{"id":1125,"date":"2025-06-30T16:36:36","date_gmt":"2025-06-30T16:36:36","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1125"},"modified":"2025-08-15T16:33:23","modified_gmt":"2025-08-15T16:33:23","slug":"ratio-and-root-tests-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/ratio-and-root-tests-learn-it-3\/","title":{"raw":"Ratio and Root Tests: Learn It 3","rendered":"Ratio and Root Tests: Learn It 3"},"content":{"raw":"<h2>Choosing a Convergence Test<\/h2>\r\nYou now have several convergence tests in your toolkit, but no single test works for every series. The key is developing a systematic strategy to choose the most effective test for each situation.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p data-type=\"title\"><strong>Problem-Solving Strategy: Choosing a Convergence Test for a Series<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When analyzing a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex], work through these steps:<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 1: Check for familiar series<\/strong> Is this a series you recognize?<\/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\">Harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] (diverges)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Alternating harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n}[\/latex] (converges)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Geometric series [latex]\\displaystyle\\sum_{n=1}^{\\infty}ar^{n-1}[\/latex] (check if [latex]|r|&lt;1[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">p-series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^p}[\/latex] (check if [latex]p&gt;1[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 2: Check for alternating series<\/strong> Does the series have terms that alternate in sign?<\/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 you only need to know whether it converges, use the <strong>alternating series test<\/strong><\/li>\r\n \t<li class=\"whitespace-normal break-words\">If you need <strong>absolute convergence<\/strong>, proceed to Step 3 using [latex]\\displaystyle\\sum_{n=1}^{\\infty}|a_{n}|[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 3: Look for comparison opportunities<\/strong> Does the series behave similarly to a p-series or geometric series?<\/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\">Try the <strong>comparison test<\/strong> or <strong>limit comparison test<\/strong><\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 4: Check for factorials or exponentials<\/strong> Do the terms contain factorials ([latex]n![\/latex]) or exponential expressions?<\/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], try the <strong>root test<\/strong> first<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Otherwise, try the <strong>ratio test<\/strong> first<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 5: Use remaining tests<\/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\">Apply the <strong>divergence test<\/strong> (if [latex]\\lim_{n\\to\\infty}a_{n} \\neq 0[\/latex], the series diverges)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If the divergence test is inconclusive, try the <strong>integral test<\/strong><\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169736685655\" data-type=\"problem\">\r\n<p id=\"fs-id1169736685660\">For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169736685666\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}\\left(3n+1\\right)}{n\\text{!}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{e}^{n}}{{n}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{3}^{n}}{{\\left(n+1\\right)}^{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1169736852464\" data-type=\"solution\">\r\n<ol id=\"fs-id1169736852466\" type=\"a\">\r\n \t<li>Step 1. The series is not a [latex]p-\\text{series}[\/latex] or geometric series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. The series is not alternating.<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. For large values of [latex]n[\/latex], we approximate the series by the expression<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736852496\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}\\approx \\frac{{n}^{2}}{{n}^{3}}=\\frac{1}{n}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, it seems reasonable to apply the comparison test or limit comparison test using the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. Using the limit comparison test, we see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736843074\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({n}^{2}+2n\\right)}{\\left({n}^{3}+3{n}^{2}+1\\right)}}{\\frac{1}{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{3}+2{n}^{2}}{{n}^{3}+3{n}^{2}+1}=1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges, this series diverges as well.<\/li>\r\n \t<li>Step 1.The series is not a familiar series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. The series is alternating. Since we are interested in absolute convergence, consider the series<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736747084\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3n}{\\left(n+1\\right)\\text{!