{"id":1723,"date":"2021-07-23T20:34:42","date_gmt":"2021-07-23T20:34:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1723"},"modified":"2022-03-21T23:06:40","modified_gmt":"2022-03-21T23:06:40","slug":"ratio-and-root-tests-3","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/ratio-and-root-tests-3\/","title":{"raw":"Ratio and Root Tests","rendered":"Ratio and Root Tests"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the ratio test to determine absolute convergence of a series<\/li>\r\n \t<li>Use the root test to determine absolute convergence of a series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1169736729033\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Ratio Test<\/h2>\r\n<p id=\"fs-id1169736618515\">Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. From our earlier discussion and examples, we know that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex] is not a sufficient condition for the series to converge. Not only do we need [latex]{a}_{n}\\to 0[\/latex], but we need [latex]{a}_{n}\\to 0[\/latex] quickly enough. For example, consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] and the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. We know that [latex]\\frac{1}{n}\\to 0[\/latex] and [latex]\\frac{1}{{n}^{2}}\\to 0[\/latex]. However, only the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges. The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges because the terms in the sequence [latex]\\left\\{\\frac{1}{n}\\right\\}[\/latex] do not approach zero fast enough as [latex]n\\to \\infty [\/latex]. Here we introduce the<strong> ratio test<\/strong>, which provides a way of measuring how fast the terms of a series approach zero.<\/p>\r\n\r\n<div id=\"fs-id1169739067523\" 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: Ratio Test<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169736661198\">Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] be a series with nonzero terms. Let<\/p>\r\n\r\n<div id=\"fs-id1169739298350\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|[\/latex].<\/div>\r\n&nbsp;\r\n<ol id=\"fs-id1169739302372\" type=\"i\">\r\n \t<li>If [latex]0\\le \\rho &lt;1[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges absolutely.<\/li>\r\n \t<li>If [latex]\\rho &gt;1[\/latex] or [latex]\\rho =\\infty [\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\r\n \t<li>If [latex]\\rho =1[\/latex], the test does not provide any information.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1169736593669\" data-depth=\"2\">\r\n<h4 data-type=\"title\">Proof<\/h4>\r\n<p id=\"fs-id1169739175440\">Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] be a series with nonzero terms.<\/p>\r\n<p id=\"fs-id1169739208935\">We begin with the proof of part i. In this case, [latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|&lt;1[\/latex]. Since [latex]0\\le \\rho &lt;1[\/latex], there exists [latex]R[\/latex] such that [latex]0\\le \\rho &lt;R&lt;1[\/latex]. Let [latex]\\epsilon =R-\\rho &gt;0[\/latex]. By the definition of limit of a sequence, there exists some integer [latex]N[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169739275391\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |&lt;\\epsilon \\text{ for all }n\\ge N[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738923259\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1169738869993\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|\\frac{{a}_{n+1}}{{a}_{n}}|&lt;\\rho +\\epsilon =R\\text{ for all }n\\ge N[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739344904\">and, thus,<\/p>\r\n\r\n<div id=\"fs-id1169738914987\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}|{a}_{N+1}|&lt;R|{a}_{N}|\\\\ |{a}_{N+2}|&lt;R|{a}_{N+1}|&lt;{R}^{2}|{a}_{N}|\\\\ |{a}_{N+3}|&lt;R|{a}_{N+2}|&lt;{R}^{2}|{a}_{N+1}|&lt;{R}^{3}|{a}_{N}|\\\\ |{a}_{N+4}|&lt;R|{a}_{N+3}|&lt;{R}^{2}|{a}_{N+2}|&lt;{R}^{3}|{a}_{N+1}|&lt;{R}^{4}|{a}_{N}|\\\\ \\vdots .\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739302830\">Since [latex]R&lt;1[\/latex], the geometric series<\/p>\r\n\r\n<div id=\"fs-id1169739212808\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738971692\">converges. Given the inequalities above, we can apply the comparison test and conclude that the series<\/p>\r\n\r\n<div id=\"fs-id1169738971694\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+|{a}_{N+4}|+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738904500\">converges. Therefore, since<\/p>\r\n\r\n<div id=\"fs-id1169739298749\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|=\\displaystyle\\sum _{n=1}^{N}|{a}_{n}|+\\displaystyle\\sum _{n=N+1}^{\\infty }|{a}_{n}|[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738971786\">where [latex]\\displaystyle\\sum _{n=1}^{N}|{a}_{n}|[\/latex] is a finite sum and [latex]\\displaystyle\\sum _{n=N+1}^{\\infty }|{a}_{n}|[\/latex] converges, we conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges.<\/p>\r\n<p id=\"fs-id1169739010004\">For part ii.