{"id":1694,"date":"2021-07-23T16:44:00","date_gmt":"2021-07-23T16:44:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1694"},"modified":"2022-03-21T23:01:58","modified_gmt":"2022-03-21T23:01:58","slug":"bounded-sequences","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/bounded-sequences\/","title":{"raw":"Bounded Sequences","rendered":"Bounded Sequences"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine the convergence or divergence of a given sequence<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169739206971\">We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.<\/p>\r\n\r\n<div id=\"fs-id1169739206976\" 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-id1169739206979\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is <strong>bounded above<\/strong> if there exists a real number [latex]M[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169736790140\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\le M[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736790156\">for all positive integers [latex]n[\/latex].<\/p>\r\n<p id=\"fs-id1169736790163\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is <strong>bounded below<\/strong> if there exists a real number [latex]M[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169736790186\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]M\\le {a}_{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736790202\">for all positive integers [latex]n[\/latex].<\/p>\r\n<p id=\"fs-id1169736790212\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a <strong>bounded sequence<\/strong> if it is bounded above and bounded below.<\/p>\r\n<p id=\"fs-id1169736790231\">If a sequence is not bounded, it is an <strong>unbounded sequence<\/strong>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169736790239\">For example, the sequence [latex]\\left\\{\\frac{1}{n}\\right\\}[\/latex] is bounded above because [latex]\\frac{1}{n}\\le 1[\/latex] for all positive integers [latex]n[\/latex]. It is also bounded below because [latex]\\frac{1}{n}\\ge 0[\/latex] for all positive integers <em data-effect=\"italics\">n.<\/em> Therefore, [latex]\\left\\{\\frac{1}{n}\\right\\}[\/latex] is a bounded sequence. On the other hand, consider the sequence [latex]\\left\\{{2}^{n}\\right\\}[\/latex]. Because [latex]{2}^{n}\\ge 2[\/latex] for all [latex]n\\ge 1[\/latex], the sequence is bounded below. However, the sequence is not bounded above. Therefore, [latex]\\left\\{{2}^{n}\\right\\}[\/latex] is an unbounded sequence.<\/p>\r\n<p id=\"fs-id1169736694672\">We now discuss the relationship between boundedness and convergence. Suppose a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms [latex]{a}_{n}[\/latex] that are arbitrarily large in magnitude as [latex]n[\/latex] gets larger. As a result, the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.<\/p>\r\n\r\n<div id=\"fs-id1169736694714\" 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: Convergent Sequences Are Bounded<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169736694721\">If a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges, then it is bounded.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169736694738\">Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence [latex]\\left\\{{\\left(-1\\right)}^{n}\\right\\}[\/latex] is bounded, but the sequence diverges because the sequence oscillates between [latex]1[\/latex] and [latex]-1[\/latex] and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.<\/p>\r\n<p id=\"fs-id1169739133146\">Consider a bounded sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex]. Suppose the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing. That is, [latex]{a}_{1}\\le {a}_{2}\\le {a}_{3}\\ldots[\/latex]. Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges. For example, consider the sequence<\/p>\r\n\r\n<div id=\"fs-id1169739133224\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{\\frac{1}{2},\\frac{2}{3},\\frac{3}{4},\\frac{4}{5}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739133271\">Since this sequence is increasing and bounded above, it converges. Next, consider the sequence<\/p>\r\n\r\n<div id=\"fs-id1169739133274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{2,0,3,0,4,0,1,-\\frac{1}{2},-\\frac{1}{3},-\\frac{1}{4}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736726568\">Even though the sequence is not increasing for all values of [latex]n[\/latex], we see that [latex]-\\frac{1}{2}&lt;\\text{-}\\frac{1}{3}&lt;\\text{-}\\frac{1}{4}&lt;\\cdots [\/latex]. Therefore, starting with the eighth term, [latex]{a}_{8}=-\\frac{1}{2}[\/latex], the sequence is increasing. In this case, we say the sequence is <em data-effect=\"italics\">eventually<\/em> increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.