{"id":1128,"date":"2025-06-30T16:36:42","date_gmt":"2025-06-30T16:36:42","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1128"},"modified":"2025-09-09T20:13:21","modified_gmt":"2025-09-09T20:13:21","slug":"ratio-and-root-tests-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/ratio-and-root-tests-fresh-take\/","title":{"raw":"Ratio and Root Tests: Fresh Take","rendered":"Ratio and Root Tests: Fresh Take"},"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<h2 data-type=\"title\">Ratio Test<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">The ratio test gives you a systematic way to determine convergence by examining how consecutive terms in your series relate to each other. Instead of hunting for comparison series, you can work directly with the series you're given.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>The Test:<\/strong> For a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex], calculate [latex]\\rho = \\lim{n\\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right|[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>The Three Outcomes:<\/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\"><strong>[latex]\\rho &lt; 1[\/latex]<\/strong>: <strong>Series converges absolutely<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho &gt; 1[\/latex]<\/strong>: <strong>Series diverges<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho = 1[\/latex]<\/strong>: <strong>Test is inconclusive<\/strong> - try a different approach<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>When to use it:<\/strong> The ratio test works particularly well for series containing factorials or exponentials because the ratios often simplify nicely. You'll find that many terms cancel out, making your calculations much cleaner than they initially appear.<\/p>\r\nDon't forget the absolute value bars when calculating [latex]\\rho[\/latex]. Even if your series has alternating signs, you need [latex]\\left|\\frac{a_{n+1}}{a_{n}}\\right|[\/latex] for the test to work properly.\r\n\r\nIf [latex]\\rho = 1[\/latex], the ratio test gives no information. This happens with [latex]p[\/latex]-series and other series where consecutive terms don't have a clear multiplicative relationship. When this occurs, consider using the integral test or comparison tests instead.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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.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 \"5.6.1\" here (opens in new window)<\/a>.<\/section>\r\n<h2 data-type=\"title\">Root Test<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The root test provides another way to test convergence by examining the nth root of your terms instead of ratios between consecutive terms. This test is particularly useful when your series terms involve expressions raised to the nth power.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Test:<\/strong> For a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex], calculate [latex]\\rho = \\lim{n\\to \\infty}\\sqrt[n]{|a_{n}|}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Three Outcomes:<\/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\"><strong>[latex]\\rho &lt; 1[\/latex]<\/strong>: <strong>Series converges absolutely<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho &gt; 1[\/latex]<\/strong>: <strong>Series diverges<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho = 1[\/latex]<\/strong>: <strong>Test is inconclusive<\/strong><\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>When the root test excels:<\/strong> Use this test when your terms have the form [latex]|a_n| = b_n^n[\/latex] for some sequence [latex]b_n[\/latex]. In these cases, [latex]\\sqrt[n]{|a_n|} = b_n[\/latex], so you only need to find [latex]\\lim_{n\\to \\infty} b_n[\/latex]. This makes the root test much simpler than the ratio test for such series.<\/p>\r\n<p class=\"whitespace-normal break-words\">The root test works because [latex]\\sqrt[n]{|a_n|} \\approx \\rho[\/latex] means [latex]|a_n| \\approx \\rho^n[\/latex] for large n. This transforms your series into something that behaves like a geometric series [latex]\\sum \\rho^n[\/latex], which converges when [latex]\\rho &lt; 1[\/latex] and diverges when [latex]\\rho &gt; 1[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Root vs. Ratio:<\/strong> Both tests give the same results when they're both conclusive, but each has its strengths:<\/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\">Use the <strong>root test<\/strong> for terms with nth powers<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use the <strong>ratio test<\/strong> for terms with factorials or simple exponentials<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Just like the ratio test, when [latex]\\rho = 1[\/latex], you get no information. This happens with p-series and other \"boundary cases\" where the test can't distinguish between convergence and divergence.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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\r\n<\/section>\r\n<h2>Choosing a Convergence Test<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">With multiple convergence tests available, success comes from having a systematic strategy rather than randomly trying different approaches. Each test has specific strengths, and knowing when to use which one saves time and effort.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 1: Check for familiar series<\/strong> - Start by looking for series you already know the behavior of: geometric series [latex]\\sum ar^{n-1}[\/latex], p-series [latex]\\sum \\frac{1}{n^p}[\/latex], or the harmonic series [latex]\\sum \\frac{1}{n}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 2: Handle alternating series<\/strong> - If your series has terms that alternate in sign, you have options. Use the alternating series test if you only need to know convergence. If you need absolute convergence, work with [latex]\\sum |a_n|[\/latex] and continue to Step 3.