{"id":1123,"date":"2025-06-30T16:36:34","date_gmt":"2025-06-30T16:36:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1123"},"modified":"2025-09-11T16:31:47","modified_gmt":"2025-09-11T16:31:47","slug":"ratio-and-root-tests-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/ratio-and-root-tests-learn-it-1\/","title":{"raw":"Ratio and Root Tests: Learn It 1","rendered":"Ratio and Root Tests: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the ratio test to check if a series converges absolutely<\/li>\r\n \t<li>Use the root test to check if a series converges absolutely<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p data-type=\"title\">You've already worked through several convergence tests, and you've likely noticed that no single test works for every series. The ratio test and root test complete our toolkit by offering particularly useful methods that don't require finding a comparable series. The ratio test will prove especially valuable when we explore power series in the next module.<\/p>\r\n\r\n<h2 data-type=\"title\">Ratio Test<\/h2>\r\n<p class=\"whitespace-normal break-words\">Consider a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex]. We know that having [latex]\\lim{n\\to \\infty}{a}_{n}=0[\/latex] is necessary for convergence, but it's not sufficient. The terms must approach zero quickly enough.<\/p>\r\n<p class=\"whitespace-normal break-words\">Think about these two familiar examples:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] (harmonic series) diverges<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^{2}}[\/latex] (p-series with [latex]p=2[\/latex]) converges<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Both have terms that approach zero, but [latex]\\frac{1}{n^{2}}[\/latex] approaches zero much faster than [latex]\\frac{1}{n}[\/latex]. The ratio test provides a systematic way to measure this \"speed\" of approach to zero.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>ratio test<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] be a series with nonzero terms. Let<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\rho =\\lim_{n\\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right|[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Then:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">If [latex]0\\le \\rho &lt;1[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] <strong>converges absolutely<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]\\rho &gt;1[\/latex] or [latex]\\rho =\\infty[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] <strong>diverges<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]\\rho =1[\/latex], the test is <strong>inconclusive\u00a0<\/strong>and does not provide any information<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\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<center>[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |&lt;\\epsilon \\text{ for all }n\\ge N[\/latex].<\/center>\r\n<p id=\"fs-id1169738923259\">Therefore,<\/p>\r\n\r\n<center>[latex]|\\frac{{a}_{n+1}}{{a}_{n}}|&lt;\\rho +\\epsilon =R\\text{ for all }n\\ge N[\/latex]<\/center>\r\n<p id=\"fs-id1169739344904\">and, thus,<\/p>\r\n\r\n<center>[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]<\/center>\r\n<p id=\"fs-id1169739302830\">Since [latex]R&lt;1[\/latex], the geometric series<\/p>\r\n\r\n<center>[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\\cdots [\/latex]<\/center>\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<center>[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+|{a}_{N+4}|+\\cdots [\/latex]<\/center>\r\n<p id=\"fs-id1169738904500\">converges. Therefore, since<\/p>\r\n\r\n<center>[latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|=\\displaystyle\\sum _{n=1}^{N}|{a}_{n}|+\\displaystyle\\sum _{n=N+1}^{\\infty }|{a}_{n}|[\/latex]<\/center>\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<center>[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|&gt;1[\/latex].<\/center>\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<center>[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |&lt;\\epsilon \\text{for all}n\\ge N[\/latex].<\/center>\r\n<p id=\"fs-id1169739006446\">Therefore,<\/p>\r\n\r\n<center>[latex]R=\\rho -\\epsilon &lt;|\\frac{{a}_{n+1}}{{a}_{n}}|\\text{for all}n\\ge N[\/latex],<\/center>\r\n<p id=\"fs-id1169738895005\">and, thus,<\/p>\r\n\r\n<center>[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]<\/center>\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<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<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-id1167793385034\">[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">When to Use the Ratio Test<\/h3>\r\n<p class=\"whitespace-normal break-words\">The ratio test works particularly well for series containing <strong>factorials<\/strong> or <strong>exponentials<\/strong>, where the ratio [latex]\\frac{a_{n+1}}{a_n}[\/latex] often simplifies nicely. This makes it convenient since you don't need to hunt for a comparison series. However, the test has limitations\u2014sometimes it provides no useful information, especially when [latex]\\rho = 1[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Before applying the ratio test, let's review some essential algebra that will make your calculations much smoother.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Simplifying Exponential Expressions<\/strong><\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For 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]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Examples:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{e^{n+1}}{e^n} = e[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{3^{2n+1}}{3^{2n+3}} = \\frac{1}{3^2} = \\frac{1}{9}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Factorial Notation<\/strong><\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">The expression [latex]n![\/latex], called \"[latex]n[\/latex] factorial,\" is defined as: [latex]n! = n(n-1)(n-2)\\ldots 3 \\cdot 2 \\cdot 1[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key facts:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]0! = 1[\/latex] (by definition)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Simplifying factorial ratios:<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\frac{10!}{8!} = \\frac{10 \\cdot 9 \\cdot 8!}{8!