<\/p>\r\n\r\n<div id=\"fs-id1169739100280\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|&gt;1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739019548\">Since [latex]\\rho &gt;1[\/latex], there exists [latex]R[\/latex] such that [latex]\\rho &gt;R&gt;1[\/latex]. Let [latex]\\epsilon =\\rho -R&gt;0[\/latex]. By the definition of the limit of a sequence, there exists an integer [latex]N[\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169739007422\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |&lt;\\epsilon \\text{for all}n\\ge N[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739006446\">Therefore,<\/p>\r\n\r\n<div id=\"fs-id1169738906387\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]R=\\rho -\\epsilon &lt;|\\frac{{a}_{n+1}}{{a}_{n}}|\\text{for all}n\\ge N[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738895005\">and, thus,<\/p>\r\n\r\n<div id=\"fs-id1169738912557\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}|{a}_{N+1}|&gt;R|{a}_{N}|\\\\ |{a}_{N+2}|&gt;R|{a}_{N+1}|&gt;{R}^{2}|{a}_{N}|\\\\ |{a}_{N+3}|&gt;R|{a}_{N+2}|&gt;{R}^{2}|{a}_{N+1}|&gt;{R}^{3}|{a}_{N}|\\\\ |{a}_{N+4}|&gt;R|{a}_{N+3}|&gt;{R}^{2}|{a}_{N+2}|&gt;{R}^{3}|{a}_{N+1}|&gt;{R}^{4}|{a}_{N}|.\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739042629\">Since [latex]R&gt;1[\/latex], the geometric series<\/p>\r\n\r\n<div id=\"fs-id1169739017227\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736661594\">diverges. Applying the comparison test, we conclude that the series<\/p>\r\n\r\n<div id=\"fs-id1169739097736\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739001004\">diverges, and therefore the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] diverges.<\/p>\r\n<p id=\"fs-id1169739020685\">For part iii. we show that the test does not provide any information if [latex]\\rho =1[\/latex] by considering the [latex]p-\\text{series}[\/latex] [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]. For any real number [latex]p[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169739043970\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{{\\left(n+1\\right)}^{p}}}{\\frac{1}{{n}^{p}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{p}}{{\\left(n+1\\right)}^{p}}=1[\/latex].<\/div>\r\n<p id=\"fs-id1169739190247\">However, we know that if [latex]p\\le 1[\/latex], the [latex]p-\\text{series}[\/latex] [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex] diverges, whereas [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex] converges if [latex]p&gt;1[\/latex].<\/p>\r\n<p id=\"fs-id1169739038269\">[latex]_\\blacksquare[\/latex]<\/p>\r\n<p id=\"fs-id1169736611442\">The ratio test is particularly useful for series whose terms contain factorials or exponentials, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.<\/p>\r\nBefore we apply the Ratio Test, we briefly review some laws of exponents and factorial notation, both of which will be of critical importance for using the Ratio Test.\r\n<div id=\"fs-id1169736845320\" data-type=\"example\">\r\n<div id=\"fs-id1169739301472\" data-type=\"exercise\">\r\n<div id=\"fs-id1169736708955\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Simplifying exponential expressions<\/h3>\r\nFor any [latex] b &gt; 0 [\/latex] and real numbers [latex] m [\/latex] and [latex] n [\/latex]: [latex] \\frac{b^m}{b^n} = b^{m-n} = \\frac{1}{b^{n-m}}[\/latex].\r\n\r\nConsider the examples below:\r\n<ul>\r\n \t<li>[latex] \\frac{e^{n+1}}{e^n} = e [\/latex]<\/li>\r\n \t<li>[latex] \\frac{3^{2n+1}}{3^{2n+3}} = \\frac{1}{3^2} = \\frac{1}{9} [\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Factorial Notation<\/h3>\r\nThe expression [latex] n! [\/latex], called <strong>[latex] n [\/latex] factorial<\/strong>, is defined as the product of the integer [latex] n [\/latex] with all positive integers less than [latex] n [\/latex].\r\n<p style=\"padding-left: 30px;\">[latex] n! = n (n-1) (n-2) \\ldots 3 \\cdot 2 \\cdot 1 [\/latex]<\/p>\r\nFor example, [latex] 5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120 [\/latex]\r\n\r\nBy definition, [latex] 0! = 1 [\/latex]\r\n\r\nNote that a ratio of factorial expressions can be simplified.\u00a0 Consider the examples below:\r\n<ul>\r\n \t<li>[latex] \\frac{10!}{8!} = \\frac{10 \\cdot 9 \\cdot 8 \\cdot 7 \\ldots 3 \\cdot 2 \\cdot 1 }{8 \\cdot 7 \\ldots 3 \\cdot 2 \\cdot 1} = 10\\cdot 9 = 90 [\/latex]<\/li>\r\n \t<li>[latex] \\frac {n!}{(n+2)!} = \\frac{n!}{(n+2)(n+1)n!} = \\frac{1}{(n+2)(n+1)} [\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Using the Ratio Test<\/h3>\r\n<div id=\"fs-id1169736708955\" data-type=\"problem\">\r\n<p id=\"fs-id1169736843900\">For each of the following series, use the ratio test to determine whether the series converges or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169738970420\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{\\left(n\\text{!}\\right)}^{2}}{\\left(2n\\right)\\text{!