{"id":885,"date":"2025-06-20T17:19:57","date_gmt":"2025-06-20T17:19:57","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=885"},"modified":"2025-08-18T14:30:50","modified_gmt":"2025-08-18T14:30:50","slug":"separation-of-variables-learn-it-3-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/separation-of-variables-learn-it-3-2\/","title":{"raw":"The Divergence and Integral Tests: Learn It 3","rendered":"The Divergence and Integral Tests: Learn It 3"},"content":{"raw":"

The p<\/em>-Series<\/h2>\r\n

The harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] and the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] are both examples of a type of series called a [latex]p[\/latex]-series<\/strong>.<\/p>\r\n\r\n

\r\n

[latex]p[\/latex]-series<\/h3>\r\n

For any real number [latex]p[\/latex], the series<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]<\/div>\r\n \r\n

is called a [latex]p[\/latex]-series<\/strong>.<\/p>\r\n\r\n<\/section>\r\n

We know the [latex]p[\/latex]-series converges when [latex]p = 2[\/latex] and diverges when [latex]p = 1[\/latex]. What happens for other values of [latex]p[\/latex]? While computing exact values of most [latex]p[\/latex]-series is difficult or impossible, we can determine their convergence behavior.<\/p>\r\n\r\n

Testing Different Values of [latex]p[\/latex]<\/h3>\r\nCase 1: [latex]p \\leq 0[\/latex]<\/strong>\r\n
    \r\n \t
  • If [latex]p < 0[\/latex], then [latex]\\frac{1}{n^p} \\to \\infty[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\r\n \t
  • If [latex]p = 0[\/latex], then [latex]\\frac{1}{n^p} = 1 \\to 1[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\r\n<\/ul>\r\n

    By the divergence test, [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] diverges if [latex]p \\leq 0[\/latex]<\/strong>.<\/p>\r\nCase 2: [latex]p > 0[\/latex]<\/strong>\r\n

    When [latex]p > 0[\/latex], the function [latex]f(x) = \\frac{1}{x^p}[\/latex] is positive, continuous, and decreasing. We can apply the integral test by comparing:<\/p>\r\n

    [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] and [latex]\\int_1^{\\infty} \\frac{1}{x^p}dx[\/latex].<\/p>\r\n

    For [latex]p > 0, p \\neq 1[\/latex]:<\/p>\r\n\r\n

    [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}{x}^{1-p}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}\\left[{b}^{1-p}-1\\right][\/latex].<\/div>\r\n

    The key insight is how [latex]b^{1-p}[\/latex] behaves:<\/p>\r\n\r\n

      \r\n \t
    • [latex]b^{1-p} \\to 0[\/latex] if [latex]p > 1[\/latex]<\/li>\r\n \t
    • [latex]b^{1-p} \\to \\infty[\/latex] if [latex]p < 1[\/latex]<\/li>\r\n<\/ul>\r\n

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

      [latex]{\\displaystyle\\int _{1}^{\\infty}} \\dfrac{1}{x^{p}}dx = \\Bigg\\{ \\begin{array}{c} \\frac{1}{p-1}\\text{ if }p>1\\\\ \\infty \\text{ if }p<1\\end{array}[\/latex]<\/div>\r\n

      This means [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] converges if [latex]p > 1[\/latex]<\/strong> and diverges if [latex]0 < p < 1[\/latex]<\/strong>.<\/p>\r\n\r\n

      \r\n

      [latex]p[\/latex]-series convergence test<\/h3>\r\nThe [latex]p[\/latex]-series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] converges if and only if the exponent [latex]p > 1[\/latex].\r\n

      [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p} \\begin{cases} \\text{converges} & \\text{if } p > 1 \\\\ \\text{diverges} & \\text{if } p \\leq 1 \\end{cases}[\/latex]<\/p>\r\n