<\/p>\r\n\r\n<div id=\"fs-id1169736726641\" 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-id1169736726644\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing for all [latex]n\\ge {n}_{0}[\/latex] if<\/p>\r\n\r\n<div id=\"fs-id1169736767089\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\le {a}_{n+1}\\text{ for all }n\\ge {n}_{0}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736767132\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is decreasing for all [latex]n\\ge {n}_{0}[\/latex] if<\/p>\r\n\r\n<div id=\"fs-id1169736767163\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\ge {a}_{n+1}\\text{ for all }n\\ge {n}_{0}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736767206\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a <strong>monotone sequence<\/strong> for all [latex]n\\ge {n}_{0}[\/latex] if it is increasing for all [latex]n\\ge {n}_{0}[\/latex] or decreasing for all [latex]n\\ge {n}_{0}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169739335676\">We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.<\/p>\r\n\r\n<div id=\"fs-id1169739335681\" 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: Monotone Convergence Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169739335687\">If [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a bounded sequence and there exists a positive integer [latex]{n}_{0}[\/latex] such that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is monotone for all [latex]n\\ge {n}_{0}[\/latex], then [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169739335753\">The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 6).<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_09_01_008\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"456\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234256\/CNX_Calc_Figure_09_01_008.jpg\" alt=\"A graph in quadrant 1 with the x and y axes labeled n and a_n, respectively. A dotted horizontal is drawn from the a_n axis into quadrant 1. Many points are plotted under the dotted line, increasing in a_n value and converging to the dotted line.\" width=\"456\" height=\"233\" data-media-type=\"image\/jpeg\" \/> Figure 6. Since the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing and bounded above, it must converge.[\/caption]<\/figure>\r\n<p id=\"fs-id1169739335792\">In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.<\/p>\r\n\r\n<div id=\"fs-id1169739335796\" data-type=\"example\">\r\n<div id=\"fs-id1169739335798\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738993730\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Using the Monotone Convergence Theorem<\/h3>\r\n<div id=\"fs-id1169738993730\" data-type=\"problem\">\r\n<p id=\"fs-id1169738993735\">For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.<\/p>\r\n\r\n<ol id=\"fs-id1169738993740\" type=\"a\">\r\n \t<li>[latex]\\left\\{\\frac{{4}^{n}}{n\\text{!}}\\right\\}[\/latex]<\/li>\r\n \t<li>[latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738993786\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1}=2\\text{ and }{a}_{n+1}=\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\text{ for all }n\\ge 2[\/latex].<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1169738993861\" data-type=\"solution\">\r\n<ol id=\"fs-id1169738993864\" type=\"a\">\r\n \t<li>Writing out the first few terms, we see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738993872\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{\\frac{{4}^{n}}{n\\text{!}}\\right\\}=\\left\\{4,8,\\frac{32}{3},\\frac{32}{3},\\frac{128}{15}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nAt first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all [latex]n\\ge 3[\/latex]. We can show this as follows.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739213903\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n+1}=\\frac{{4}^{n+1}}{\\left(n+1\\right)\\text{!}}=\\frac{4}{n+1}\\cdot \\frac{{4}^{n}}{n\\text{!}}=\\frac{4}{n+1}\\cdot {a}_{n}\\le {a}_{n}\\text{ if }n\\ge 3[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, the sequence is decreasing for all [latex]n\\ge 3[\/latex]. Further, the sequence is bounded below by [latex]0[\/latex] because [latex]\\frac{{4}^{n}}{n\\text{!}}\\ge 0[\/latex] for all positive integers [latex]n[\/latex]. Therefore, by the Monotone Convergence Theorem, the sequence converges.<span data-type=\"newline\">\r\n<\/span>\r\nTo find the limit, we use the fact that the sequence converges and let [latex]L=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. Now note this important observation. Consider [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}[\/latex]. Since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736790081\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{{a}_{n+1}\\right\\}=\\left\\{{a}_{2,}{a}_{3},{a}_{4}\\text{,}\\ldots\\right\\}[\/latex],<\/div>\r\n&nbsp;\r\n\r\nthe only difference between the sequences [latex]\\left\\{{a}_{n+1}\\right\\}[\/latex] and [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is that [latex]\\left\\{{a}_{n+1}\\right\\}[\/latex] omits the first term. Since a finite number of terms does not affect the convergence of a sequence,\r\n\r\n<span data-type=\"newline\">\u00a0<\/span>\r\n<div id=\"fs-id1169739110746\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=L[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nCombining this fact with the equation<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739110806\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n+1}=\\frac{4}{n+1}{a}_{n}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nand taking the limit of both sides of the equation<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739195686\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{4}{n+1}{a}_{n}[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwe