{"id":884,"date":"2025-06-20T17:19:54","date_gmt":"2025-06-20T17:19:54","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=884"},"modified":"2025-09-11T16:15:57","modified_gmt":"2025-09-11T16:15:57","slug":"separation-of-variables-learn-it-2-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/separation-of-variables-learn-it-2-2\/","title":{"raw":"The Divergence and Integral Tests: Learn It 2","rendered":"The Divergence and Integral Tests: Learn It 2"},"content":{"raw":"

Integral Test<\/h2>\r\n

We've already proved that the harmonic series diverges using partial sums. Now we'll use a different approach that's powerful enough to handle many other series: the integral test<\/strong>. The integral test compares an infinite sum to an improper integral. This comparison works because both represent ways of \"adding up\" infinitely many values.<\/p>\r\n\r\n

\r\n

The integral test only applies to series with all positive terms<\/strong>.<\/p>\r\n\r\n<\/section>Let's see how this works with the harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n}[\/latex].\r\n

In Figure 1, we represent each term of the series as a rectangle with area [latex]1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\ldots[\/latex] These rectangles sit above the curve [latex]f(x) = \\frac{1}{x}[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]\"This Figure 1. The sum of the areas of the rectangles is greater than the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is infinite (as calculated by an improper integral), the sum of the areas of the rectangles is also infinite.[\/caption]From the graph, we can see that the total area of the rectangles is greater than the area under the curve:\r\n

[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots +\\frac{1}{k}>{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx[\/latex].<\/div>\r\nLet's evaluate that integral:\r\n
[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}>{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx=\\text{ln}x{|}_{1}^{k+1}=\\text{ln}\\left(k+1\\right)-\\text{ln}\\left(1\\right)=\\text{ln}\\left(k+1\\right)[\/latex].<\/div>\r\n

Since [latex]\\underset{k\\to \\infty}{\\text{lim}} \\ln(k+1) = \\infty[\/latex], the partial sums [latex]{S_k}[\/latex] are unbounded.<\/p>\r\n

Therefore, [latex]{S_k}[\/latex] diverges, which means the harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n}[\/latex] also diverges.<\/p>\r\n\r\n

When the improper integral diverges to infinity, it forces the series to diverge too. The rectangles have more area than the region under the curve, so if the curve's area is infinite, the series must also be infinite.<\/section>Now let's see how the integral test can prove convergence. Consider the series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex].\r\n\r\nIn Figure 2, we sketch rectangles with areas [latex]1, \\frac{1}{2^2}, \\frac{1}{3^2}, \\ldots[\/latex] along with the function [latex]f(x) = \\frac{1}{x^2}[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"263\"]\"This Figure 2. The sum of the areas of the rectangles is less than the sum of the area of the first rectangle and the area between the curve [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is finite, the sum of the areas of the rectangles is also finite.[\/caption] \r\n
<\/figcaption>\r\n
<\/div><\/figure>\r\n

Notice that the rectangles sit below the curve, so the total area of the rectangles is less than the area under the curve:<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}=1+\\frac{1}{{2}^{2}}+\\frac{1}{{3}^{2}}+\\cdots +\\frac{1}{{k}^{2}}<1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\r\nLet's evaluate this integral:\r\n
[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}<1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx=1-{\\frac{1}{x}|}_{1}^{k}=1-\\frac{1}{k}+1=2-\\frac{1}{k}<2[\/latex].<\/div>\r\nThis shows that the partial sums [latex]{S_k}[\/latex] are bounded above by [latex]2[\/latex].\r\n\r\nSince [latex]S_k = S_{k-1} + \\frac{1}{k^2}[\/latex] for [latex]k \\geq 2[\/latex], the sequence [latex]{S_k}[\/latex] is also increasing. A sequence that is both increasing and bounded must converge. Since [latex]{S_k}[\/latex] is increasing and bounded, it converges. Therefore, the series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex] converges.\r\n

General Integral Test<\/h3>\r\nWe can extend this approach to test many different series. The key is finding the right function to compare with your series. For a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] with positive terms, we need a continuous, positive, decreasing function [latex]f[\/latex] where [latex]f(n) = a_n[\/latex] for all positive integers, then:\r\n

For convergence<\/strong> (Figure 3a), the [latex]k[\/latex]th partial sum [latex]S_k[\/latex] satisfies:<\/p>\r\n

[latex]S_k = a_1 + a_2 + a_3 + \\cdots + a_k < a_1 + \\int_1^k f(x)dx < 1 + \\int_1^{\\infty} f(x)dx[\/latex]<\/p>\r\n

If [latex]\\int_1^{\\infty} f(x)dx[\/latex] converges, then [latex]{S_k}[\/latex] is bounded and increasing, so the series converges.<\/p>\r\n

For divergence<\/strong> (Figure 3b), the [latex]k[\/latex]th partial sum satisfies:<\/p>\r\n