}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1169739094061\" data-type=\"solution\">\r\n<ol id=\"fs-id1169738905677\" type=\"a\">\r\n \t<li>From the ratio test, we can see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739097521\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{{2}^{n+1}}{\\left(n+1\\right)\\text{!}}}{\\frac{{2}^{n}}{n\\text{!}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{{2}^{n}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\left(n+1\\right)\\text{!}=\\left(n+1\\right)\\cdot n\\text{!}[\/latex], <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738890904\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{2}{n+1}=0[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\rho &lt;1[\/latex], the series converges.<\/li>\r\n \t<li>We can see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736738161\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\rho \\hfill &amp; =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{{\\left(n+1\\right)}^{n+1}}{\\left(n+1\\right)\\text{!}}}{\\frac{{n}^{n}}{n\\text{!}}}\\hfill \\\\ &amp; =\\underset{n\\to \\infty }{\\text{lim}}\\frac{{\\left(n+1\\right)}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{{n}^{n}}\\hfill \\\\ &amp; =\\underset{n\\to \\infty }{\\text{lim}}{\\left(\\frac{n+1}{n}\\right)}^{n}=\\underset{n\\to \\infty }{\\text{lim}}{\\left(1+\\frac{1}{n}\\right)}^{n}=e.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\rho &gt;1[\/latex], the series diverges.<\/li>\r\n \t<li>Since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739205416\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill |\\frac{\\frac{{\\left(-1\\right)}^{n+1}{\\left(\\left(n+1\\right)\\text{!}\\right)}^{2}}{\\left(2\\left(n+1\\right)\\right)\\text{!}}}{\\frac{{\\left(-1\\right)}^{n}{\\left(n\\text{!}\\right)}^{2}}{\\left(2n\\right)\\text{!}}}|&amp; =\\frac{\\left(n+1\\right)\\text{!}\\left(n+1\\right)\\text{!}}{\\left(2n+2\\right)\\text{!}}\\cdot \\frac{\\left(2n\\right)\\text{!}}{n\\text{!}n\\text{!}}\\hfill \\\\ &amp; =\\frac{\\left(n+1\\right)\\left(n+1\\right)}{\\left(2n+2\\right)\\left(2n+1\\right)}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwe see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739093131\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\left(n+1\\right)\\left(n+1\\right)}{\\left(2n+2\\right)\\left(2n+1\\right)}=\\frac{1}{4}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\rho &lt;1[\/latex], the series converges.<\/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-id1169736850668\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169736850672\" data-type=\"exercise\">\r\n<div id=\"fs-id1169739274678\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169739274678\" data-type=\"problem\">\r\n<p id=\"fs-id1169739274680\">Use the ratio test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{3}}{{3}^{n}}[\/latex] converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1169739191528\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169736846183\">Evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{\\left(n+1\\right)}^{3}}{{3}^{n+1}}\\cdot \\frac{{3}^{n}}{{n}^{3}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1169739029192\" data-type=\"solution\">\r\n<p id=\"fs-id1169739029194\">The series converges.<\/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\/bQ2aQNS-mjU?controls=0&amp;start=606&amp;end=698&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\/5.6.1_606to698_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.1\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]49923[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1169736607627\" data-depth=\"2\">\r\n<h2 data-type=\"title\">Root Test<\/h2>\r\n<p id=\"fs-id1169739033678\">The approach of the <span data-type=\"term\">root test<\/span> is similar to that of the ratio test. Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] such that [latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|{a}_{n}|}=\\rho [\/latex] for some real number [latex]\\rho [\/latex]. Then for [latex]N[\/latex] sufficiently large, [latex]|{a}_{N}|\\approx {\\rho }^{N}[\/latex]. Therefore, we can approximate [latex]\\displaystyle\\sum _{n=N}^{\\infty }|{a}_{n}|[\/latex] by writing<\/p>\r\n\r\n<div id=\"fs-id1169739040737\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{a}_{N}|+|{a}_{N+1}|+|{a}_{N+2}|+\\cdots \\approx {\\rho }^{N}+{\\rho }^{N+1}+{\\rho }^{N+2}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738907603\">The expression on the right-hand side is a geometric series. As in the ratio test, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges absolutely if [latex]0\\le \\rho &lt;1[\/latex] and the series diverges if [latex]\\rho \\ge 1[\/latex]. If [latex]\\rho =1[\/latex], the test does not provide any information. For example, for any <em data-effect=\"italics\">p<\/em>-series, [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1169736852758\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|\\frac{1}{{n}^{p}}|}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{1}{{n}^{{p}{n}}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736844025\">To