}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. The series is not similar to a <em data-effect=\"italics\">p<\/em>-series or geometric series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 4. Since each term contains a factorial, apply the ratio test. We see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736643942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left(3\\left(n+1\\right)\\right)}{\\left(n+1\\right)\\text{!}}}{\\frac{\\left(3n+1\\right)}{n\\text{!}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{3n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\left(3n+1\\right)}=0[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.<\/li>\r\n \t<li>Step 1. The series is not a familiar series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. It is not an alternating series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.<span data-type=\"newline\">\r\n<\/span>\r\nStep 5. To apply the divergence test, we calculate that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736843793\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{e}^{n}}{{n}^{3}}=\\infty [\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, by the divergence test, the series diverges.<\/li>\r\n \t<li>Step 1. This series is not a familiar series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. It is not an alternating series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 4. Since each term is a power of [latex]n[\/latex], we can apply the root test. Since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739077365\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{{\\left(\\frac{3}{n+1}\\right)}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3}{n+1}=0[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nby the root test, we conclude that 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]230818[\/ohm_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]230819[\/ohm_question]<\/section>\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Summary of Convergence Tests<\/h2>\r\n<p class=\"whitespace-normal break-words\">The table below summarizes all convergence tests and their applications. Remember that the comparison test, limit comparison test, and integral test require nonnegative terms. If your series has negative terms, apply these tests to [latex]\\displaystyle\\sum_{n=1}^{\\infty}|a_{n}|[\/latex] to check for absolute convergence.<\/p>\r\n\r\n<table class=\"bg-bg-100 min-w-full border-separate border-spacing-0 text-sm leading-[1.88888] whitespace-normal\">\r\n<thead class=\"border-b-border-100\/50 border-b-[0.5px] text-left\">\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Test<\/strong><\/th>\r\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>When to Use<\/strong><\/th>\r\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Conclusions<\/strong><\/th>\r\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Key Notes<\/strong><\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Divergence Test<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Any series [latex]\\displaystyle\\sum_{n=1}^{\\infty}a_{n}[\/latex]<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\r\n<ul>\r\n \t<li>If [latex]\\lim_{n\\to \\infty}a_{n} \\neq 0[\/latex], series <strong>diverges<\/strong><\/li>\r\n \t<li>If [latex]\\lim_{n\\to \\infty}a_{n} = 0[\/latex], test is <strong>inconclusive<\/strong><\/li>\r\n<\/ul>\r\n<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Cannot prove convergence<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Geometric Series<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}ar^{n-1}[\/latex]<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\r\n<ul>\r\n \t<li>If [latex]|r| &lt; 1[\/latex], converges to [latex]\\frac{a}{1-r}[\/latex]<\/li>\r\n \t<li>If [latex]|r| \\geq 1[\/latex], <strong>diverges<\/strong><\/li>\r\n<\/ul>\r\n<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Look for constant ratio between terms<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>p-Series<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^p}[\/latex]<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\r\n<ul>\r\n \t<li>If [latex]p &gt; 1[\/latex], <strong>converges<\/strong><\/li>\r\n \t<li>If [latex]p \\leq 1[\/latex], <strong>diverges<\/strong><\/li>\r\n<\/ul>\r\n<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]p = 1[\/latex] gives harmonic series<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Comparison Test<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Nonnegative terms, similar to known series<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\r\n<ul>\r\n \t<li>If [latex]a_n \\leq b_n[\/latex] and [latex]\\sum b_n[\/latex] converges, then [latex]\\sum a_n[\/latex] <strong>converges<\/strong><\/li>\r\n \t<li>If [latex]a_n \\geq b_n[\/latex] and [latex]\\sum b_n[\/latex] diverges, then [latex]\\sum a_n[\/latex] <strong>diverges<\/strong><\/li>\r\n<\/ul>\r\n<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Need to find good comparison series<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Limit Comparison Test<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Positive terms, similar to known series<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\r\n<ul>\r\n \t<li>If [latex]L = \\lim_{n\\to\\infty}\\frac{a_n}{b_n}[\/latex] exists and [latex]L \\neq 0[\/latex], both series have same behavior<\/li>\r\n \t<li>Special cases: [latex]L = 0[\/latex] or [latex]L = \\infty[\/latex]<\/li>\r\n<\/ul>\r\n<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Often easier than comparison test<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Integral Test<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Positive, continuous, decreasing function [latex]f[\/latex] where [latex]a_n = f(n)[\/latex]<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\int_N^{\\infty}f(x)dx[\/latex] and [latex]\\sum a_n[\/latex] both converge or both diverge<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Limited to easily integrable functions<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Alternating Series Test<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^{n+1}b_n[\/latex] or [latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^n b_n[\/latex]<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">If [latex]b_{n+1} \\leq b_n[\/latex] and [latex]b_n \\to 0[\/latex], series <strong>converges<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Only for alternating series<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Ratio Test<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Series with factorials or exponentials<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Let [latex]\\rho = \\lim_{n\\to\\infty}\\left|\\frac{a_{n+1}}{a_n}\\right|[\/latex]\r\n<ul>\r\n \t<li>If [latex]\\rho &lt; 1[\/latex], <strong>converges absolutely<\/strong><\/li>\r\n \t<li>If [latex]\\rho &gt; 1[\/latex], <strong>diverges<\/strong><\/li>\r\n \t<li>If [latex]\\rho = 1[\/latex], <strong>inconclusive<\/strong><\/li>\r\n<\/ul>\r\n<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Great for factorials<\/td>\r\n<\/tr>\r\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Root Test<\/strong><\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Series where [latex]|a_n| = b_n^n[\/latex]<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Let [latex]\\rho = \\lim_{n\\to\\infty}\\sqrt[n]{|a_n|}[\/latex]\r\n<ul>\r\n \t<li>If [latex]\\rho &lt; 1[\/latex], <strong>converges absolutely<\/strong><\/li>\r\n \t<li>If [latex]\\rho &gt; 1[\/latex], <strong>diverges<\/strong><\/li>\r\n \t<li>If [latex]\\rho = 1[\/latex], <strong>inconclusive<\/strong><\/li>\r\n<\/ul>\r\n<\/td>\r\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Best for [latex]n[\/latex]th powers<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>","rendered":"<h2>Choosing a Convergence Test<\/h2>\n<p>You now have several convergence tests in your toolkit, but no single test works for every series. The key is developing a systematic strategy to choose the most effective test for each situation.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p data-type=\"title\"><strong>Problem-Solving Strategy: Choosing a Convergence Test for a Series<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When analyzing a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex], work through these steps:<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 1: Check for familiar series<\/strong> Is this a series you recognize?<\/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\">Harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] (diverges)<\/li>\n<li class=\"whitespace-normal break-words\">Alternating harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n}[\/latex] (converges)<\/li>\n<li class=\"whitespace-normal break-words\">Geometric series [latex]\\displaystyle\\sum_{n=1}^{\\infty}ar^{n-1}[\/latex] (check if [latex]|r|<1[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">p-series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^p}[\/latex] (check if [latex]p>1[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Step 2: Check for alternating series<\/strong> Does the series have terms that alternate in sign?<\/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 you only need to know whether it converges, use the <strong>alternating series test<\/strong><\/li>\n<li class=\"whitespace-normal break-words\">If you need <strong>absolute convergence<\/strong>, proceed to Step 3 using [latex]\\displaystyle\\sum_{n=1}^{\\infty}|a_{n}|[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Step 3: Look for comparison opportunities<\/strong> Does the series behave similarly to a p-series or geometric series?<\/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\">Try the <strong>comparison test<\/strong> or <strong>limit comparison test<\/strong><\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Step 4: Check for factorials or exponentials<\/strong> Do the terms contain factorials ([latex]n![\/latex]) or exponential expressions?<\/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], try the <strong>root test<\/strong> first<\/li>\n<li class=\"whitespace-normal break-words\">Otherwise, try the <strong>ratio test<\/strong> first<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Step 5: Use remaining tests<\/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\">Apply the <strong>divergence test<\/strong> (if [latex]\\lim_{n\\to\\infty}a_{n} \\neq 0[\/latex], the series diverges)<\/li>\n<li class=\"whitespace-normal break-words\">If the divergence test is inconclusive, try the <strong>integral test<\/strong><\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169736685655\" data-type=\"problem\">\n<p id=\"fs-id1169736685660\">For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.