can conclude that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739195752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L=0\\cdot L=0[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Writing out the first several terms,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739195780\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{2,\\frac{5}{4},\\frac{41}{40},\\frac{3281}{3280}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwe can conjecture that the sequence is decreasing and bounded below by [latex]1[\/latex]. To show that the sequence is bounded below by [latex]1[\/latex], we can show that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739341340\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\ge 1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTo show this, first rewrite<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739341380\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}=\\frac{{a}_{n}^{2}+1}{2{a}_{n}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]{a}_{1}&gt;0[\/latex] and [latex]{a}_{2}[\/latex] is defined as a sum of positive terms, [latex]{a}_{2}&gt;0[\/latex]. Similarly, all terms [latex]{a}_{n}&gt;0[\/latex]. Therefore,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739341490\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}^{2}+1}{2{a}_{n}}\\ge 1[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nif and only if<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736708125\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}^{2}+1\\ge 2{a}_{n}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nRewriting the inequality [latex]{a}_{n}^{2}+1\\ge 2{a}_{n}[\/latex] as [latex]{a}_{n}^{2}-2{a}_{n}+1\\ge 0[\/latex], and using the fact that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736708214\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}^{2}-2{a}_{n}+1={\\left({a}_{n}-1\\right)}^{2}\\ge 0[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nbecause the square of any real number is nonnegative, we can conclude that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736842868\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\ge 1[\/latex].<\/div>\r\n&nbsp;\r\n<p style=\"text-align: left;\"><span data-type=\"newline\">\r\n<\/span>\r\nTo show that the sequence is decreasing, we must show that [latex]{a}_{n+1}\\le {a}_{n}[\/latex] for all [latex]n\\ge 1[\/latex]. Since [latex]1\\le {a}_{n}^{2}[\/latex], it follows that<span data-type=\"newline\">\r\n<\/span><\/p>\r\n\r\n<div id=\"fs-id1169736842957\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}^{2}+1\\le 2{a}_{n}^{2}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nDividing both sides by [latex]2{a}_{n}[\/latex], we obtain<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736843005\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\le {a}_{n}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nUsing the definition of [latex]{a}_{n+1}[\/latex], we conclude that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736705845\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n+1}=\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\le {a}_{n}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is bounded below and decreasing, by the Monotone Convergence Theorem, it converges.<span data-type=\"newline\">\r\n<\/span>\r\nTo find the limit, let [latex]L=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. Then using the recurrence relation and the fact that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}[\/latex], we have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736702887\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\left(\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\right)[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nand therefore<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736702967\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L=\\frac{L}{2}+\\frac{1}{2L}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nMultiplying both sides of this equation by [latex]2L[\/latex], we arrive at the equation<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736703009\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2{L}^{2}={L}^{2}+1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSolving this equation for [latex]L[\/latex], we conclude that [latex]{L}^{2}=1[\/latex], which implies [latex]L=\\pm 1[\/latex]. Since all the terms are positive, the limit [latex]L=1[\/latex].<\/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-id1169739202289\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169739202292\" data-type=\"exercise\">\r\n<div id=\"fs-id1169739202294\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169739202294\" data-type=\"problem\">\r\n<p id=\"fs-id1169739202296\">Consider the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that [latex]{a}_{1}=1[\/latex], [latex]{a}_{n}=\\frac{{a}_{n - 1}}{2}[\/latex]. Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558599\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558599\"]\r\n<div id=\"fs-id1169739202370\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739202376\">Show the sequence is decreasing and bounded below.