evaluate this limit, we use the natural logarithm function. Doing so, we see that<\/p>\r\n\r\n<div id=\"fs-id1169736853081\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}\\rho =\\text{ln}\\left(\\underset{n\\to \\infty }{\\text{lim}}\\frac{1}{{n}^{\\frac{p}{n}}}\\right)=\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}{\\left(\\frac{1}{n}\\right)}^{\\frac{p}{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{p}{n}\\cdot \\text{ln}\\left(\\frac{1}{n}\\right)=\\underset{n\\to \\infty }{\\text{lim}}\\frac{p\\text{ln}\\left(\\frac{1}{n}\\right)}{n}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736852760\">Using L\u2019H\u00f4pital\u2019s rule, it follows that [latex]\\text{ln}\\rho =0[\/latex], and therefore [latex]\\rho =1[\/latex] for all [latex]p[\/latex]. However, we know that the <em data-effect=\"italics\">p<\/em>-series only converges if [latex]p&gt;1[\/latex] and diverges if [latex]p&lt;1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169736852039\" 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: Root Test<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169736852046\">Consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. Let<\/p>\r\n\r\n<div id=\"fs-id1169739260282\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|{a}_{n}|}[\/latex].<\/div>\r\n&nbsp;\r\n<ol id=\"fs-id1169739352105\" type=\"i\">\r\n \t<li>If [latex]0\\le \\rho &lt;1[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges absolutely.<\/li>\r\n \t<li>If [latex]\\rho &gt;1[\/latex] or [latex]\\rho =\\infty [\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\r\n \t<li>If [latex]\\rho =1[\/latex], the test does not provide any information.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169736583976\">The root test is useful for series whose terms involve exponentials. In particular, for a series whose terms [latex]{a}_{n}[\/latex] satisfy [latex]|{a}_{n}|={b}_{n}^{n}[\/latex], then [latex]\\sqrt[n]{|{a}_{n}|}={b}_{n}[\/latex] and we need only evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169736615350\" data-type=\"example\">\r\n<div id=\"fs-id1169736615352\" data-type=\"exercise\">\r\n<div id=\"fs-id1169736858393\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using the Root Test<\/h3>\r\n<div id=\"fs-id1169736858393\" data-type=\"problem\">\r\n<p id=\"fs-id1169736858398\">For each of the following series, use the root test to determine whether the series converges or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169736858402\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left({n}^{2}+3n\\right)}^{n}}{{\\left(4{n}^{2}+5\\right)}^{n}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{n}}{{\\left(\\text{ln}\\left(n\\right)\\right)}^{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1169739110614\" data-type=\"solution\">\r\n<ol id=\"fs-id1169739110616\" type=\"a\">\r\n \t<li>To apply the root test, we compute<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739110625\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{{\\left({n}^{2}+3n\\right)}^{n}}{{\\left(4{n}^{2}+5\\right)}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{2}+3n}{4{n}^{2}+5}=\\frac{1}{4}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\rho &lt;1[\/latex], the series converges absolutely.<\/li>\r\n \t<li>We have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736636617\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{{n}^{n}}{{\\left(\\text{ln}n\\right)}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{n}{\\text{ln}n}=\\infty \\text{by L'H\u00f4pital's rule}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\rho =\\infty [\/latex], 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-id1169736852676\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169736852679\" data-type=\"exercise\">\r\n<div id=\"fs-id1169736852681\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169736852681\" data-type=\"problem\">\r\n<p id=\"fs-id1169736852683\">Use the root test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{n}}[\/latex] converges or diverges.<\/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-id1169736852722\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169736592424\">Evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{1}{{n}^{n}}}[\/latex] using L\u2019H\u00f4pital\u2019s rule.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169736852715\" data-type=\"solution\">\r\n<p id=\"fs-id1169736852717\">The series converges.