<\/p>\n<ol id=\"fs-id1169736685666\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}\\left(3n+1\\right)}{n\\text{!}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{e}^{n}}{{n}^{3}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{3}^{n}}{{\\left(n+1\\right)}^{n}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736852464\" data-type=\"solution\">\n<ol id=\"fs-id1169736852466\" type=\"a\">\n<li>Step 1. The series is not a [latex]p-\\text{series}[\/latex] or geometric series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. The series is not alternating.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. For large values of [latex]n[\/latex], we approximate the series by the expression<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736852496\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}\\approx \\frac{{n}^{2}}{{n}^{3}}=\\frac{1}{n}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, it seems reasonable to apply the comparison test or limit comparison test using the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. Using the limit comparison test, we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736843074\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({n}^{2}+2n\\right)}{\\left({n}^{3}+3{n}^{2}+1\\right)}}{\\frac{1}{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{3}+2{n}^{2}}{{n}^{3}+3{n}^{2}+1}=1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges, this series diverges as well.<\/li>\n<li>Step 1.The series is not a familiar series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. The series is alternating. Since we are interested in absolute convergence, consider the series<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736747084\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3n}{\\left(n+1\\right)\\text{!}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. The series is not similar to a <em data-effect=\"italics\">p<\/em>-series or geometric series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 4. Since each term contains a factorial, apply the ratio test. We see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736643942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left(3\\left(n+1\\right)\\right)}{\\left(n+1\\right)\\text{!}}}{\\frac{\\left(3n+1\\right)}{n\\text{!}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{3n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\left(3n+1\\right)}=0[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.<\/li>\n<li>Step 1. The series is not a familiar series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. It is not an alternating series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 5. To apply the divergence test, we calculate that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736843793\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{e}^{n}}{{n}^{3}}=\\infty[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, by the divergence test, the series diverges.<\/li>\n<li>Step 1. This series is not a familiar series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. It is not an alternating series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 4. Since each term is a power of [latex]n[\/latex], we can apply the root test. Since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739077365\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{{\\left(\\frac{3}{n+1}\\right)}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3}{n+1}=0[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nby the root test, we conclude that 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=\"ohm230818\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=230818&theme=lumen&iframe_resize_id=ohm230818&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm230819\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=230819&theme=lumen&iframe_resize_id=ohm230819&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Summary of Convergence Tests<\/h2>\n<p class=\"whitespace-normal break-words\">The table below summarizes all convergence tests and their applications. Remember that the comparison test, limit comparison test, and integral test require nonnegative terms. If your series has negative terms, apply these tests to [latex]\\displaystyle\\sum_{n=1}^{\\infty}|a_{n}|[\/latex] to check for absolute convergence.