      <\/p>\r\n\r\n<\/section>\r\n

      \r\n
      \r\n

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

        \r\n \t
      1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex]<\/li>\r\n \t
      2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{2}{3}}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558884\"]\r\n
        \r\n
          \r\n \t
        1. This is a p<\/em>-series with [latex]p=4>1[\/latex], so the series converges.<\/li>\r\n \t
        2. Since [latex]p=\\frac{2}{3}<1[\/latex], the series diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"

          The p<\/em>-Series<\/h2>\n

          The harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] and the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] are both examples of a type of series called a [latex]p[\/latex]-series<\/strong>.<\/p>\n

          \n

          [latex]p[\/latex]-series<\/h3>\n

          For any real number [latex]p[\/latex], the series<\/p>\n

          [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]<\/div>\n

           <\/p>\n

          is called a [latex]p[\/latex]-series<\/strong>.<\/p>\n<\/section>\n

          We know the [latex]p[\/latex]-series converges when [latex]p = 2[\/latex] and diverges when [latex]p = 1[\/latex]. What happens for other values of [latex]p[\/latex]? While computing exact values of most [latex]p[\/latex]-series is difficult or impossible, we can determine their convergence behavior.<\/p>\n

          Testing Different Values of [latex]p[\/latex]<\/h3>\n

          Case 1: [latex]p \\leq 0[\/latex]<\/strong><\/p>\n

            \n
          • If [latex]p < 0[\/latex], then [latex]\\frac{1}{n^p} \\to \\infty[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\n
          • If [latex]p = 0[\/latex], then [latex]\\frac{1}{n^p} = 1 \\to 1[\/latex] as [latex]n \\to \\infty[\/latex].<\/li>\n<\/ul>\n

            By the divergence test, [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] diverges if [latex]p \\leq 0[\/latex]<\/strong>.<\/p>\n

            Case 2: [latex]p > 0[\/latex]<\/strong><\/p>\n

            When [latex]p > 0[\/latex], the function [latex]f(x) = \\frac{1}{x^p}[\/latex] is positive, continuous, and decreasing. We can apply the integral test by comparing:<\/p>\n

            [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] and [latex]\\int_1^{\\infty} \\frac{1}{x^p}dx[\/latex].<\/p>\n

            For [latex]p > 0, p \\neq 1[\/latex]:<\/p>\n

            [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}{x}^{1-p}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}\\left[{b}^{1-p}-1\\right][\/latex].<\/div>\n

            The key insight is how [latex]b^{1-p}[\/latex] behaves:<\/p>\n

              \n
            • [latex]b^{1-p} \\to 0[\/latex] if [latex]p > 1[\/latex]<\/li>\n
            • [latex]b^{1-p} \\to \\infty[\/latex] if [latex]p < 1[\/latex]<\/li>\n<\/ul>\n

              Therefore:<\/p>\n

              [latex]{\\displaystyle\\int _{1}^{\\infty}} \\dfrac{1}{x^{p}}dx = \\Bigg\\{ \\begin{array}{c} \\frac{1}{p-1}\\text{ if }p>1\\\\ \\infty \\text{ if }p<1\\end{array}[\/latex]<\/div>\n

              This means [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] converges if [latex]p > 1[\/latex]<\/strong> and diverges if [latex]0 < p < 1[\/latex]<\/strong>.<\/p>\n

              \n

              [latex]p[\/latex]-series convergence test<\/h3>\n

              The [latex]p[\/latex]-series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p}[\/latex] converges if and only if the exponent [latex]p > 1[\/latex].<\/p>\n

              [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^p} \\begin{cases} \\text{converges} & \\text{if } p > 1 \\\\ \\text{diverges} & \\text{if } p \\leq 1 \\end{cases}[\/latex]<\/p>\n

              \n<\/section>\n

              \n
              \n
              \n

              For each of the following series, determine whether it converges or diverges.<\/p>\n

                \n
              1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex]<\/li>\n
              2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{2}{3}}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n