]a=-3[\/latex] and the ratio [latex]r=-\\frac{3}{4}[\/latex]. Since [latex]|r|=\\frac{3}{4}&lt;1[\/latex], the series converges to<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737269768\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{9}{1-\\left(-\\frac{3}{4}\\right)}=\\frac{9}{\\frac{7}{4}}=\\frac{36}{7}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Writing this series as<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738077637\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{2}\\displaystyle\\sum _{n=1}^{\\infty }{\\left({e}^{2}\\right)}^{n - 1}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwe can see that this is a geometric series where [latex]r={e}^{2}&gt;1[\/latex]. Therefore, the series diverges.<\/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-id1169737168529\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169737168532\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737168534\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/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\/05JxlmgOCkM?controls=0&amp;start=0&amp;end=29&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/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.\r\n\r\nYou 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>.\r\n<p id=\"fs-id1169737363857\">We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.<\/p>\r\n\r\n<div id=\"fs-id1169737363861\" data-type=\"example\">\r\n<div id=\"fs-id1169737363863\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737363866\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Writing Repeating Decimals as Fractions of Integers<\/h3>\r\n<div id=\"fs-id1169737363866\" data-type=\"problem\">\r\n<p id=\"fs-id1169737910133\">Use a geometric series to write [latex]3.\\overline{26}[\/latex] as a fraction of integers.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1169737910152\" data-type=\"solution\">\r\n<p id=\"fs-id1169737910154\">Since [latex]3.\\overline{26}=3.262626\\ldots [\/latex], first we write<\/p>\r\n\r\n<div id=\"fs-id1169737153641\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill 3.262626\\ldots&amp; =3+\\frac{26}{100}+\\frac{26}{1000}+\\frac{26}{100,000}+\\cdots \\hfill \\\\ &amp; =3+\\frac{26}{{10}^{2}}+\\frac{26}{{10}^{4}}+\\frac{26}{{10}^{6}}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737933437\">Ignoring the term 3, the rest of this expression is a geometric series with initial term [latex]a=\\frac{26}{{10}^{2}}[\/latex] and ratio [latex]r=\\frac{1}{{10}^{2}}[\/latex]. Therefore, the sum of this series is<\/p>\r\n\r\n<div id=\"fs-id1169737299443\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{26}{{10}^{2}}}{1-\\left(\\frac{1}{{10}^{2}}\\right)}=\\frac{\\frac{26}{{10}^{2}}}{\\frac{99}{{10}^{2}}}=\\frac{26}{99}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737934722\">Thus,<\/p>\r\n\r\n<div id=\"fs-id1169737934725\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3.262626\\ldots =3+\\frac{26}{99}=\\frac{323}{99}[\/latex].<\/div>\r\n&nbsp;\r\n\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-id1169738115123\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738115128\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738115130\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1169737201473\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the sum of a geometric series<\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">The Harmonic Series<\/h2>\n<p id=\"fs-id1169737935177\">A useful series to know about is the <span data-type=\"term\">harmonic series<\/span>. The harmonic series is defined as<\/p>\n<div id=\"fs-id1169738057265\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737393217\">This series is interesting because it diverges, but it diverges very slowly. By this we mean that the terms in the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] approach infinity, but do so very slowly. We will show that the series diverges, but first we illustrate the slow growth of the terms in the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] in the following table.<\/p>\n<table id=\"fs-id1169738150006\" class=\"unnumbered\" summary=\"This is a table with two rows and seven 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]10[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]100[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]1000[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]10,000[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]100,000[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]1,000,000[\/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]2.92897[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]5.18738[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]7.48547[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]9.78761[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]12.09015[\/latex]<\/td>\n<td data-valign=\"top\" data-align=\"left\">[latex]14.39273[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1169737910299\">Even after [latex]1,000,000[\/latex] terms, the partial sum is still relatively small. From this table, it is not clear that this series actually diverges. However, we can show analytically that the sequence of partial sums diverges, and therefore the series diverges.