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n<div id=\"fs-id1169739202358\" data-type=\"solution\">\r\n<p id=\"fs-id1169739202360\">[latex]0[\/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\/V-B05oiNKFk?controls=0&amp;start=713&amp;end=822&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\/5.1.3_713to822_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.3\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5808[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169739202383\" class=\"project\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox tryit\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">ACTIVITY: Fibonacci Numbers<\/h3>\r\n<p id=\"fs-id1169739202389\">The <span class=\"no-emphasis\" data-type=\"term\">Fibonacci numbers<\/span> are defined recursively by the sequence [latex]\\left\\{{F}_{n}\\right\\}[\/latex] where [latex]{F}_{0}=0[\/latex], [latex]{F}_{1}=1[\/latex] and for [latex]n\\ge 2[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169739344161\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739344200\">Here we look at properties of the Fibonacci numbers.<\/p>\r\n\r\n<ol id=\"fs-id1169739344204\" type=\"1\">\r\n \t<li>Write out the first twenty Fibonacci numbers.<\/li>\r\n \t<li>Find a closed formula for the Fibonacci sequence by using the following steps.\r\n<ol id=\"fs-id1169739344213\" type=\"a\">\r\n \t<li>Consider the recursively defined sequence [latex]\\left\\{{x}_{n}\\right\\}[\/latex] where [latex]{x}_{o}=c[\/latex] and [latex]{x}_{n+1}=a{x}_{n}[\/latex]. Show that this sequence can be described by the closed formula [latex]{x}_{n}=c{a}^{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/li>\r\n \t<li>Using the result from part a. as motivation, look for a solution of the equation<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739344309\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nof the form [latex]{F}_{n}=c{\\lambda }^{n}[\/latex]. Determine what two values for [latex]\\lambda [\/latex] will allow [latex]{F}_{n}[\/latex] to satisfy this equation.<\/li>\r\n \t<li>Consider the two solutions from part b.: [latex]{\\lambda }_{1}[\/latex] and [latex]{\\lambda }_{2}[\/latex]. Let [latex]{F}_{n}={c}_{1}{\\lambda }_{1}{}^{n}+{c}_{2}{\\lambda }_{2}{}^{n}[\/latex]. Use the initial conditions [latex]{F}_{0}[\/latex] and [latex]{F}_{1}[\/latex] to determine the values for the constants [latex]{c}_{1}[\/latex] and [latex]{c}_{2}[\/latex] and write the closed formula [latex]{F}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Use the answer in 2 c. to show that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736856912\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{F}_{n+1}}{{F}_{n}}=\\frac{1+\\sqrt{5}}{2}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe number [latex]\\varphi =\\frac{\\left(1+\\sqrt{5}\\right)}{2}[\/latex] is known as the <span class=\"no-emphasis\" data-type=\"term\">golden ratio<\/span> (Figures 7 and 8).<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_09_01_004\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234259\/CNX_Calc_Figure_09_01_004.jpg\" alt=\"This is a photo of a sunflower, particularly the curves of the seeds at its middle. The number of spirals in each direction is always a Fibonacci number.\" width=\"325\" height=\"244\" data-media-type=\"image\/jpeg\" \/> Figure 7. The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number\u2014always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)[\/caption]<\/figure>\r\n<span data-type=\"newline\">\u00a0<\/span>\r\n<figure id=\"CNX_Calc_Figure_09_01_005\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234302\/CNX_Calc_Figure_09_01_005.jpg\" alt=\"This is a photo of the Parthenon, an ancient Greek temple that was designed with the proportions of the Golden Rule. The entire temple\u2019s front side fits perfectly into a rectangle with those proportions, as do the columns, the level between the columns and the roof, and a portion of the trim below the roof.\" width=\"488\" height=\"300\" data-media-type=\"image\/jpeg\" \/> Figure 8. The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon was designed with these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)[\/caption]<\/figure>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1169736857064\" class=\"key-concepts\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the convergence or divergence of a given sequence<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169739206971\">We now turn our attention to one of the most important theorems involving sequences: the Monotone Convergence Theorem. Before stating the theorem, we need to introduce some terminology and motivation. We begin by defining what it means for a sequence to be bounded.<\/p>\n<div id=\"fs-id1169739206976\" 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-id1169739206979\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is <strong>bounded above<\/strong> if there exists a real number [latex]M[\/latex] such that<\/p>\n<div id=\"fs-id1169736790140\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\le M[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736790156\">for all positive integers [latex]n[\/latex].