<\/p>\n<table class=\"bg-bg-100 min-w-full border-separate border-spacing-0 text-sm leading-[1.88888] whitespace-normal\">\n<thead class=\"border-b-border-100\/50 border-b-[0.5px] text-left\">\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Test<\/strong><\/th>\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>When to Use<\/strong><\/th>\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Conclusions<\/strong><\/th>\n<th class=\"text-text-000 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] font-400 px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Key Notes<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Divergence Test<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Any series [latex]\\displaystyle\\sum_{n=1}^{\\infty}a_{n}[\/latex]<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\n<ul>\n<li>If [latex]\\lim_{n\\to \\infty}a_{n} \\neq 0[\/latex], series <strong>diverges<\/strong><\/li>\n<li>If [latex]\\lim_{n\\to \\infty}a_{n} = 0[\/latex], test is <strong>inconclusive<\/strong><\/li>\n<\/ul>\n<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Cannot prove convergence<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Geometric Series<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}ar^{n-1}[\/latex]<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\n<ul>\n<li>If [latex]|r| < 1[\/latex], converges to [latex]\\frac{a}{1-r}[\/latex]<\/li>\n<li>If [latex]|r| \\geq 1[\/latex], <strong>diverges<\/strong><\/li>\n<\/ul>\n<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Look for constant ratio between terms<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>p-Series<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^p}[\/latex]<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\n<ul>\n<li>If [latex]p > 1[\/latex], <strong>converges<\/strong><\/li>\n<li>If [latex]p \\leq 1[\/latex], <strong>diverges<\/strong><\/li>\n<\/ul>\n<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]p = 1[\/latex] gives harmonic series<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Comparison Test<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Nonnegative terms, similar to known series<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\n<ul>\n<li>If [latex]a_n \\leq b_n[\/latex] and [latex]\\sum b_n[\/latex] converges, then [latex]\\sum a_n[\/latex] <strong>converges<\/strong><\/li>\n<li>If [latex]a_n \\geq b_n[\/latex] and [latex]\\sum b_n[\/latex] diverges, then [latex]\\sum a_n[\/latex] <strong>diverges<\/strong><\/li>\n<\/ul>\n<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Need to find good comparison series<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Limit Comparison Test<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Positive terms, similar to known series<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">\n<ul>\n<li>If [latex]L = \\lim_{n\\to\\infty}\\frac{a_n}{b_n}[\/latex] exists and [latex]L \\neq 0[\/latex], both series have same behavior<\/li>\n<li>Special cases: [latex]L = 0[\/latex] or [latex]L = \\infty[\/latex]<\/li>\n<\/ul>\n<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Often easier than comparison test<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Integral Test<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Positive, continuous, decreasing function [latex]f[\/latex] where [latex]a_n = f(n)[\/latex]<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\int_N^{\\infty}f(x)dx[\/latex] and [latex]\\sum a_n[\/latex] both converge or both diverge<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Limited to easily integrable functions<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Alternating Series Test<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">[latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^{n+1}b_n[\/latex] or [latex]\\displaystyle\\sum_{n=1}^{\\infty}(-1)^n b_n[\/latex]<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">If [latex]b_{n+1} \\leq b_n[\/latex] and [latex]b_n \\to 0[\/latex], series <strong>converges<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Only for alternating series<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Ratio Test<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Series with factorials or exponentials<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Let [latex]\\rho = \\lim_{n\\to\\infty}\\left|\\frac{a_{n+1}}{a_n}\\right|[\/latex]<\/p>\n<ul>\n<li>If [latex]\\rho < 1[\/latex], <strong>converges absolutely<\/strong><\/li>\n<li>If [latex]\\rho > 1[\/latex], <strong>diverges<\/strong><\/li>\n<li>If [latex]\\rho = 1[\/latex], <strong>inconclusive<\/strong><\/li>\n<\/ul>\n<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Great for factorials<\/td>\n<\/tr>\n<tr class=\"[tbody&gt;&amp;]:odd:bg-bg-500\/10\">\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\"><strong>Root Test<\/strong><\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Series where [latex]|a_n| = b_n^n[\/latex]<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Let [latex]\\rho = \\lim_{n\\to\\infty}\\sqrt[n]{|a_n|}[\/latex]<\/p>\n<ul>\n<li>If [latex]\\rho < 1[\/latex], <strong>converges absolutely<\/strong><\/li>\n<li>If [latex]\\rho > 1[\/latex], <strong>diverges<\/strong><\/li>\n<li>If [latex]\\rho = 1[\/latex], <strong>inconclusive<\/strong><\/li>\n<\/ul>\n<\/td>\n<td class=\"border-t-border-100\/50 [&amp;:not(:first-child)]:-x-[hsla(var(--border-100) \/ 0.5)] border-t-[0.5px] px-2 [&amp;:not(:first-child)]:border-l-[0.5px]\">Best for [latex]n[\/latex]th powers<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n","protected":false},"author":15,"menu_order":35,"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 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