[latex]S_k = a_1 + a_2 + a_3 + \\cdots + a_k > \\int_1^{k+1} f(x)dx[\/latex]<\/p>\r\n

If [latex]\\int_1^{\\infty} f(x)dx[\/latex] diverges, then [latex]\\underset{k\\to \\infty}{\\text{lim}} \\int_1^{k+1} f(x)dx = \\infty[\/latex], making [latex]{S_k}[\/latex] unbounded. Therefore the series diverges.<\/p>\r\n\r\n

<\/figure>\r\n
[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"This Figure 3. (a) If we can inscribe rectangles inside a region bounded by a curve [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] -axis, and the area bounded by those curves for [latex]x\\ge 1[\/latex] is finite, then the sum of the areas of the rectangles is also finite. (b) If a set of rectangles circumscribes the region bounded by [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] axis for [latex]x\\ge 1[\/latex] and the region has infinite area, then the sum of the areas of the rectangles is also infinite.[\/caption]<\/figure>\r\n
\r\n

integral test<\/h3>\r\n

Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a series with positive terms [latex]{a}_{n}[\/latex]. Suppose there exists a function [latex]f[\/latex] and a positive integer [latex]N[\/latex] such that the following three conditions are satisfied:<\/p>\r\n\r\n

    \r\n \t
  1. [latex]f[\/latex] is continuous,<\/li>\r\n \t
  2. [latex]f[\/latex] is decreasing, and<\/li>\r\n \t
  3. [latex]f\\left(n\\right)={a}_{n}[\/latex] for all integers [latex]n\\ge N[\/latex].<\/li>\r\n<\/ol>\r\nThen [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}\\text{ and }{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex]both converge or both diverge (see Figure 3).\r\n\r\n<\/section>Although convergence of [latex]\\int_N^{\\infty} f(x)dx[\/latex] implies convergence of the related series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex], it does not imply that the value of the integral and the series are the same.\r\n

    Consider this example: The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{n}=\\frac{1}{e}+{\\left(\\frac{1}{e}\\right)}^{2}+{\\left(\\frac{1}{e}\\right)}^{3}+\\cdots [\/latex] is a geometric series with initial term [latex]a=\\frac{1}{e}[\/latex] and ratio [latex]r=\\frac{1}{e}[\/latex].<\/p>\r\nThe series converges to:\r\n

    [latex]\\frac{\\frac{1}{e}}{1-\\left(\\frac{1}{e}\\right)}=\\frac{\\frac{1}{e}}{\\frac{\\left(e - 1\\right)}{e}}=\\frac{1}{e - 1}[\/latex].<\/div>\r\n

    However, the related integral [latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx[\/latex] gives us:<\/p>\r\n\r\n

    [latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx={\\displaystyle\\int }_{1}^{\\infty }{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}-{e}^{\\text{-}x}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\left[\\text{-}{e}^{\\text{-}b}+{e}^{-1}\\right]=\\frac{1}{e}[\/latex].<\/div>\r\n
    So the series sums to [latex]\\frac{1}{e-1}[\/latex] while the integral equals [latex]\\frac{1}{e}[\/latex]\u2014different values, but both finite.<\/div>\r\n
    \r\n\r\nIn the following examples, we explore how to use the integral test.\u00a0 Before doing so, we should note that since one of the conditions of the test is that the function is decreasing, we can use calculus to verify that this condition is met.\r\n\r\n<\/div>\r\n
    A differentiable function [latex] f(x) [\/latex] is decreasing on an interval [latex] \\left( a,b \\right) [\/latex] if [latex] f'(x) < 0 [\/latex] for all [latex] x\\in\\left( a,b \\right) [\/latex].<\/section>
    \r\n
    \r\n

    For each of the following series, use the integral 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{1}{{n}^{3}}[\/latex]<\/li>\r\n \t
    2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}[\/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\r\n
        [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}\\text{and}{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\r\n\r\n<\/span>\r\nWe have\r\n<\/span>\r\n
        [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{{x}^{3}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\left[-\\frac{1}{2{x}^{2}}{|}_{1}^{b}\\right]=\\underset{b\\to \\infty }{\\text{lim}}\\left[-\\frac{1}{2{b}^{2}}+\\frac{1}{2}\\right]=\\frac{1}{2}[\/latex].<\/div>\r\n\r\n<\/span>\r\nThus the integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex] converges, and therefore so does the series\r\n<\/span>\r\n
        [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex].<\/div>\r\n \r\n
          \r\n \t
        1. Compare\r\n<\/span><\/li>\r\n<\/ol>\r\n \r\n
          [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}\\text{ and }{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince\r\n<\/span>\r\n
          [latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx& =\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{\\sqrt{2x - 1}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\sqrt{2x - 1}{|}_{1}^{b}\\hfill \\\\ & =\\underset{b\\to \\infty }{\\text{lim}}\\left[\\sqrt{2b - 1}-1\\right]=\\infty ,\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nthe integral [latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{\\sqrt{2x - 1}}dx[\/latex] diverges, and therefore\r\n<\/span>\r\n
          [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{2n - 1}}[\/latex]<\/div>\r\n\r\n<\/span>\r\ndiverges.\r\n\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section>
          [ohm_question hide_question_numbers=1]311397[\/ohm_question]<\/section>","rendered":"