<\/p>\n<p id=\"fs-id1169736790163\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is <strong>bounded below<\/strong> if there exists a real number [latex]M[\/latex] such that<\/p>\n<div id=\"fs-id1169736790186\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]M\\le {a}_{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736790202\">for all positive integers [latex]n[\/latex].<\/p>\n<p id=\"fs-id1169736790212\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a <strong>bounded sequence<\/strong> if it is bounded above and bounded below.<\/p>\n<p id=\"fs-id1169736790231\">If a sequence is not bounded, it is an <strong>unbounded sequence<\/strong>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169736790239\">For example, the sequence [latex]\\left\\{\\frac{1}{n}\\right\\}[\/latex] is bounded above because [latex]\\frac{1}{n}\\le 1[\/latex] for all positive integers [latex]n[\/latex]. It is also bounded below because [latex]\\frac{1}{n}\\ge 0[\/latex] for all positive integers <em data-effect=\"italics\">n.<\/em> Therefore, [latex]\\left\\{\\frac{1}{n}\\right\\}[\/latex] is a bounded sequence. On the other hand, consider the sequence [latex]\\left\\{{2}^{n}\\right\\}[\/latex]. Because [latex]{2}^{n}\\ge 2[\/latex] for all [latex]n\\ge 1[\/latex], the sequence is bounded below. However, the sequence is not bounded above. Therefore, [latex]\\left\\{{2}^{n}\\right\\}[\/latex] is an unbounded sequence.<\/p>\n<p id=\"fs-id1169736694672\">We now discuss the relationship between boundedness and convergence. Suppose a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is unbounded. Then it is not bounded above, or not bounded below, or both. In either case, there are terms [latex]{a}_{n}[\/latex] that are arbitrarily large in magnitude as [latex]n[\/latex] gets larger. As a result, the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] cannot converge. Therefore, being bounded is a necessary condition for a sequence to converge.<\/p>\n<div id=\"fs-id1169736694714\" 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: Convergent Sequences Are Bounded<\/h3>\n<hr \/>\n<p id=\"fs-id1169736694721\">If a sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges, then it is bounded.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169736694738\">Note that a sequence being bounded is not a sufficient condition for a sequence to converge. For example, the sequence [latex]\\left\\{{\\left(-1\\right)}^{n}\\right\\}[\/latex] is bounded, but the sequence diverges because the sequence oscillates between [latex]1[\/latex] and [latex]-1[\/latex] and never approaches a finite number. We now discuss a sufficient (but not necessary) condition for a bounded sequence to converge.<\/p>\n<p id=\"fs-id1169739133146\">Consider a bounded sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex]. Suppose the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing. That is, [latex]{a}_{1}\\le {a}_{2}\\le {a}_{3}\\ldots[\/latex]. Since the sequence is increasing, the terms are not oscillating. Therefore, there are two possibilities. The sequence could diverge to infinity, or it could converge. However, since the sequence is bounded, it is bounded above and the sequence cannot diverge to infinity. We conclude that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges. For example, consider the sequence<\/p>\n<div id=\"fs-id1169739133224\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{\\frac{1}{2},\\frac{2}{3},\\frac{3}{4},\\frac{4}{5}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739133271\">Since this sequence is increasing and bounded above, it converges. Next, consider the sequence<\/p>\n<div id=\"fs-id1169739133274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{2,0,3,0,4,0,1,-\\frac{1}{2},-\\frac{1}{3},-\\frac{1}{4}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736726568\">Even though the sequence is not increasing for all values of [latex]n[\/latex], we see that [latex]-\\frac{1}{2}<\\text{-}\\frac{1}{3}<\\text{-}\\frac{1}{4}<\\cdots[\/latex]. Therefore, starting with the eighth term, [latex]{a}_{8}=-\\frac{1}{2}[\/latex], the sequence is increasing. In this case, we say the sequence is <em data-effect=\"italics\">eventually<\/em> increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.<\/p>\n<div id=\"fs-id1169736726641\" 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-id1169736726644\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing for all [latex]n\\ge {n}_{0}[\/latex] if<\/p>\n<div id=\"fs-id1169736767089\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\le {a}_{n+1}\\text{ for all }n\\ge {n}_{0}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736767132\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is decreasing for all [latex]n\\ge {n}_{0}[\/latex] if<\/p>\n<div id=\"fs-id1169736767163\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}\\ge {a}_{n+1}\\text{ for all }n\\ge {n}_{0}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736767206\">A sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a <strong>monotone sequence<\/strong> for all [latex]n\\ge {n}_{0}[\/latex] if it is increasing for all [latex]n\\ge {n}_{0}[\/latex] or decreasing for all [latex]n\\ge {n}_{0}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169739335676\">We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.