{"id":896,"date":"2025-06-20T17:20:37","date_gmt":"2025-06-20T17:20:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=896"},"modified":"2025-08-18T15:05:34","modified_gmt":"2025-08-18T15:05:34","slug":"first-order-linear-equations-and-applications-learn-it-2-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/first-order-linear-equations-and-applications-learn-it-2-2\/","title":{"raw":"Comparison Tests: Learn It 2","rendered":"Comparison Tests: Learn It 2"},"content":{"raw":"

Limit Comparison Test<\/h2>\r\n
\r\n
\r\n

The comparison test works well when we can find a suitable comparison series. However, sometimes finding an appropriate series that satisfies the required inequalities can be challenging.<\/p>\r\n

Consider the series [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}[\/latex]. <\/span>It is natural to compare this series with the convergent series [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

However, we can't use the regular comparison test because:<\/p>\r\n\r\n

[latex]\\frac{1}{{n}^{2}-1}>\\frac{1}{{n}^{2}}[\/latex]<\/div>\r\nfor all integers [latex]n \\geq 2[\/latex]. Since our series has larger terms than the convergent comparison series, the comparison test doesn't give us useful information.\r\n\r\nWe could search for a different comparison series, but there's a better approach: the limit comparison test<\/strong>. This test allows us to compare [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2 - 1}[\/latex] and [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2}[\/latex] even when the inequality goes the \"wrong\" way.\r\n\r\n
\r\n

The limit comparison test is particularly useful when:<\/p>\r\n\r\n

    \r\n \t
  • The terms of your series and a known series have similar behavior for large [latex]n[\/latex]<\/li>\r\n \t
  • The regular comparison test fails because the inequality goes the wrong direction<\/li>\r\n \t
  • Both series have the same \"dominant\" behavior in their numerator and denominator<\/li>\r\n<\/ul>\r\n<\/section>\r\n

    The limit comparison test examines how two series relate to each other asymptotically. For two series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] with positive terms, we evaluate: [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}[\/latex].<\/p>\r\n

    Case 1: Finite, Non-zero Limit<\/strong><\/p>\r\nIf [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{a_n}{b_n} = L \\neq 0[\/latex], then for sufficiently large [latex]n[\/latex], we have [latex]a_n \\approx Lb_n[\/latex]. This means the series behave similarly\u2014either both converge or both diverge.\r\n

    Let's apply this to our example. For [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2-1}[\/latex] and [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2}[\/latex]:<\/p>\r\n\r\n

    [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left({n}^{2}-1\\right)}}{\\frac{1}{{n}^{2}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{2}}{{n}^{2}-1}=1[\/latex].<\/div>\r\nSince [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2}[\/latex] converges, we conclude that [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2-1}[\/latex] converges.\r\n\r\nCase 2: Limit is Zero<\/strong>\r\n\r\nIf [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{a_n}{b_n} = 0[\/latex], then [latex]{\\frac{a_n}{b_n}}[\/latex] is bounded. This means there exists a constant [latex]M[\/latex] such that [latex]a_n \\leq Mb_n[\/latex].\r\n

    Therefore, if [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] converges, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] converges by the comparison test.<\/p>\r\n

    Case 3: Limit is Infinite<\/strong><\/p>\r\n

    If [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{a_n}{b_n} = \\infty[\/latex], then [latex]{{a_n}{b_n}}[\/latex] is unbounded. For every constant [latex]M[\/latex], there exists an integer [latex]N[\/latex] such that [latex]a_n \\geq Mb_n[\/latex] for all [latex]n \\geq N[\/latex].<\/p>\r\n

    Therefore, if [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] diverges, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] also diverges.<\/p>\r\n\r\n

    \r\n

    limit comparison test<\/h3>\r\n

    Let [latex]{a}_{n},{b}_{n}\\ge 0[\/latex] for all [latex]n\\ge 1[\/latex].<\/p>\r\n\r\n

      \r\n \t
    1. If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=L\\ne 0[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] both converge or both diverge<\/strong>.<\/li>\r\n \t
    2. If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges<\/strong>.<\/li>\r\n \t
    3. If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=\\infty [\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges<\/strong>.<\/li>\r\n<\/ol>\r\n<\/section>
      \r\n

      Understanding what each limit case means can help you apply the test more effectively:<\/p>\r\n\r\n