<\/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\/Y_kfrSKCjZk?controls=0&amp;start=255&amp;end=351&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.6.2_255to351_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.6.2\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question] 20315[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><\/section><section id=\"fs-id1169736592463\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the ratio test to determine absolute convergence of a series<\/li>\n<li>Use the root test to determine absolute convergence of a series<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1169736729033\" data-depth=\"1\">\n<h2 data-type=\"title\">Ratio Test<\/h2>\n<p id=\"fs-id1169736618515\">Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. From our earlier discussion and examples, we know that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex] is not a sufficient condition for the series to converge. Not only do we need [latex]{a}_{n}\\to 0[\/latex], but we need [latex]{a}_{n}\\to 0[\/latex] quickly enough. For example, consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] and the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. We know that [latex]\\frac{1}{n}\\to 0[\/latex] and [latex]\\frac{1}{{n}^{2}}\\to 0[\/latex]. However, only the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges. The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges because the terms in the sequence [latex]\\left\\{\\frac{1}{n}\\right\\}[\/latex] do not approach zero fast enough as [latex]n\\to \\infty[\/latex]. Here we introduce the<strong> ratio test<\/strong>, which provides a way of measuring how fast the terms of a series approach zero.<\/p>\n<div id=\"fs-id1169739067523\" 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: Ratio Test<\/h3>\n<hr \/>\n<p id=\"fs-id1169736661198\">Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] be a series with nonzero terms. Let<\/p>\n<div id=\"fs-id1169739298350\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<ol id=\"fs-id1169739302372\" type=\"i\">\n<li>If [latex]0\\le \\rho <1[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges absolutely.<\/li>\n<li>If [latex]\\rho >1[\/latex] or [latex]\\rho =\\infty[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\n<li>If [latex]\\rho =1[\/latex], the test does not provide any information.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169736593669\" data-depth=\"2\">\n<h4 data-type=\"title\">Proof<\/h4>\n<p id=\"fs-id1169739175440\">Let [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] be a series with nonzero terms.<\/p>\n<p id=\"fs-id1169739208935\">We begin with the proof of part i. In this case, [latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|<1[\/latex]. Since [latex]0\\le \\rho <1[\/latex], there exists [latex]R[\/latex] such that [latex]0\\le \\rho <R<1[\/latex]. Let [latex]\\epsilon =R-\\rho >0[\/latex]. By the definition of limit of a sequence, there exists some integer [latex]N[\/latex] such that<\/p>\n<div id=\"fs-id1169739275391\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |<\\epsilon \\text{ for all }n\\ge N[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738923259\">Therefore,<\/p>\n<div id=\"fs-id1169738869993\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|\\frac{{a}_{n+1}}{{a}_{n}}|<\\rho +\\epsilon =R\\text{ for all }n\\ge N[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739344904\">and, thus,<\/p>\n<div id=\"fs-id1169738914987\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}|{a}_{N+1}|<R|{a}_{N}|\\\\ |{a}_{N+2}|<R|{a}_{N+1}|<{R}^{2}|{a}_{N}|\\\\ |{a}_{N+3}|<R|{a}_{N+2}|<{R}^{2}|{a}_{N+1}|<{R}^{3}|{a}_{N}|\\\\ |{a}_{N+4}|<R|{a}_{N+3}|<{R}^{2}|{a}_{N+2}|<{R}^{3}|{a}_{N+1}|<{R}^{4}|{a}_{N}|\\\\ \\vdots .\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739302830\">Since [latex]R<1[\/latex], the geometric series<\/p>\n<div id=\"fs-id1169739212808\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738971692\">converges. Given the inequalities above, we can apply the comparison test and conclude that the series<\/p>\n<div id=\"fs-id1169738971694\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+|{a}_{N+4}|+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738904500\">converges. Therefore, since<\/p>\n<div id=\"fs-id1169739298749\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|=\\displaystyle\\sum _{n=1}^{N}|{a}_{n}|+\\displaystyle\\sum _{n=N+1}^{\\infty }|{a}_{n}|[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738971786\">where [latex]\\displaystyle\\sum _{n=1}^{N}|{a}_{n}|[\/latex] is a finite sum and [latex]\\displaystyle\\sum _{n=N+1}^{\\infty }|{a}_{n}|[\/latex] converges, we conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges.<\/p>\n<p id=\"fs-id1169739010004\">For part ii.<\/p>\n<div id=\"fs-id1169739100280\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|>1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739019548\">Since [latex]\\rho >1[\/latex], there exists [latex]R[\/latex] such that [latex]\\rho >R>1[\/latex]. Let [latex]\\epsilon =\\rho -R>0[\/latex]. By the definition of the limit of a sequence, there exists an integer [latex]N[\/latex] such that<\/p>\n<div id=\"fs-id1169739007422\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |<\\epsilon \\text{for all}n\\ge N[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739006446\">Therefore,<\/p>\n<div id=\"fs-id1169738906387\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]R=\\rho -\\epsilon <|\\frac{{a}_{n+1}}{{a}_{n}}|\\text{for all}n\\ge N[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738895005\">and, thus,<\/p>\n<div id=\"fs-id1169738912557\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}|{a}_{N+1}|>R|{a}_{N}|\\\\ |{a}_{N+2}|>R|{a}_{N+1}|>{R}^{2}|{a}_{N}|\\\\ |{a}_{N+3}|>R|{a}_{N+2}|>{R}^{2}|{a}_{N+1}|>{R}^{3}|{a}_{N}|\\\\ |{a}_{N+4}|>R|{a}_{N+3}|>{R}^{2}|{a}_{N+2}|>{R}^{3}|{a}_{N+1}|>{R}^{4}|{a}_{N}|.