<\/p>\n<p id=\"fs-id1169737168935\">To show that the sequence of partial sums diverges, we show that the sequence of partial sums is unbounded. We begin by writing the first several partial sums:<\/p>\n<div id=\"fs-id1169737292630\" 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}\\hfill \\\\ {S}_{3}=1+\\frac{1}{2}+\\frac{1}{3}\\hfill \\\\ {S}_{4}=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738078962\">Notice that for the last two terms in [latex]{S}_{4}[\/latex],<\/p>\n<div id=\"fs-id1169738078297\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{3}+\\frac{1}{4}>\\frac{1}{4}+\\frac{1}{4}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737818894\">Therefore, we conclude that<\/p>\n<div id=\"fs-id1169738250171\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{4}>1+\\frac{1}{2}+\\left(\\frac{1}{4}+\\frac{1}{4}\\right)=1+\\frac{1}{2}+\\frac{1}{2}=1+2\\left(\\frac{1}{2}\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737265736\">Using the same idea for [latex]{S}_{8}[\/latex], we see that<\/p>\n<div id=\"fs-id1169737809241\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {S}_{8}& =1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}>1+\\frac{1}{2}+\\left(\\frac{1}{4}+\\frac{1}{4}\\right)+\\left(\\frac{1}{8}+\\frac{1}{8}+\\frac{1}{8}+\\frac{1}{8}\\right)\\hfill \\\\ & =1+\\frac{1}{2}+\\frac{1}{2}+\\frac{1}{2}=1+3\\left(\\frac{1}{2}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738165396\">From this pattern, we see that [latex]{S}_{1}=1[\/latex], [latex]{S}_{2}=1+\\frac{1}{2}[\/latex], [latex]{S}_{4}>1+2\\left(\\frac{1}{2}\\right)[\/latex], and [latex]{S}_{8}>1+3\\left(\\frac{1}{2}\\right)[\/latex]. More generally, it can be shown that [latex]{S}_{{2}^{j}}>1+j\\left(\\frac{1}{2}\\right)[\/latex] for all [latex]j>1[\/latex]. Since [latex]1+j\\left(\\frac{1}{2}\\right)\\to \\infty[\/latex], we conclude that the sequence [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded and therefore diverges. In the previous section, we stated that convergent sequences are bounded. Consequently, since [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is unbounded, it diverges. Thus, the harmonic series diverges.<\/p>\n<section id=\"fs-id1169738056466\" data-depth=\"2\">\n<h2 data-type=\"title\">Algebraic Properties of Convergent Series<\/h2>\n<p id=\"fs-id1169738188266\">Since the sum of a convergent infinite series is defined as a limit of a sequence, the algebraic properties for series listed below follow directly from the algebraic properties for sequences.<\/p>\n<div id=\"fs-id1169738189943\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Theorem: Algebraic Properties of Convergent Series<\/h3>\n<hr \/>\n<p id=\"fs-id1169737179327\">Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] be convergent series. Then the following algebraic properties hold.<\/p>\n<ol id=\"fs-id1169737947446\" type=\"i\">\n<li>The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}+\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. (Sum Rule)<\/li>\n<li>The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}-{b}_{n}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. (Difference Rule)<\/li>\n<li>For any real number [latex]c[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}[\/latex] converges and [latex]\\displaystyle\\sum _{n=1}^{\\infty }c{a}_{n}=c\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. (Constant Multiple Rule)<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738055547\" data-type=\"example\">\n<div id=\"fs-id1169738055550\" data-type=\"exercise\">\n<div id=\"fs-id1169738148599\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Using Algebraic Properties of Convergent Series<\/h3>\n<div id=\"fs-id1169738148599\" data-type=\"problem\">\n<p id=\"fs-id1169738058908\">Evaluate<\/p>\n<div id=\"fs-id1169738058911\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right][\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738042181\" data-type=\"solution\">\n<p id=\"fs-id1169738042184\">We showed earlier that<\/p>\n<div id=\"fs-id1169737949111\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737297603\">and<\/p>\n<div id=\"fs-id1169738189920\" 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-id1169737209441\">Since both of those series converge, we can apply the Alegraic Properties of Convergent Series to evaluate<\/p>\n<div id=\"fs-id1169737355719\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right][\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738146272\">Using the sum rule, write<\/p>\n<div id=\"fs-id1169738146275\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{3}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{n - 2}\\right]=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3}{n\\left(n+1\\right)}\\underset{n=1}{\\overset{\\infty }{+\\displaystyle\\sum }}{\\left(\\frac{1}{2}\\right)}^{n - 2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738248902\">Then, using the constant multiple rule and the sums above, we can conclude that<\/p>\n<div id=\"fs-id1169738086671\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ll}{\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3}{n\\left(n+1\\right)}+\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 2}}\\hfill & =3{\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}+{\\left(\\frac{1}{2}\\right)}^{-1}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}}\\hfill \\\\ & =3\\left(1\\right)+{\\left(\\frac{1}{2}\\right)}^{-1}\\left(2\\right)=3+2\\left(2\\right)=7.