<\/p>\n<div id=\"fs-id1169739335681\" 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: Monotone Convergence Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1169739335687\">If [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a bounded sequence and there exists a positive integer [latex]{n}_{0}[\/latex] such that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is monotone for all [latex]n\\ge {n}_{0}[\/latex], then [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169739335753\">The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 6).<\/p>\n<figure id=\"CNX_Calc_Figure_09_01_008\"><figcaption><\/figcaption><div style=\"width: 466px\" 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\/11234256\/CNX_Calc_Figure_09_01_008.jpg\" alt=\"A graph in quadrant 1 with the x and y axes labeled n and a_n, respectively. A dotted horizontal is drawn from the a_n axis into quadrant 1. Many points are plotted under the dotted line, increasing in a_n value and converging to the dotted line.\" width=\"456\" height=\"233\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Since the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing and bounded above, it must converge.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1169739335792\">In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.<\/p>\n<div id=\"fs-id1169739335796\" data-type=\"example\">\n<div id=\"fs-id1169739335798\" data-type=\"exercise\">\n<div id=\"fs-id1169738993730\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Using the Monotone Convergence Theorem<\/h3>\n<div id=\"fs-id1169738993730\" data-type=\"problem\">\n<p id=\"fs-id1169738993735\">For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.<\/p>\n<ol id=\"fs-id1169738993740\" type=\"a\">\n<li>[latex]\\left\\{\\frac{{4}^{n}}{n\\text{!}}\\right\\}[\/latex]<\/li>\n<li>[latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738993786\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{a}_{1}=2\\text{ and }{a}_{n+1}=\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\text{ for all }n\\ge 2[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558799\">Show Solution<\/span><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738993861\" data-type=\"solution\">\n<ol id=\"fs-id1169738993864\" type=\"a\">\n<li>Writing out the first few terms, we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738993872\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{\\frac{{4}^{n}}{n\\text{!}}\\right\\}=\\left\\{4,8,\\frac{32}{3},\\frac{32}{3},\\frac{128}{15}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nAt first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all [latex]n\\ge 3[\/latex]. We can show this as follows.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739213903\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n+1}=\\frac{{4}^{n+1}}{\\left(n+1\\right)\\text{!}}=\\frac{4}{n+1}\\cdot \\frac{{4}^{n}}{n\\text{!}}=\\frac{4}{n+1}\\cdot {a}_{n}\\le {a}_{n}\\text{ if }n\\ge 3[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, the sequence is decreasing for all [latex]n\\ge 3[\/latex]. Further, the sequence is bounded below by [latex]0[\/latex] because [latex]\\frac{{4}^{n}}{n\\text{!}}\\ge 0[\/latex] for all positive integers [latex]n[\/latex]. Therefore, by the Monotone Convergence Theorem, the sequence converges.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo find the limit, we use the fact that the sequence converges and let [latex]L=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. Now note this important observation. Consider [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}[\/latex]. Since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736790081\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{{a}_{n+1}\\right\\}=\\left\\{{a}_{2,}{a}_{3},{a}_{4}\\text{,}\\ldots\\right\\}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p>the only difference between the sequences [latex]\\left\\{{a}_{n+1}\\right\\}[\/latex] and [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is that [latex]\\left\\{{a}_{n+1}\\right\\}[\/latex] omits the first term. Since a finite number of terms does not affect the convergence of a sequence,<\/p>\n<p><span data-type=\"newline\">\u00a0<\/span><\/p>\n<div id=\"fs-id1169739110746\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=L[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nCombining this fact with the equation<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739110806\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n+1}=\\frac{4}{n+1}{a}_{n}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nand taking the limit of both sides of the equation<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739195686\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{4}{n+1}{a}_{n}[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwe can conclude that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739195752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L=0\\cdot L=0[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Writing out the first several terms,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739195780\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left\\{2,\\frac{5}{4},\\frac{41}{40},\\frac{3281}{3280}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwe can conjecture that the sequence is decreasing and bounded below by [latex]1[\/latex]. To show that the sequence is bounded below by [latex]1[\/latex], we can show that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739341340\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\ge 1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo show this, first rewrite<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739341380\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}=\\frac{{a}_{n}^{2}+1}{2{a}_{n}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]{a}_{1}>0[\/latex] and [latex]{a}_{2}[\/latex] is defined as a sum of positive terms, [latex]{a}_{2}>0[\/latex]. Similarly, all terms [latex]{a}_{n}>0[\/latex]. Therefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739341490\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}^{2}+1}{2{a}_{n}}\\ge 1[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nif and only if<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736708125\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}^{2}+1\\ge 2{a}_{n}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nRewriting the inequality [latex]{a}_{n}^{2}+1\\ge 2{a}_{n}[\/latex] as [latex]{a}_{n}^{2}-2{a}_{n}+1\\ge 0[\/latex], and using the fact that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736708214\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}^{2}-2{a}_{n}+1={\\left({a}_{n}-1\\right)}^{2}\\ge 0[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nbecause the square of any real number is nonnegative, we can conclude that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736842868\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\ge 1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: left;\"><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo show that the sequence is decreasing, we must show that [latex]{a}_{n+1}\\le {a}_{n}[\/latex] for all [latex]n\\ge 1[\/latex]. Since [latex]1\\le {a}_{n}^{2}[\/latex], it follows that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736842957\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n}^{2}+1\\le 2{a}_{n}^{2}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nDividing both sides by [latex]2{a}_{n}[\/latex], we obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736843005\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\le {a}_{n}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nUsing the definition of [latex]{a}_{n+1}[\/latex], we conclude that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736705845\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{a}_{n+1}=\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\le {a}_{n}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is bounded below and decreasing, by the Monotone Convergence Theorem, it converges.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo find the limit, let [latex]L=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. Then using the recurrence relation and the fact that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}[\/latex], we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736702887\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\left(\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\right)[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nand therefore<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736702967\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L=\\frac{L}{2}+\\frac{1}{2L}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nMultiplying both sides of this equation by [latex]2L[\/latex], we arrive at the equation<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736703009\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]2{L}^{2}={L}^{2}+1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSolving this equation for [latex]L[\/latex], we conclude that [latex]{L}^{2}=1[\/latex], which implies [latex]L=\\pm 1[\/latex]. Since all the terms are positive, the limit [latex]L=1[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739202289\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169739202292\" data-type=\"exercise\">\n<div id=\"fs-id1169739202294\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169739202294\" data-type=\"problem\">\n<p id=\"fs-id1169739202296\">Consider the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that [latex]{a}_{1}=1[\/latex], [latex]{a}_{n}=\\frac{{a}_{n - 1}}{2}[\/latex]. Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558599\">Hint<\/span><\/p>\n<div id=\"q44558599\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739202370\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739202376\">Show the sequence is decreasing and bounded below.