\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739042629\">Since [latex]R>1[\/latex], the geometric series<\/p>\n<div id=\"fs-id1169739017227\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736661594\">diverges. Applying the comparison test, we conclude that the series<\/p>\n<div id=\"fs-id1169739097736\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739001004\">diverges, and therefore the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] diverges.<\/p>\n<p id=\"fs-id1169739020685\">For part iii. we show that the test does not provide any information if [latex]\\rho =1[\/latex] by considering the [latex]p-\\text{series}[\/latex] [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]. For any real number [latex]p[\/latex],<\/p>\n<div id=\"fs-id1169739043970\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{{\\left(n+1\\right)}^{p}}}{\\frac{1}{{n}^{p}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{p}}{{\\left(n+1\\right)}^{p}}=1[\/latex].<\/div>\n<p id=\"fs-id1169739190247\">However, we know that if [latex]p\\le 1[\/latex], the [latex]p-\\text{series}[\/latex] [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex] diverges, whereas [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex] converges if [latex]p>1[\/latex].<\/p>\n<p id=\"fs-id1169739038269\">[latex]_\\blacksquare[\/latex]<\/p>\n<p id=\"fs-id1169736611442\">The ratio test is particularly useful for series whose terms contain factorials or exponentials, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.<\/p>\n<p>Before we apply the Ratio Test, we briefly review some laws of exponents and factorial notation, both of which will be of critical importance for using the Ratio Test.<\/p>\n<div id=\"fs-id1169736845320\" data-type=\"example\">\n<div id=\"fs-id1169739301472\" data-type=\"exercise\">\n<div id=\"fs-id1169736708955\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3>Recall: Simplifying exponential expressions<\/h3>\n<p>For any [latex]b > 0[\/latex] and real numbers [latex]m[\/latex] and [latex]n[\/latex]: [latex]\\frac{b^m}{b^n} = b^{m-n} = \\frac{1}{b^{n-m}}[\/latex].<\/p>\n<p>Consider the examples below:<\/p>\n<ul>\n<li>[latex]\\frac{e^{n+1}}{e^n} = e[\/latex]<\/li>\n<li>[latex]\\frac{3^{2n+1}}{3^{2n+3}} = \\frac{1}{3^2} = \\frac{1}{9}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Factorial Notation<\/h3>\n<p>The expression [latex]n![\/latex], called <strong>[latex]n[\/latex] factorial<\/strong>, is defined as the product of the integer [latex]n[\/latex] with all positive integers less than [latex]n[\/latex].<\/p>\n<p style=\"padding-left: 30px;\">[latex]n! = n (n-1) (n-2) \\ldots 3 \\cdot 2 \\cdot 1[\/latex]<\/p>\n<p>For example, [latex]5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120[\/latex]<\/p>\n<p>By definition, [latex]0! = 1[\/latex]<\/p>\n<p>Note that a ratio of factorial expressions can be simplified.\u00a0 Consider the examples below:<\/p>\n<ul>\n<li>[latex]\\frac{10!}{8!} = \\frac{10 \\cdot 9 \\cdot 8 \\cdot 7 \\ldots 3 \\cdot 2 \\cdot 1 }{8 \\cdot 7 \\ldots 3 \\cdot 2 \\cdot 1} = 10\\cdot 9 = 90[\/latex]<\/li>\n<li>[latex]\\frac {n!}{(n+2)!} = \\frac{n!}{(n+2)(n+1)n!} = \\frac{1}{(n+2)(n+1)}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Using the Ratio Test<\/h3>\n<div id=\"fs-id1169736708955\" data-type=\"problem\">\n<p id=\"fs-id1169736843900\">For each of the following series, use the ratio test to determine whether the series converges or diverges.<\/p>\n<ol id=\"fs-id1169738970420\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{n\\text{!}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{n}}{n\\text{!}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{\\left(n\\text{!}\\right)}^{2}}{\\left(2n\\right)\\text{!}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739094061\" data-type=\"solution\">\n<ol id=\"fs-id1169738905677\" type=\"a\">\n<li>From the ratio test, we can see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739097521\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{{2}^{n+1}}{\\left(n+1\\right)\\text{!}}}{\\frac{{2}^{n}}{n\\text{!}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{{2}^{n}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\left(n+1\\right)\\text{!}=\\left(n+1\\right)\\cdot n\\text{!}[\/latex], <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738890904\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{2}{n+1}=0[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\rho <1[\/latex], the series converges.