          Integral Test<\/h2>\n

          We’ve already proved that the harmonic series diverges using partial sums. Now we’ll use a different approach that’s powerful enough to handle many other series: the integral test<\/strong>. The integral test compares an infinite sum to an improper integral. This comparison works because both represent ways of “adding up” infinitely many values.<\/p>\n

          \n

          The integral test only applies to series with all positive terms<\/strong>.<\/p>\n<\/section>\n

          Let’s see how this works with the harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n}[\/latex].<\/p>\n

          In Figure 1, we represent each term of the series as a rectangle with area [latex]1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\ldots[\/latex] These rectangles sit above the curve [latex]f(x) = \\frac{1}{x}[\/latex].<\/p>\n

          \"This
          Figure 1. The sum of the areas of the rectangles is greater than the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is infinite (as calculated by an improper integral), the sum of the areas of the rectangles is also infinite.<\/figcaption><\/figure>\n

          From the graph, we can see that the total area of the rectangles is greater than the area under the curve:<\/p>\n

          [latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}=1+\\frac{1}{2}+\\frac{1}{3}+\\cdots +\\frac{1}{k}>{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx[\/latex].<\/div>\n

          Let’s evaluate that integral:<\/p>\n

          [latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}>{\\displaystyle\\int }_{1}^{k+1}\\frac{1}{x}dx=\\text{ln}x{|}_{1}^{k+1}=\\text{ln}\\left(k+1\\right)-\\text{ln}\\left(1\\right)=\\text{ln}\\left(k+1\\right)[\/latex].<\/div>\n

          Since [latex]\\underset{k\\to \\infty}{\\text{lim}} \\ln(k+1) = \\infty[\/latex], the partial sums [latex]{S_k}[\/latex] are unbounded.<\/p>\n

          Therefore, [latex]{S_k}[\/latex] diverges, which means the harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n}[\/latex] also diverges.<\/p>\n

          When the improper integral diverges to infinity, it forces the series to diverge too. The rectangles have more area than the region under the curve, so if the curve’s area is infinite, the series must also be infinite.<\/section>\n

          Now let’s see how the integral test can prove convergence. Consider the series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex].<\/p>\n

          In Figure 2, we sketch rectangles with areas [latex]1, \\frac{1}{2^2}, \\frac{1}{3^2}, \\ldots[\/latex] along with the function [latex]f(x) = \\frac{1}{x^2}[\/latex].<\/p>\n

          \"This
          Figure 2. The sum of the areas of the rectangles is less than the sum of the area of the first rectangle and the area between the curve [latex]f\\left(x\\right)=\\frac{1}{{x}^{2}}[\/latex] and the [latex]x[\/latex] -axis for [latex]x\\ge 1[\/latex]. Since the area bounded by the curve is finite, the sum of the areas of the rectangles is also finite.<\/figcaption><\/figure>\n

           <\/p>\n

          <\/figcaption>
          <\/div>\n<\/figure>\n

          Notice that the rectangles sit below the curve, so the total area of the rectangles is less than the area under the curve:<\/p>\n

          [latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}=1+\\frac{1}{{2}^{2}}+\\frac{1}{{3}^{2}}+\\cdots +\\frac{1}{{k}^{2}}<1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\n

          Let’s evaluate this integral:<\/p>\n

          [latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{{n}^{2}}<1+{\\displaystyle\\int }_{1}^{k}\\frac{1}{{x}^{2}}dx=1-{\\frac{1}{x}|}_{1}^{k}=1-\\frac{1}{k}+1=2-\\frac{1}{k}<2[\/latex].<\/div>\n

          This shows that the partial sums [latex]{S_k}[\/latex] are bounded above by [latex]2[\/latex].<\/p>\n

          Since [latex]S_k = S_{k-1} + \\frac{1}{k^2}[\/latex] for [latex]k \\geq 2[\/latex], the sequence [latex]{S_k}[\/latex] is also increasing. A sequence that is both increasing and bounded must converge. Since [latex]{S_k}[\/latex] is increasing and bounded, it converges. Therefore, the series [latex]\\displaystyle\\sum_{n=1}^{\\infty} \\frac{1}{n^2}[\/latex] converges.<\/p>\n