} = 10 \\cdot 9 = 90[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[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<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311399[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the ratio test to check if a series converges absolutely<\/li>\n<li>Use the root test to check if a series converges absolutely<\/li>\n<\/ul>\n<\/section>\n<p data-type=\"title\">You&#8217;ve already worked through several convergence tests, and you&#8217;ve likely noticed that no single test works for every series. The ratio test and root test complete our toolkit by offering particularly useful methods that don&#8217;t require finding a comparable series. The ratio test will prove especially valuable when we explore power series in the next module.<\/p>\n<h2 data-type=\"title\">Ratio Test<\/h2>\n<p class=\"whitespace-normal break-words\">Consider a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex]. We know that having [latex]\\lim{n\\to \\infty}{a}_{n}=0[\/latex] is necessary for convergence, but it&#8217;s not sufficient. The terms must approach zero quickly enough.<\/p>\n<p class=\"whitespace-normal break-words\">Think about these two familiar examples:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] (harmonic series) diverges<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^{2}}[\/latex] (p-series with [latex]p=2[\/latex]) converges<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Both have terms that approach zero, but [latex]\\frac{1}{n^{2}}[\/latex] approaches zero much faster than [latex]\\frac{1}{n}[\/latex]. The ratio test provides a systematic way to measure this &#8220;speed&#8221; of approach to zero.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>ratio test<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] be a series with nonzero terms. Let<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\rho =\\lim_{n\\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right|[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Then:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">If [latex]0\\le \\rho <1[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] <strong>converges absolutely<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">If [latex]\\rho >1[\/latex] or [latex]\\rho =\\infty[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] <strong>diverges<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">If [latex]\\rho =1[\/latex], the test is <strong>inconclusive\u00a0<\/strong>and does not provide any information<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\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 style=\"text-align: center;\">[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |<\\epsilon \\text{ for all }n\\ge N[\/latex].<\/div>\n<p id=\"fs-id1169738923259\">Therefore,<\/p>\n<div style=\"text-align: center;\">[latex]|\\frac{{a}_{n+1}}{{a}_{n}}|<\\rho +\\epsilon =R\\text{ for all }n\\ge N[\/latex]<\/div>\n<p id=\"fs-id1169739344904\">and, thus,<\/p>\n<div style=\"text-align: center;\">[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 id=\"fs-id1169739302830\">Since [latex]R<1[\/latex], the geometric series<\/p>\n<div style=\"text-align: center;\">[latex]R|{a}_{N}|+{R}^{2}|{a}_{N}|+{R}^{3}|{a}_{N}|+\\cdots[\/latex]<\/div>\n<p id=\"fs-id1169738971692\">converges. Given the inequalities above, we can apply the comparison test and conclude that the series<\/p>\n<div style=\"text-align: center;\">[latex]|{a}_{N+1}|+|{a}_{N+2}|+|{a}_{N+3}|+|{a}_{N+4}|+\\cdots[\/latex]<\/div>\n<p id=\"fs-id1169738904500\">converges. Therefore, since<\/p>\n<div style=\"text-align: center;\">[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 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 style=\"text-align: center;\">[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|>1[\/latex].<\/div>\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 style=\"text-align: center;\">[latex]||\\frac{{a}_{n+1}}{{a}_{n}}|-\\rho |<\\epsilon \\text{for all}n\\ge N[\/latex].<\/div>\n<p id=\"fs-id1169739006446\">Therefore,<\/p>\n<div style=\"text-align: center;\">[latex]R=\\rho -\\epsilon <|\\frac{{a}_{n+1}}{{a}_{n}}|\\text{for all}n\\ge N[\/latex],<\/div>\n<p id=\"fs-id1169738895005\">and, thus,<\/p>\n<div style=\"text-align: center;\">[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 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 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 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-id1167793385034\">[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">When to Use the Ratio Test<\/h3>\n<p class=\"whitespace-normal break-words\">The ratio test works particularly well for series containing <strong>factorials<\/strong> or <strong>exponentials<\/strong>, where the ratio [latex]\\frac{a_{n+1}}{a_n}[\/latex] often simplifies nicely. This makes it convenient since you don&#8217;t need to hunt for a comparison series. However, the test has limitations\u2014sometimes it provides no useful information, especially when [latex]\\rho = 1[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Before applying the ratio test, let&#8217;s review some essential algebra that will make your calculations much smoother.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Simplifying Exponential Expressions<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">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 class=\"whitespace-normal break-words\"><strong>Examples:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\frac{e^{n+1}}{e^n} = e[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{3^{2n+1}}{3^{2n+3}} = \\frac{1}{3^2} = \\frac{1}{9}[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Factorial Notation<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">The expression [latex]n![\/latex], called &#8220;[latex]n[\/latex] factorial,&#8221; is defined as: [latex]n! = n(n-1)(n-2)\\ldots 3 \\cdot 2 \\cdot 1[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key facts:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]5! = 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1 = 120[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]0! = 1[\/latex] (by definition)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Simplifying factorial ratios:<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\frac{10!}{8!} = \\frac{10 \\cdot 9 \\cdot 8!}{8!} = 10 \\cdot 9 = 90[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\frac{n!}{(n+2)!} = \\frac{n!}{(n+2)(n+1)n!} = \\frac{1}{(n+2)(n+1)}[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-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<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311399\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311399&theme=lumen&iframe_resize_id=ohm311399&source=tnh\" width=\"100%\" 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