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 3: Look for comparison opportunities<\/strong> - When your series behaves similarly to a p-series or geometric series for large n, try the comparison test or limit comparison test. This works well when you can approximate your terms.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 4: Identify factorials and exponentials<\/strong> - These are your cues for specific tests:<\/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]a_n = b_n^n[\/latex], use the <strong>root test<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">If you see factorials or other exponentials, use the <strong>ratio test<\/strong><\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 5: Use your remaining tools<\/strong> - Apply the divergence test (if [latex]\\lim_{n\\to\\infty} a_n \\neq 0[\/latex], you're done - the series diverges). If that's inconclusive, try the integral test.<\/p>\r\n<p class=\"whitespace-normal break-words\">The comparison and limit comparison tests often require the most mathematical intuition - you need to \"guess\" what your series behaves like. The ratio and root tests are more mechanical but work best on specific types of series.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Remember:<\/strong> When dealing with series that have negative terms, most tests (except the alternating series test and divergence test) require you to work with [latex]\\sum |a_n|[\/latex] to test for absolute convergence.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"https:\/\/www.csusm.edu\/mathlab\/documents\/seriesconvergediverge.pdf\" target=\"_blank\" rel=\"noopener\">this website for more information on testing series for convergence<\/a>, plus general information on sequences and series.<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169739077454\" data-type=\"problem\">\r\n<p id=\"fs-id1169739077456\">For the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{{3}^{n}+n}[\/latex], determine which convergence test is the best to use and explain why.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1169739041818\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739041825\">The series is similar to the geometric series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1169739041763\" data-type=\"solution\">\r\n<p id=\"fs-id1169739041765\">The comparison test because [latex]\\frac{{2}^{n}}{\\left({3}^{n}+n\\right)}&lt;\\frac{{2}^{n}}{{3}^{n}}[\/latex] for all positive integers [latex]n[\/latex]. The limit comparison test could also be used.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/YpHtcijxlzw?controls=0&amp;start=813&amp;end=789&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/MixedConvergenceTests813to789_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"Mixed Convergence Tests\" here (opens in new window)<\/a>.\r\n\r\n<\/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<h2 data-type=\"title\">Ratio Test<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">The ratio test gives you a systematic way to determine convergence by examining how consecutive terms in your series relate to each other. Instead of hunting for comparison series, you can work directly with the series you&#8217;re given.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>The Test:<\/strong> For a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex], calculate [latex]\\rho = \\lim{n\\to \\infty}\\left|\\frac{a_{n+1}}{a_{n}}\\right|[\/latex].<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>The Three Outcomes:<\/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\"><strong>[latex]\\rho < 1[\/latex]<\/strong>: <strong>Series converges absolutely<\/strong><\/li>\n<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho > 1[\/latex]<\/strong>: <strong>Series diverges<\/strong><\/li>\n<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho = 1[\/latex]<\/strong>: <strong>Test is inconclusive<\/strong> &#8211; try a different approach<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>When to use it:<\/strong> The ratio test works particularly well for series containing factorials or exponentials because the ratios often simplify nicely. You&#8217;ll find that many terms cancel out, making your calculations much cleaner than they initially appear.<\/p>\n<p>Don&#8217;t forget the absolute value bars when calculating [latex]\\rho[\/latex]. Even if your series has alternating signs, you need [latex]\\left|\\frac{a_{n+1}}{a_{n}}\\right|[\/latex] for the test to work properly.<\/p>\n<p>If [latex]\\rho = 1[\/latex], the ratio test gives no information. This happens with [latex]p[\/latex]-series and other series where consecutive terms don&#8217;t have a clear multiplicative relationship. When this occurs, consider using the integral test or comparison tests instead.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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.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>.<\/section>\n<h2 data-type=\"title\">Root Test<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The root test provides another way to test convergence by examining the nth root of your terms instead of ratios between consecutive terms. This test is particularly useful when your series terms involve expressions raised to the nth power.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Test:<\/strong> For a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex], calculate [latex]\\rho = \\lim{n\\to \\infty}\\sqrt[n]{|a_{n}|}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Three Outcomes:<\/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\"><strong>[latex]\\rho < 1[\/latex]<\/strong>: <strong>Series converges absolutely<\/strong><\/li>\n<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho > 1[\/latex]<\/strong>: <strong>Series diverges<\/strong><\/li>\n<li class=\"whitespace-normal break-words\"><strong>[latex]\\rho = 1[\/latex]<\/strong>: <strong>Test is inconclusive<\/strong><\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>When the root test excels:<\/strong> Use this test when your terms have the form [latex]|a_n| = b_n^n[\/latex] for some sequence [latex]b_n[\/latex]. In these cases, [latex]\\sqrt[n]{|a_n|} = b_n[\/latex], so you only need to find [latex]\\lim_{n\\to \\infty} b_n[\/latex]. This makes the root test much simpler than the ratio test for such series.