<\/li>\n<li>We can see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736738161\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\rho \\hfill & =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{{\\left(n+1\\right)}^{n+1}}{\\left(n+1\\right)\\text{!}}}{\\frac{{n}^{n}}{n\\text{!}}}\\hfill \\\\ & =\\underset{n\\to \\infty }{\\text{lim}}\\frac{{\\left(n+1\\right)}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{{n}^{n}}\\hfill \\\\ & =\\underset{n\\to \\infty }{\\text{lim}}{\\left(\\frac{n+1}{n}\\right)}^{n}=\\underset{n\\to \\infty }{\\text{lim}}{\\left(1+\\frac{1}{n}\\right)}^{n}=e.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\rho >1[\/latex], the series diverges.<\/li>\n<li>Since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739205416\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill |\\frac{\\frac{{\\left(-1\\right)}^{n+1}{\\left(\\left(n+1\\right)\\text{!}\\right)}^{2}}{\\left(2\\left(n+1\\right)\\right)\\text{!}}}{\\frac{{\\left(-1\\right)}^{n}{\\left(n\\text{!}\\right)}^{2}}{\\left(2n\\right)\\text{!}}}|& =\\frac{\\left(n+1\\right)\\text{!}\\left(n+1\\right)\\text{!}}{\\left(2n+2\\right)\\text{!}}\\cdot \\frac{\\left(2n\\right)\\text{!}}{n\\text{!}n\\text{!}}\\hfill \\\\ & =\\frac{\\left(n+1\\right)\\left(n+1\\right)}{\\left(2n+2\\right)\\left(2n+1\\right)}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwe see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739093131\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\left(n+1\\right)\\left(n+1\\right)}{\\left(2n+2\\right)\\left(2n+1\\right)}=\\frac{1}{4}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\rho <1[\/latex], the series converges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736850668\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169736850672\" data-type=\"exercise\">\n<div id=\"fs-id1169739274678\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169739274678\" data-type=\"problem\">\n<p id=\"fs-id1169739274680\">Use the ratio test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{3}}{{3}^{n}}[\/latex] converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739191528\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169736846183\">Evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{\\left(n+1\\right)}^{3}}{{3}^{n+1}}\\cdot \\frac{{3}^{n}}{{n}^{3}}[\/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=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739029192\" data-type=\"solution\">\n<p id=\"fs-id1169739029194\">The series converges.<\/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\/bQ2aQNS-mjU?controls=0&amp;start=606&amp;end=698&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\/5.6.1_606to698_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.1&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm49923\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=49923&theme=oea&iframe_resize_id=ohm49923&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1169736607627\" data-depth=\"2\">\n<h2 data-type=\"title\">Root Test<\/h2>\n<p id=\"fs-id1169739033678\">The approach of the <span data-type=\"term\">root test<\/span> is similar to that of the ratio test. Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] such that [latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|{a}_{n}|}=\\rho[\/latex] for some real number [latex]\\rho[\/latex]. Then for [latex]N[\/latex] sufficiently large, [latex]|{a}_{N}|\\approx {\\rho }^{N}[\/latex]. Therefore, we can approximate [latex]\\displaystyle\\sum _{n=N}^{\\infty }|{a}_{n}|[\/latex] by writing<\/p>\n<div id=\"fs-id1169739040737\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|{a}_{N}|+|{a}_{N+1}|+|{a}_{N+2}|+\\cdots \\approx {\\rho }^{N}+{\\rho }^{N+1}+{\\rho }^{N+2}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738907603\">The expression on the right-hand side is a geometric series. As in the ratio test, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges absolutely if [latex]0\\le \\rho <1[\/latex] and the series diverges if [latex]\\rho \\ge 1[\/latex]. If [latex]\\rho =1[\/latex], the test does not provide any information. For example, for any <em data-effect=\"italics\">p<\/em>-series, [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex], we see that<\/p>\n<div id=\"fs-id1169736852758\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|\\frac{1}{{n}^{p}}|}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{1}{{n}^{{p}{n}}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736844025\">To evaluate this limit, we use the natural logarithm function. Doing so, we see that<\/p>\n<div id=\"fs-id1169736853081\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}\\rho =\\text{ln}\\left(\\underset{n\\to \\infty }{\\text{lim}}\\frac{1}{{n}^{\\frac{p}{n}}}\\right)=\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}{\\left(\\frac{1}{n}\\right)}^{\\frac{p}{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{p}{n}\\cdot \\text{ln}\\left(\\frac{1}{n}\\right)=\\underset{n\\to \\infty }{\\text{lim}}\\frac{p\\text{ln}\\left(\\frac{1}{n}\\right)}{n}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736852760\">Using L\u2019H\u00f4pital\u2019s rule, it follows that [latex]\\text{ln}\\rho =0[\/latex], and therefore [latex]\\rho =1[\/latex] for all [latex]p[\/latex]. However, we know that the <em data-effect=\"italics\">p<\/em>-series only converges if [latex]p>1[\/latex] and diverges if [latex]p<1[\/latex].