        \r\n \t
      • Finite limit<\/strong>: The series are \"the same size\" asymptotically<\/li>\r\n \t
      • Limit is 0<\/strong>: The [latex]a_n[\/latex] series shrinks much faster than [latex]b_n[\/latex]<\/li>\r\n \t
      • Limit is \u221e<\/strong>: The [latex]a_n[\/latex] series grows much faster than [latex]b_n[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>\r\n

        When the Test Is Inconclusive<\/h3>\r\n

        The limit comparison test doesn't always provide information. It gives no conclusion<\/strong> when:<\/p>\r\n

        [latex]\\frac{a_n}{b_n} \\to 0[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] diverges<\/p>\r\n

        [latex]\\frac{a_n}{b_n} \\to \\infty[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] converges<\/p>\r\n

        Consider the [latex]p[\/latex]-series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{\\sqrt{n}}[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex].<\/p>\r\n

        We know that:<\/p>\r\n

        [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{\\sqrt{n}}[\/latex] diverges (since [latex]p = \\frac{1}{2} \\leq 1[\/latex])<\/p>\r\n

        [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex] converges (since [latex]p = 2 > 1[\/latex])<\/p>\r\n

        Now suppose we try to use the convergent [latex]p[\/latex]-series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^3}[\/latex] as our comparison series:<\/p>\r\n\r\n

          \r\n \t
        • For the divergent series: [latex]\\frac{\\frac{1}{\\sqrt{n}}}{\\frac{1}{n^3}} = \\frac{n^3}{\\sqrt{n}} = n^{5\/2} \\to \\infty[\/latex] as [latex]n \\to \\infty[\/latex]<\/li>\r\n \t
        • For the convergent series: [latex]\\frac{\\frac{1}{n^2}}{\\frac{1}{n^3}} = n \\to \\infty[\/latex] as [latex]n \\to \\infty[\/latex]<\/li>\r\n<\/ul>\r\n

          In both cases, we get [latex]\\frac{a_n}{b_n} \\to \\infty[\/latex] with a convergent comparison series, so the limit comparison test provides no information about either series.<\/p>\r\n\r\n

          When the limit comparison test is inconclusive, try a different comparison series. For [latex]\\frac{1}{\\sqrt{n}}[\/latex], compare with [latex]\\frac{1}{n}[\/latex] instead of [latex]\\frac{1}{n^3}[\/latex] to get a useful result.<\/section>
          \r\n
          \r\n