          General Integral Test<\/h3>\n

          We can extend this approach to test many different series. The key is finding the right function to compare with your series. For a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] with positive terms, we need a continuous, positive, decreasing function [latex]f[\/latex] where [latex]f(n) = a_n[\/latex] for all positive integers, then:<\/p>\n

          For convergence<\/strong> (Figure 3a), the [latex]k[\/latex]th partial sum [latex]S_k[\/latex] satisfies:<\/p>\n

          [latex]S_k = a_1 + a_2 + a_3 + \\cdots + a_k < a_1 + \\int_1^k f(x)dx < 1 + \\int_1^{\\infty} f(x)dx[\/latex]<\/p>\n

          If [latex]\\int_1^{\\infty} f(x)dx[\/latex] converges, then [latex]{S_k}[\/latex] is bounded and increasing, so the series converges.<\/p>\n

          For divergence<\/strong> (Figure 3b), the [latex]k[\/latex]th partial sum satisfies:<\/p>\n

          [latex]S_k = a_1 + a_2 + a_3 + \\cdots + a_k > \\int_1^{k+1} f(x)dx[\/latex]<\/p>\n

          If [latex]\\int_1^{\\infty} f(x)dx[\/latex] diverges, then [latex]\\underset{k\\to \\infty}{\\text{lim}} \\int_1^{k+1} f(x)dx = \\infty[\/latex], making [latex]{S_k}[\/latex] unbounded. Therefore the series diverges.<\/p>\n

          <\/figure>\n
          \n
          \"This
          Figure 3. (a) If we can inscribe rectangles inside a region bounded by a curve [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] -axis, and the area bounded by those curves for [latex]x\\ge 1[\/latex] is finite, then the sum of the areas of the rectangles is also finite. (b) If a set of rectangles circumscribes the region bounded by [latex]y=f\\left(x\\right)[\/latex] and the [latex]x[\/latex] axis for [latex]x\\ge 1[\/latex] and the region has infinite area, then the sum of the areas of the rectangles is also infinite.<\/figcaption><\/figure>\n<\/figure>\n
          \n

          integral test<\/h3>\n

          Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a series with positive terms [latex]{a}_{n}[\/latex]. Suppose there exists a function [latex]f[\/latex] and a positive integer [latex]N[\/latex] such that the following three conditions are satisfied:<\/p>\n

            \n
          1. [latex]f[\/latex] is continuous,<\/li>\n
          2. [latex]f[\/latex] is decreasing, and<\/li>\n
          3. [latex]f\\left(n\\right)={a}_{n}[\/latex] for all integers [latex]n\\ge N[\/latex].<\/li>\n<\/ol>\n

            Then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}\\text{ and }{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex]both converge or both diverge (see Figure 3).<\/p>\n<\/section>\n

            Although convergence of [latex]\\int_N^{\\infty} f(x)dx[\/latex] implies convergence of the related series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex], it does not imply that the value of the integral and the series are the same.<\/p>\n

            Consider this example: The series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{n}=\\frac{1}{e}+{\\left(\\frac{1}{e}\\right)}^{2}+{\\left(\\frac{1}{e}\\right)}^{3}+\\cdots[\/latex] is a geometric series with initial term [latex]a=\\frac{1}{e}[\/latex] and ratio [latex]r=\\frac{1}{e}[\/latex].<\/p>\n

            The series converges to:<\/p>\n

            [latex]\\frac{\\frac{1}{e}}{1-\\left(\\frac{1}{e}\\right)}=\\frac{\\frac{1}{e}}{\\frac{\\left(e - 1\\right)}{e}}=\\frac{1}{e - 1}[\/latex].<\/div>\n

            However, the related integral [latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx[\/latex] gives us:<\/p>\n

            [latex]{\\displaystyle\\int }_{1}^{\\infty }{\\left(\\frac{1}{e}\\right)}^{x}dx={\\displaystyle\\int }_{1}^{\\infty }{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}{e}^{\\text{-}x}dx=\\underset{b\\to \\infty }{\\text{lim}}-{e}^{\\text{-}x}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\left[\\text{-}{e}^{\\text{-}b}+{e}^{-1}\\right]=\\frac{1}{e}[\/latex].<\/div>\n
            So the series sums to [latex]\\frac{1}{e-1}[\/latex] while the integral equals [latex]\\frac{1}{e}[\/latex]\u2014different values, but both finite.<\/div>\n
            \n

            In the following examples, we explore how to use the integral test.\u00a0 Before doing so, we should note that since one of the conditions of the test is that the function is decreasing, we can use calculus to verify that this condition is met.<\/p>\n<\/div>\n

            A differentiable function [latex]f(x)[\/latex] is decreasing on an interval [latex]\\left( a,b \\right)[\/latex] if [latex]f'(x) < 0[\/latex] for all [latex]x\\in\\left( a,b \\right)[\/latex].<\/section>\n
            \n
            \n

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

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