<\/p>\n<div id=\"fs-id1169736852039\" 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: Root Test<\/h3>\n<hr \/>\n<p id=\"fs-id1169736852046\">Consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. Let<\/p>\n<div id=\"fs-id1169739260282\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|{a}_{n}|}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<ol id=\"fs-id1169739352105\" type=\"i\">\n<li>If [latex]0\\le \\rho <1[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges absolutely.<\/li>\n<li>If [latex]\\rho >1[\/latex] or [latex]\\rho =\\infty[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\n<li>If [latex]\\rho =1[\/latex], the test does not provide any information.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169736583976\">The root test is useful for series whose terms involve exponentials. In particular, for a series whose terms [latex]{a}_{n}[\/latex] satisfy [latex]|{a}_{n}|={b}_{n}^{n}[\/latex], then [latex]\\sqrt[n]{|{a}_{n}|}={b}_{n}[\/latex] and we need only evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}{b}_{n}[\/latex].<\/p>\n<div id=\"fs-id1169736615350\" data-type=\"example\">\n<div id=\"fs-id1169736615352\" data-type=\"exercise\">\n<div id=\"fs-id1169736858393\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Using the Root Test<\/h3>\n<div id=\"fs-id1169736858393\" data-type=\"problem\">\n<p id=\"fs-id1169736858398\">For each of the following series, use the root test to determine whether the series converges or diverges.<\/p>\n<ol id=\"fs-id1169736858402\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left({n}^{2}+3n\\right)}^{n}}{{\\left(4{n}^{2}+5\\right)}^{n}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{n}}{{\\left(\\text{ln}\\left(n\\right)\\right)}^{n}}[\/latex]<\/li>\n<\/ol>\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-id1169739110614\" data-type=\"solution\">\n<ol id=\"fs-id1169739110616\" type=\"a\">\n<li>To apply the root test, we compute<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739110625\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{{\\left({n}^{2}+3n\\right)}^{n}}{{\\left(4{n}^{2}+5\\right)}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{2}+3n}{4{n}^{2}+5}=\\frac{1}{4}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\rho <1[\/latex], the series converges absolutely.<\/li>\n<li>We have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736636617\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{{n}^{n}}{{\\left(\\text{ln}n\\right)}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{n}{\\text{ln}n}=\\infty \\text{by L'H\u00f4pital's rule}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\rho =\\infty[\/latex], 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-id1169736852676\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169736852679\" data-type=\"exercise\">\n<div id=\"fs-id1169736852681\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169736852681\" data-type=\"problem\">\n<p id=\"fs-id1169736852683\">Use the root test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{n}}[\/latex] converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736852722\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169736592424\">Evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{1}{{n}^{n}}}[\/latex] using L\u2019H\u00f4pital\u2019s rule.<\/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=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736852715\" data-type=\"solution\">\n<p id=\"fs-id1169736852717\">The series converges.<\/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\/Y_kfrSKCjZk?controls=0&amp;start=255&amp;end=351&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.6.2_255to351_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.6.2&#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=\"ohm20315\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20315&theme=oea&iframe_resize_id=ohm20315&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<\/section>\n<section id=\"fs-id1169736592463\" 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-1723\">\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>5.6.1. <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>5.6.2. <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 <\/section>","protected":false},"author":17533,"menu_order":25,"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\":\"5.6.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"5.6.2\",\"author\":\"Ryan 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