<\/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=\"q44558699\">Show Solution<\/span><\/p>\n<div id=\"q44558699\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739202358\" data-type=\"solution\">\n<p id=\"fs-id1169739202360\">[latex]0[\/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\/V-B05oiNKFk?controls=0&amp;start=713&amp;end=822&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\/5.1.3_713to822_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.3&#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=\"ohm5808\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5808&theme=oea&iframe_resize_id=ohm5808&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1169739202383\" class=\"project\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox tryit\">\n<h3 style=\"text-align: center;\" data-type=\"title\">ACTIVITY: Fibonacci Numbers<\/h3>\n<p id=\"fs-id1169739202389\">The <span class=\"no-emphasis\" data-type=\"term\">Fibonacci numbers<\/span> are defined recursively by the sequence [latex]\\left\\{{F}_{n}\\right\\}[\/latex] where [latex]{F}_{0}=0[\/latex], [latex]{F}_{1}=1[\/latex] and for [latex]n\\ge 2[\/latex],<\/p>\n<div id=\"fs-id1169739344161\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739344200\">Here we look at properties of the Fibonacci numbers.<\/p>\n<ol id=\"fs-id1169739344204\" type=\"1\">\n<li>Write out the first twenty Fibonacci numbers.<\/li>\n<li>Find a closed formula for the Fibonacci sequence by using the following steps.\n<ol id=\"fs-id1169739344213\" type=\"a\">\n<li>Consider the recursively defined sequence [latex]\\left\\{{x}_{n}\\right\\}[\/latex] where [latex]{x}_{o}=c[\/latex] and [latex]{x}_{n+1}=a{x}_{n}[\/latex]. Show that this sequence can be described by the closed formula [latex]{x}_{n}=c{a}^{n}[\/latex] for all [latex]n\\ge 0[\/latex].<\/li>\n<li>Using the result from part a. as motivation, look for a solution of the equation<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739344309\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{F}_{n}={F}_{n - 1}+{F}_{n - 2}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nof the form [latex]{F}_{n}=c{\\lambda }^{n}[\/latex]. Determine what two values for [latex]\\lambda[\/latex] will allow [latex]{F}_{n}[\/latex] to satisfy this equation.<\/li>\n<li>Consider the two solutions from part b.: [latex]{\\lambda }_{1}[\/latex] and [latex]{\\lambda }_{2}[\/latex]. Let [latex]{F}_{n}={c}_{1}{\\lambda }_{1}{}^{n}+{c}_{2}{\\lambda }_{2}{}^{n}[\/latex]. Use the initial conditions [latex]{F}_{0}[\/latex] and [latex]{F}_{1}[\/latex] to determine the values for the constants [latex]{c}_{1}[\/latex] and [latex]{c}_{2}[\/latex] and write the closed formula [latex]{F}_{n}[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Use the answer in 2 c. to show that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736856912\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{F}_{n+1}}{{F}_{n}}=\\frac{1+\\sqrt{5}}{2}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe number [latex]\\varphi =\\frac{\\left(1+\\sqrt{5}\\right)}{2}[\/latex] is known as the <span class=\"no-emphasis\" data-type=\"term\">golden ratio<\/span> (Figures 7 and 8).<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_09_01_004\">\n<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\/11234259\/CNX_Calc_Figure_09_01_004.jpg\" alt=\"This is a photo of a sunflower, particularly the curves of the seeds at its middle. The number of spirals in each direction is always a Fibonacci number.\" width=\"325\" height=\"244\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. The seeds in a sunflower exhibit spiral patterns curving to the left and to the right. The number of spirals in each direction is always a Fibonacci number\u2014always. (credit: modification of work by Esdras Calderan, Wikimedia Commons)<\/p>\n<\/div>\n<\/figure>\n<p><span data-type=\"newline\">\u00a0<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_09_01_005\">\n<div style=\"width: 498px\" 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\/11234302\/CNX_Calc_Figure_09_01_005.jpg\" alt=\"This is a photo of the Parthenon, an ancient Greek temple that was designed with the proportions of the Golden Rule. The entire temple\u2019s front side fits perfectly into a rectangle with those proportions, as do the columns, the level between the columns and the roof, and a portion of the trim below the roof.\" width=\"488\" height=\"300\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. The proportion of the golden ratio appears in many famous examples of art and architecture. The ancient Greek temple known as the Parthenon was designed with these proportions, and the ratio appears again in many of the smaller details. (credit: modification of work by TravelingOtter, Flickr)<\/p>\n<\/div>\n<\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169736857064\" class=\"key-concepts\" 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-1694\">\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>5.1.3. <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":5,"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\":\"5.1.3\",\"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-1694","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1694","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":7,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1694\/revisions"}],"predecessor-version":[{"id":2240,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1694\/revisions\/2240"}],"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\/1694\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1694"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1694"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1694"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1694"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}