<\/p>\n<p class=\"whitespace-normal break-words\">The root test works because [latex]\\sqrt[n]{|a_n|} \\approx \\rho[\/latex] means [latex]|a_n| \\approx \\rho^n[\/latex] for large n. This transforms your series into something that behaves like a geometric series [latex]\\sum \\rho^n[\/latex], which converges when [latex]\\rho < 1[\/latex] and diverges when [latex]\\rho > 1[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Root vs. Ratio:<\/strong> Both tests give the same results when they&#8217;re both conclusive, but each has its strengths:<\/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\">Use the <strong>root test<\/strong> for terms with nth powers<\/li>\n<li class=\"whitespace-normal break-words\">Use the <strong>ratio test<\/strong> for terms with factorials or simple exponentials<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Just like the ratio test, when [latex]\\rho = 1[\/latex], you get no information. This happens with p-series and other &#8220;boundary cases&#8221; where the test can&#8217;t distinguish between convergence and divergence.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Hint<\/button><\/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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<h2>Choosing a Convergence Test<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">With multiple convergence tests available, success comes from having a systematic strategy rather than randomly trying different approaches. Each test has specific strengths, and knowing when to use which one saves time and effort.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 1: Check for familiar series<\/strong> &#8211; Start by looking for series you already know the behavior of: geometric series [latex]\\sum ar^{n-1}[\/latex], p-series [latex]\\sum \\frac{1}{n^p}[\/latex], or the harmonic series [latex]\\sum \\frac{1}{n}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 2: Handle alternating series<\/strong> &#8211; If your series has terms that alternate in sign, you have options. Use the alternating series test if you only need to know convergence. If you need absolute convergence, work with [latex]\\sum |a_n|[\/latex] and continue to Step 3.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 3: Look for comparison opportunities<\/strong> &#8211; When your series behaves similarly to a p-series or geometric series for large n, try the comparison test or limit comparison test. This works well when you can approximate your terms.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 4: Identify factorials and exponentials<\/strong> &#8211; These are your cues for specific tests:<\/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]a_n = b_n^n[\/latex], use the <strong>root test<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">If you see factorials or other exponentials, use the <strong>ratio test<\/strong><\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Step 5: Use your remaining tools<\/strong> &#8211; Apply the divergence test (if [latex]\\lim_{n\\to\\infty} a_n \\neq 0[\/latex], you&#8217;re done &#8211; the series diverges). If that&#8217;s inconclusive, try the integral test.<\/p>\n<p class=\"whitespace-normal break-words\">The comparison and limit comparison tests often require the most mathematical intuition &#8211; you need to &#8220;guess&#8221; what your series behaves like. The ratio and root tests are more mechanical but work best on specific types of series.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Remember:<\/strong> When dealing with series that have negative terms, most tests (except the alternating series test and divergence test) require you to work with [latex]\\sum |a_n|[\/latex] to test for absolute convergence.<\/p>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"https:\/\/www.csusm.edu\/mathlab\/documents\/seriesconvergediverge.pdf\" target=\"_blank\" rel=\"noopener\">this website for more information on testing series for convergence<\/a>, plus general information on sequences and series.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169739077454\" data-type=\"problem\">\n<p id=\"fs-id1169739077456\">For the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{{3}^{n}+n}[\/latex], determine which convergence test is the best to use and explain why.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558890\">Hint<\/button><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739041818\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739041825\">The series is similar to the geometric series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558891\">Show Solution<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739041763\" data-type=\"solution\">\n<p id=\"fs-id1169739041765\">The comparison test because [latex]\\frac{{2}^{n}}{\\left({3}^{n}+n\\right)}<\\frac{{2}^{n}}{{3}^{n}}[\/latex] for all positive integers [latex]n[\/latex]. The limit comparison test could also be used.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/YpHtcijxlzw?controls=0&amp;start=813&amp;end=789&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/MixedConvergenceTests813to789_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;Mixed Convergence Tests&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":37,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1128"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1128\/revisions"}],"predecessor-version":[{"id":2270,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1128\/revisions\/2270"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1128\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1128"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1128"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1128"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1128"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}