{"id":1124,"date":"2025-06-30T16:36:35","date_gmt":"2025-06-30T16:36:35","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1124"},"modified":"2025-09-11T16:38:42","modified_gmt":"2025-09-11T16:38:42","slug":"ratio-and-root-tests-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/ratio-and-root-tests-learn-it-2\/","title":{"raw":"Ratio and Root Tests: Learn It 2","rendered":"Ratio and Root Tests: Learn It 2"},"content":{"raw":"

Root Test<\/h2>\r\n

The root test uses a similar approach to the ratio test, but focuses on the [latex]n[\/latex]th root of terms rather than their ratios.<\/p>\r\n

Consider a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex] where [latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|a_{n}|}=\\rho[\/latex] for some real number [latex]\\rho[\/latex]. For sufficiently large [latex]N[\/latex], we have [latex]|a_{N}|\\approx \\rho^{N}[\/latex].<\/p>\r\n

This means we can approximate the tail of our series:<\/p>\r\n

[latex]|a_{N}|+|a_{N+1}|+|a_{N+2}|+\\cdots \\approx \\rho^{N}+\\rho^{N+1}+\\rho^{N+2}+\\cdots[\/latex]<\/p>\r\n

The right side is a geometric series, which gives us the same convergence pattern as the ratio test.<\/p>\r\n\r\n

\r\n

root test<\/h3>\r\n

Consider the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex]. Let<\/p>\r\n

[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|a_{n}|}[\/latex]<\/p>\r\n

Then:<\/p>\r\n\r\n

    \r\n \t
  1. If [latex]0\\leq \\rho <1[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] converges absolutely<\/strong><\/li>\r\n \t
  2. If [latex]\\rho >1[\/latex] or [latex]\\rho =\\infty[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] diverges<\/strong><\/li>\r\n \t
  3. If [latex]\\rho =1[\/latex], the test is inconclusive<\/strong> and does not provide any information<\/li>\r\n<\/ol>\r\n<\/section>
    \r\n

    Why the Root Test Sometimes Fails<\/strong><\/p>\r\n

    For any p-series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^p}[\/latex], we get [latex]\\rho = 1[\/latex] regardless of the value of [latex]p[\/latex]. Yet we know p-series converge when [latex]p > 1[\/latex] and diverge when [latex]p \\leq 1[\/latex]. This shows why [latex]\\rho = 1[\/latex] makes the test inconclusive.<\/p>\r\n\r\n<\/section>\r\n

    When to Use the Root Test<\/h3>\r\n

    The root test works particularly well for series whose terms involve exponentials, especially when [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 evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}b_{n}[\/latex].<\/p>\r\n\r\n

    \r\n
    \r\n

    For each of the following series, use the root test to determine whether the series converges or diverges.<\/p>\r\n\r\n

      \r\n \t
    1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left({n}^{2}+3n\\right)}^{n}}{{\\left(4{n}^{2}+5\\right)}^{n}}[\/latex]<\/li>\r\n \t
    2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{n}}{{\\left(\\text{ln}\\left(n\\right)\\right)}^{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n
      \r\n
        \r\n \t
      1. To apply the root test, we compute\r\n<\/span>\r\n
        [latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{{\\left({n}^{2}+3n\\right)}^{n}}{{\\left(4{n}^{2}+5\\right)}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{2}+3n}{4{n}^{2}+5}=\\frac{1}{4}[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince [latex]\\rho <1[\/latex], the series converges absolutely.<\/li>\r\n \t
      2. We have\r\n<\/span>\r\n
        [latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{\\frac{{n}^{n}}{{\\left(\\text{ln}n\\right)}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{n}{\\text{ln}n}=\\infty \\text{by L'H\u00f4pital's rule}[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince [latex]\\rho =\\infty [\/latex], the series diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        [ohm_question hide_question_numbers=1]311401[\/ohm_question]<\/section>","rendered":"

        Root Test<\/h2>\n

        The root test uses a similar approach to the ratio test, but focuses on the [latex]n[\/latex]th root of terms rather than their ratios.<\/p>\n

        Consider a series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}{n}[\/latex] where [latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|a_{n}|}=\\rho[\/latex] for some real number [latex]\\rho[\/latex]. For sufficiently large [latex]N[\/latex], we have [latex]|a_{N}|\\approx \\rho^{N}[\/latex].<\/p>\n

        This means we can approximate the tail of our series:<\/p>\n

        [latex]|a_{N}|+|a_{N+1}|+|a_{N+2}|+\\cdots \\approx \\rho^{N}+\\rho^{N+1}+\\rho^{N+2}+\\cdots[\/latex]<\/p>\n

        The right side is a geometric series, which gives us the same convergence pattern as the ratio test.<\/p>\n

        \n

        root test<\/h3>\n

        Consider the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex]. Let<\/p>\n

        [latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|a_{n}|}[\/latex]<\/p>\n

        Then:<\/p>\n

          \n
        1. If [latex]0\\leq \\rho <1[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] converges absolutely<\/strong><\/li>\n
        2. If [latex]\\rho >1[\/latex] or [latex]\\rho =\\infty[\/latex], the series [latex]\\displaystyle\\sum_{n=1}^{\\infty}{a}_{n}[\/latex] diverges<\/strong><\/li>\n
        3. If [latex]\\rho =1[\/latex], the test is inconclusive<\/strong> and does not provide any information<\/li>\n<\/ol>\n<\/section>\n
          \n

          Why the Root Test Sometimes Fails<\/strong><\/p>\n

          For any p-series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n^p}[\/latex], we get [latex]\\rho = 1[\/latex] regardless of the value of [latex]p[\/latex]. Yet we know p-series converge when [latex]p > 1[\/latex] and diverge when [latex]p \\leq 1[\/latex]. This shows why [latex]\\rho = 1[\/latex] makes the test inconclusive.<\/p>\n<\/section>\n

          When to Use the Root Test<\/h3>\n

          The root test works particularly well for series whose terms involve exponentials, especially when [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 evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}b_{n}[\/latex].<\/p>\n

          \n
          \n

          For each of the following series, use the root test to determine whether the series converges or diverges.<\/p>\n

            \n
          1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left({n}^{2}+3n\\right)}^{n}}{{\\left(4{n}^{2}+5\\right)}^{n}}[\/latex]<\/li>\n
          2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{n}}{{\\left(\\text{ln}\\left(n\\right)\\right)}^{n}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n