          For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.<\/p>\r\n\r\n

            \r\n \t
          1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}+1}[\/latex]<\/li>\r\n \t
          2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}+1}{{3}^{n}}[\/latex]<\/li>\r\n \t
          3. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}\\left(n\\right)}{{n}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n
            \r\n
              \r\n \t
            1. Compare this series to [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex]. Calculate\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left(\\sqrt{n}+1\\right)}}{\\frac{1}{\\sqrt{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\begin{array}{c}\\frac{\\sqrt{n}}{\\sqrt{n}+1}\\\\ \\end{array}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\sqrt{n}}}{1+\\frac{1}{\\sqrt{n}}}=1[\/latex].<\/div>\r\nBy the limit comparison test, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}+1}[\/latex] diverges.<\/li>\r\n \t
            2. Compare this series to [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex]. We see that\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({2}^{n}+1\\right)}{{3}^{n}}}{\\frac{{2}^{n}}{{3}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n}+1}{{3}^{n}}\\cdot \\frac{{3}^{n}}{{2}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n}+1}{{2}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\left[1+{\\left(\\frac{1}{2}\\right)}^{n}\\right]=1[\/latex].<\/div>\r\n\r\n<\/span>\r\nTherefore,\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({2}^{n}+1\\right)}{{3}^{n}}}{\\frac{{2}^{n}}{{3}^{n}}}=1[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex] converges, we conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}+1}{{3}^{n}}[\/latex] converges.<\/li>\r\n \t
            3. Since [latex]\\text{ln}n<n[\/latex], compare with [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. We see that\r\n<\/span>\r\n
              [latex]{\\underset{n\\to\\infty}\\lim} \\text{ln}\\frac{n}{n^2}\\frac{1}{n}= {\\underset{n\\to\\infty}\\lim} \\frac{\\text{ln}n}{n^2}\\cdot \\frac{n}{1}= {\\underset{n\\to\\infty}\\lim} \\frac{\\text{ln}n}{n}[\/latex]<\/div>\r\n\r\n<\/span>\r\nIn order to evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{n}[\/latex], evaluate the limit as [latex]x\\to \\infty [\/latex] of the real-valued function [latex]\\text{ln}\\frac{\\left(x\\right)}{x}[\/latex]. These two limits are equal, and making this change allows us to use L\u2019H\u00f4pital\u2019s rule. We obtain\r\n<\/span>\r\n
              [latex]\\underset{x\\to \\infty }{\\text{lim}}\\frac{\\text{ln}x}{x}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{1}{x}=0[\/latex].<\/div>\r\n\r\n<\/span>\r\nTherefore, [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{n}=0[\/latex], and, consequently,\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{n}}=0[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince the limit is [latex]0[\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges, the limit comparison test does not provide any information.\r\n<\/span>\r\nCompare with [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] instead. In this case,\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{{n}^{2}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{{n}^{2}}\\cdot \\frac{{n}^{2}}{1}=\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}n=\\infty [\/latex].<\/div>\r\n\r\n<\/span>\r\nSince the limit is [latex]\\infty [\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, the test still does not provide any information.\r\n<\/span>\r\nSo now we try a series between the two we already tried. Choosing the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{3}{2}}}[\/latex], we see that\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{{n}^{\\frac{3}{2}}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{{n}^{2}}\\cdot \\frac{{n}^{\\frac{3}{2}}}{1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{\\sqrt{n}}[\/latex].<\/div>\r\n\r\n<\/span>\r\nAs above, in order to evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{\\sqrt{n}}[\/latex], evaluate the limit as [latex]x\\to \\infty [\/latex] of the real-valued function [latex]\\text{ln}\\frac{x}{\\sqrt{x}}[\/latex]. Using L\u2019H\u00f4pital\u2019s rule,\r\n<\/span>\r\n
              [latex]\\underset{x\\to \\infty }{\\text{lim}}\\frac{\\text{ln}x}{\\sqrt{x}}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{2\\sqrt{x}}{x}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{2}{\\sqrt{x}}=0[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince the limit is [latex]0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{3}{2}}}[\/latex] converges, we can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}n}{{n}^{2}}[\/latex] converges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
              [ohm_question hide_question_numbers=1]230758[\/ohm_question]<\/section>","rendered":"

              Limit Comparison Test<\/h2>\n
              \n
              \n

              The comparison test works well when we can find a suitable comparison series. However, sometimes finding an appropriate series that satisfies the required inequalities can be challenging.<\/p>\n

              Consider the series [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}[\/latex]. <\/span>It is natural to compare this series with the convergent series [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/p>\n<\/div>\n<\/div>\n

              However, we can’t use the regular comparison test because:<\/p>\n

              [latex]\\frac{1}{{n}^{2}-1}>\\frac{1}{{n}^{2}}[\/latex]<\/div>\n

              for all integers [latex]n \\geq 2[\/latex]. Since our series has larger terms than the convergent comparison series, the comparison test doesn’t give us useful information.<\/p>\n

              We could search for a different comparison series, but there’s a better approach: the limit comparison test<\/strong>. This test allows us to compare [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2 - 1}[\/latex] and [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2}[\/latex] even when the inequality goes the “wrong” way.<\/p>\n

              \n

              The limit comparison test is particularly useful when:<\/p>\n

                \n
              • The terms of your series and a known series have similar behavior for large [latex]n[\/latex]<\/li>\n
              • The regular comparison test fails because the inequality goes the wrong direction<\/li>\n
              • Both series have the same “dominant” behavior in their numerator and denominator<\/li>\n<\/ul>\n<\/section>\n

                The limit comparison test examines how two series relate to each other asymptotically. For two series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] with positive terms, we evaluate: [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}[\/latex].<\/p>\n

                Case 1: Finite, Non-zero Limit<\/strong><\/p>\n

                If [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{a_n}{b_n} = L \\neq 0[\/latex], then for sufficiently large [latex]n[\/latex], we have [latex]a_n \\approx Lb_n[\/latex]. This means the series behave similarly\u2014either both converge or both diverge.<\/p>\n

                Let’s apply this to our example. For [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2-1}[\/latex] and [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2}[\/latex]:<\/p>\n

                [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left({n}^{2}-1\\right)}}{\\frac{1}{{n}^{2}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{2}}{{n}^{2}-1}=1[\/latex].<\/div>\n