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738201806\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738201809\" data-type=\"exercise\">\n<div id=\"fs-id1169738201811\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/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<\/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=1056&amp;end=1095&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><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 id=\"fs-id1169737395366\" data-depth=\"2\">\n<h2 data-type=\"title\">Geometric Series<\/h2>\n<p id=\"fs-id1169737395484\">A <strong>geometric series<\/strong> is any series that we can write in the form<\/p>\n<div id=\"fs-id1169738080256\" style=\"text-align: center;\" data-type=\"equation\">[latex]a+ar+a{r}^{2}+a{r}^{3}+\\cdots =\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737796160\">Because the ratio of each term in this series to the previous term is <em data-effect=\"italics\">r<\/em>, the number <em data-effect=\"italics\">r<\/em> is called the ratio. We refer to <em data-effect=\"italics\">a<\/em> as the initial term because it is the first term in the series. For example, the series<\/p>\n<div id=\"fs-id1169737179889\" 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}+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737227806\">is a geometric series with initial term [latex]a=1[\/latex] and ratio [latex]r=\\frac{1}{2}[\/latex].<\/p>\n<p id=\"fs-id1169738234421\">In general, when does a geometric series converge? Consider the geometric series<\/p>\n<div id=\"fs-id1169738234424\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738046261\">when [latex]a>0[\/latex]. Its sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is given by<\/p>\n<div id=\"fs-id1169737934871\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}a{r}^{n - 1}=a+ar+a{r}^{2}+\\cdots +a{r}^{k - 1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737394756\">Consider the case when [latex]r=1[\/latex]. In that case,<\/p>\n<div id=\"fs-id1169738080270\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=a+a\\left(1\\right)+a{\\left(1\\right)}^{2}+\\cdots +a{\\left(1\\right)}^{k - 1}=ak[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737934742\">Since [latex]a>0[\/latex], we know [latex]ak\\to \\infty[\/latex] as [latex]k\\to \\infty[\/latex]. Therefore, the sequence of partial sums is unbounded and thus diverges. Consequently, the infinite series diverges for [latex]r=1[\/latex]. For [latex]r\\ne 1[\/latex], to find the limit of [latex]\\left\\{{S}_{k}\\right\\}[\/latex], multiply the geometric series general equation by [latex]1-r[\/latex]. Doing so, we see that<\/p>\n<div id=\"fs-id1169738171219\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\left(1-r\\right){S}_{k}& =a\\left(1-r\\right)\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)\\hfill \\\\ & =a\\left[\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)-\\left(r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k}\\right)\\right]\\hfill \\\\ & =a\\left(1-{r}^{k}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737168821\">All the other terms cancel out.<\/p>\n<p id=\"fs-id1169737168824\">Therefore,<\/p>\n<div id=\"fs-id1169738228450\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\frac{a\\left(1-{r}^{k}\\right)}{1-r}\\text{for }r\\ne 1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737178277\">From our discussion in the previous section, we know that the geometric sequence [latex]{r}^{k}\\to 0[\/latex] if [latex]|r|<1[\/latex] and that [latex]{r}^{k}[\/latex] diverges if [latex]|r|>1[\/latex] or [latex]r= \\pm{1}[\/latex]. Therefore, for [latex]|r|<1[\/latex], [latex]{S}_{k}\\to \\frac{a}{\\left(1-r\\right)}[\/latex] and we have<\/p>\n<div id=\"fs-id1169737234419\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}\\text{ if }|r|<1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737178260\">If [latex]|r|\\ge 1[\/latex], [latex]{S}_{k}[\/latex] diverges, and therefore<\/p>\n<div id=\"fs-id1169738168112\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}\\text{ diverges if }|r|\\ge 1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169738166344\" 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-id1169738166347\">A geometric series is a series of the form<\/p>\n<div id=\"fs-id1169738166350\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=a+ar+a{r}^{2}+a{r}^{3}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737168122\">If [latex]|r|<1[\/latex], the series converges, and<\/p>\n<div id=\"fs-id1169737934790\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}\\text{ for }|r|<1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738127497\">If [latex]|r|\\ge 1[\/latex], the series diverges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738171242\">Geometric series sometimes appear in slightly different forms. For example, sometimes the index begins at a value other than [latex]n=1[\/latex] or the exponent involves a linear expression for [latex]n[\/latex] other than [latex]n - 1[\/latex]. As long as we can rewrite the series in the form given by the harmonic series general equation, it is a geometric series. For example, consider the series<\/p>\n<div id=\"fs-id1169737360116\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737392660\">To see that this is a geometric series, we write out the first several terms:<\/p>\n<div id=\"fs-id1169737392664\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}& ={\\left(\\frac{2}{3}\\right)}^{2}+{\\left(\\frac{2}{3}\\right)}^{3}+{\\left(\\frac{2}{3}\\right)}^{4}+\\cdots \\hfill \\\\ & =\\frac{4}{9}+\\frac{4}{9}\\cdot \\left(\\frac{2}{3}\\right)+\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738153470\">We see that the initial term is [latex]a=\\frac{4}{9}[\/latex] and the ratio is [latex]r=\\frac{2}{3}[\/latex]. Therefore, the series can be written as<\/p>\n<div id=\"fs-id1169737392509\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738216526\">Since [latex]r=\\frac{2}{3}<1[\/latex], this series converges, and its sum is given by<\/p>\n<div id=\"fs-id1169738191148\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}=\\frac{\\frac{4}{9}}{1 - \\frac{2}{3}}=\\frac{4}{3}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169738194429\" data-type=\"example\">\n<div id=\"fs-id1169738194431\" data-type=\"exercise\">\n<div id=\"fs-id1169738194433\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Determining Convergence or Divergence of a Geometric Series<\/h3>\n<div id=\"fs-id1169738194433\" data-type=\"problem\">\n<p id=\"fs-id1169738194438\">Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.<\/p>\n<ol id=\"fs-id1169738194443\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-3\\right)}^{n+1}}{{4}^{n - 1}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{2n}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737162785\" data-type=\"solution\">\n<ol id=\"fs-id1169737162787\" type=\"a\">\n<li>Writing out the first several terms in the series, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737162796\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-3\\right)}^{n+1}}{{4}^{n - 1}}& =\\frac{{\\left(-3\\right)}^{2}}{{4}^{0}}+\\frac{{\\left(-3\\right)}^{3}}{4}+\\frac{{\\left(-3\\right)}^{4}}{{4}^{2}}+\\cdots \\hfill \\\\ & ={\\left(-3\\right)}^{2}+{\\left(-3\\right)}^{2}\\cdot \\left(\\frac{-3}{4}\\right)+{\\left(-3\\right)}^{2}\\cdot {\\left(\\frac{-3}{4}\\right)}^{2}+\\cdots \\hfill \\\\ & =9+9\\cdot \\left(\\frac{-3}{4}\\right)+9\\cdot {\\left(\\frac{-3}{4}\\right)}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe initial term [latex]a=-3[\/latex] and the ratio [latex]r=-\\frac{3}{4}[\/latex]. Since [latex]|r|=\\frac{3}{4}<1[\/latex], the series converges to<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737269768\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{9}{1-\\left(-\\frac{3}{4}\\right)}=\\frac{9}{\\frac{7}{4}}=\\frac{36}{7}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Writing this series as<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738077637\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{2}\\displaystyle\\sum _{n=1}^{\\infty }{\\left({e}^{2}\\right)}^{n - 1}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwe can see that this is a geometric series where [latex]r={e}^{2}>1[\/latex]. Therefore, the series diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737168529\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169737168532\" data-type=\"exercise\">\n<div id=\"fs-id1169737168534\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/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<\/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\/05JxlmgOCkM?controls=0&amp;start=0&amp;end=29&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/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.<\/p>\n<p>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>.<\/p>\n<p id=\"fs-id1169737363857\">We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.<\/p>\n<div id=\"fs-id1169737363861\" data-type=\"example\">\n<div id=\"fs-id1169737363863\" data-type=\"exercise\">\n<div id=\"fs-id1169737363866\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Writing Repeating Decimals as Fractions of Integers<\/h3>\n<div id=\"fs-id1169737363866\" data-type=\"problem\">\n<p id=\"fs-id1169737910133\">Use a geometric series to write [latex]3.\\overline{26}[\/latex] as a fraction of integers.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Show Solution<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737910152\" data-type=\"solution\">\n<p id=\"fs-id1169737910154\">Since [latex]3.\\overline{26}=3.262626\\ldots[\/latex], first we write<\/p>\n<div id=\"fs-id1169737153641\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill 3.262626\\ldots& =3+\\frac{26}{100}+\\frac{26}{1000}+\\frac{26}{100,000}+\\cdots \\hfill \\\\ & =3+\\frac{26}{{10}^{2}}+\\frac{26}{{10}^{4}}+\\frac{26}{{10}^{6}}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737933437\">Ignoring the term 3, the rest of this expression is a geometric series with initial term [latex]a=\\frac{26}{{10}^{2}}[\/latex] and ratio [latex]r=\\frac{1}{{10}^{2}}[\/latex]. Therefore, the sum of this series is<\/p>\n<div id=\"fs-id1169737299443\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{26}{{10}^{2}}}{1-\\left(\\frac{1}{{10}^{2}}\\right)}=\\frac{\\frac{26}{{10}^{2}}}{\\frac{99}{{10}^{2}}}=\\frac{26}{99}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737934722\">Thus,<\/p>\n<div id=\"fs-id1169737934725\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3.262626\\ldots =3+\\frac{26}{99}=\\frac{323}{99}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738115123\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738115128\" data-type=\"exercise\">\n<div id=\"fs-id1169738115130\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Hint<\/span><\/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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1169737201473\" data-depth=\"1\"><\/section>\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-1703\">\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><li>5.2.2. <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 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