                Since [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2}[\/latex] converges, we conclude that [latex]\\displaystyle\\sum_{n=2}^{\\infty} \\frac{1}{n^2-1}[\/latex] converges.<\/p>\n

                Case 2: Limit is Zero<\/strong><\/p>\n

                If [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{a_n}{b_n} = 0[\/latex], then [latex]{\\frac{a_n}{b_n}}[\/latex] is bounded. This means there exists a constant [latex]M[\/latex] such that [latex]a_n \\leq Mb_n[\/latex].<\/p>\n

                Therefore, if [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] converges, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] converges by the comparison test.<\/p>\n

                Case 3: Limit is Infinite<\/strong><\/p>\n

                If [latex]\\underset{n\\to \\infty}{\\text{lim}} \\frac{a_n}{b_n} = \\infty[\/latex], then [latex]{{a_n}{b_n}}[\/latex] is unbounded. For every constant [latex]M[\/latex], there exists an integer [latex]N[\/latex] such that [latex]a_n \\geq Mb_n[\/latex] for all [latex]n \\geq N[\/latex].<\/p>\n

                Therefore, if [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] diverges, then [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] also diverges.<\/p>\n

                \n

                limit comparison test<\/h3>\n

                Let [latex]{a}_{n},{b}_{n}\\ge 0[\/latex] for all [latex]n\\ge 1[\/latex].<\/p>\n

                  \n
                1. If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=L\\ne 0[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] both converge or both diverge<\/strong>.<\/li>\n
                2. If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges<\/strong>.<\/li>\n
                3. If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=\\infty[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges<\/strong>.<\/li>\n<\/ol>\n<\/section>\n
                  \n

                  Understanding what each limit case means can help you apply the test more effectively:<\/p>\n

                    \n
                  • Finite limit<\/strong>: The series are “the same size” asymptotically<\/li>\n
                  • Limit is 0<\/strong>: The [latex]a_n[\/latex] series shrinks much faster than [latex]b_n[\/latex]<\/li>\n
                  • Limit is \u221e<\/strong>: The [latex]a_n[\/latex] series grows much faster than [latex]b_n[\/latex]<\/li>\n<\/ul>\n<\/section>\n

                    When the Test Is Inconclusive<\/h3>\n

                    The limit comparison test doesn’t always provide information. It gives no conclusion<\/strong> when:<\/p>\n

                    [latex]\\frac{a_n}{b_n} \\to 0[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] diverges<\/p>\n

                    [latex]\\frac{a_n}{b_n} \\to \\infty[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} b_n[\/latex] converges<\/p>\n

                    Consider the [latex]p[\/latex]-series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{\\sqrt{n}}[\/latex] and [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex].<\/p>\n

                    We know that:<\/p>\n

                    [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{\\sqrt{n}}[\/latex] diverges (since [latex]p = \\frac{1}{2} \\leq 1[\/latex])<\/p>\n

                    [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex] converges (since [latex]p = 2 > 1[\/latex])<\/p>\n

                    Now suppose we try to use the convergent [latex]p[\/latex]-series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^3}[\/latex] as our comparison series:<\/p>\n

                      \n
                    • For the divergent series: [latex]\\frac{\\frac{1}{\\sqrt{n}}}{\\frac{1}{n^3}} = \\frac{n^3}{\\sqrt{n}} = n^{5\/2} \\to \\infty[\/latex] as [latex]n \\to \\infty[\/latex]<\/li>\n
                    • For the convergent series: [latex]\\frac{\\frac{1}{n^2}}{\\frac{1}{n^3}} = n \\to \\infty[\/latex] as [latex]n \\to \\infty[\/latex]<\/li>\n<\/ul>\n

                      In both cases, we get [latex]\\frac{a_n}{b_n} \\to \\infty[\/latex] with a convergent comparison series, so the limit comparison test provides no information about either series.<\/p>\n

                      When the limit comparison test is inconclusive, try a different comparison series. For [latex]\\frac{1}{\\sqrt{n}}[\/latex], compare with [latex]\\frac{1}{n}[\/latex] instead of [latex]\\frac{1}{n^3}[\/latex] to get a useful result.<\/section>\n
                      \n
                      \n

                      For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.<\/p>\n

                        \n
                      1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}+1}[\/latex]<\/li>\n
                      2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}+1}{{3}^{n}}[\/latex]<\/li>\n
                      3. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}\\left(n\\right)}{{n}^{2}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n