{"id":886,"date":"2025-06-20T17:19:59","date_gmt":"2025-06-20T17:19:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=886"},"modified":"2025-08-18T14:36:56","modified_gmt":"2025-08-18T14:36:56","slug":"separation-of-variables-learn-it-4-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/separation-of-variables-learn-it-4-2\/","title":{"raw":"The Divergence and Integral Tests: Learn It 4","rendered":"The Divergence and Integral Tests: Learn It 4"},"content":{"raw":"

Estimating the Value of a Series<\/h2>\r\n

When we know a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] converges, we often want to estimate its sum. We can approximate this sum using any finite partial sum [latex]\\displaystyle\\sum_{n=1}^{N} a_n[\/latex] where [latex]N[\/latex] is a positive integer.<\/p>\r\n

The key question is: How good is this approximation?<\/strong><\/p>\r\nWhen we approximate an infinite series with its [latex]N[\/latex]th partial sum, we define the remainder<\/strong> as:\r\n

[latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex]<\/div>\r\n

This remainder [latex]R_N[\/latex] represents the \"error\" in our approximation\u2014how much we're missing by stopping at the [latex]N[\/latex]th term instead of continuing infinitely.<\/p>\r\n

For series that satisfy the conditions of the integral test, we can estimate how large this remainder is.<\/p>\r\n\r\n

\r\n

remainder estimate from the integral test<\/h3>\r\n

Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a convergent series with positive terms. Suppose there exists a function [latex]f[\/latex] satisfying the following three conditions:<\/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 1[\/latex].<\/li>\r\n<\/ol>\r\n

    Let [latex]S_N[\/latex] be the [latex]N[\/latex]th partial sum. Then:<\/p>\r\n\r\n

    [latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}<{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n \r\n

    Remainder Estimate<\/strong>: The remainder [latex]R_N = \\displaystyle\\sum_{n=1}^{\\infty} a_n - S_N[\/latex] satisfies:<\/p>\r\n\r\n

    [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n \r\n

    This gives us upper and lower bounds on how much error we make when approximating the series with [latex]S_N[\/latex].<\/p>\r\n\r\n<\/section>\r\n

    Visualizing the Remainder Estimate<\/h3>\r\n

    Figure 4 illustrates how the remainder estimate works.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"This Figure 4. Given a continuous, positive, decreasing function [latex]f[\/latex] and a sequence of positive terms [latex]{a}_{n}[\/latex] such that [latex]{a}_{n}=f\\left(n\\right)[\/latex] for all positive integers [latex]n[\/latex], (a) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots <{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex], or (b) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots >{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]. Therefore, the integral is either an overestimate or an underestimate of the error.[\/caption]\r\n

    We can represent the remainder [latex]R_N = a_{N+1} + a_{N+2} + a_{N+3} + \\cdots[\/latex] as the sum of areas of rectangles.<\/p>\r\n

    From the visual comparison, we see that the total area of these rectangles is:<\/p>\r\n\r\n

      \r\n \t
    • Bounded above<\/strong> by [latex]\\int_N^{\\infty} f(x)dx[\/latex]<\/li>\r\n \t
    • Bounded below<\/strong> by [latex]\\int_{N+1}^{\\infty} f(x)dx[\/latex]<\/li>\r\n<\/ul>\r\n

      This gives us the inequalities:<\/p>\r\n\r\n

      [latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots >{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]<\/div>\r\n

      and<\/p>\r\n\r\n

      [latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots <{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n

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

      [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n \r\n

      Since the infinite series equals the partial sum plus the remainder:<\/p>\r\n\r\n

      [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={S}_{N}+{R}_{N}[\/latex],<\/div>\r\n \r\n

      We can substitute our bounds for [latex]R_N[\/latex] to get bounds for the entire series:<\/p>\r\n\r\n

      [latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}<{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n
      This result gives you a \"sandwich\" estimate for your series. Calculate [latex]S_N[\/latex] (which you can compute exactly), then add the lower and upper integral bounds to get a range where the true sum must lie. The integrals tell you either an overestimate or underestimate of your approximation error.<\/section><\/div>\r\n
      \r\n

      Consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex].<\/p>\r\n\r\n

        \r\n \t
      1. Calculate [latex]{S}_{10}=\\displaystyle\\sum _{n=1}^{10}\\frac{1}{{n}^{3}}[\/latex] and estimate the error.<\/li>\r\n \t
      2. Determine the least value of [latex]N[\/latex] necessary such that [latex]{S}_{N}[\/latex] will estimate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] to within [latex]0.001[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"44558882\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558882\"]\r\n
          \r\n \t
        1. Using a calculating utility, we have\r\n<\/span><\/li>\r\n<\/ol>\r\n
          [latex]{S}_{10}=1+\\frac{1}{{2}^{3}}+\\frac{1}{{3}^{3}}+\\frac{1}{{4}^{3}}+\\cdots +\\frac{1}{{10}^{3}}\\approx 1.19753[\/latex].<\/div>\r\n\r\n<\/span>\r\nBy the remainder estimate, we know\r\n<\/span>\r\n
          [latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\r\n\r\n<\/span>\r\nWe have\r\n<\/span>\r\n
          [latex]{\\displaystyle\\int _{10}^{\\infty }}\\frac{1}{{x}^{3}}dx= {\\underset{b\\to \\infty }\\lim} {\\displaystyle\\int _{10}^{b}} \\frac{1}{{x}^{3}}dx= {\\underset{b\\to\\infty}\\lim} \\left[ -\\frac{1}{2x^2} \\right] _{N}^{b}= {\\underset{b\\to \\infty }\\lim} \\left[-\\frac{1}{2{b}^{2}} + \\frac{1}{2{N}^{2}}\\right]= \\frac{1}{2{N}^{2}}[\/latex].<\/div>\r\n\r\n<\/span>\r\nTherefore, the error is [latex]{R}_{10}<\\frac{1}{2{\\left(10\\right)}^{2}}=0.005[\/latex].\r\n
            \r\n \t
          • Find [latex]N[\/latex] such that [latex]{R}_{N}<0.001[\/latex]. In part a. we showed that [latex]{R}_{N}<\\frac{1}{2{N}^{2}}[\/latex]. Therefore, the remainder [latex]{R}_{N}<0.001[\/latex] as long as [latex]\\frac{1}{2{N}^{2}}<0.001[\/latex]. That is, we need [latex]2{N}^{2}>1000[\/latex]. Solving this inequality for [latex]N[\/latex], we see that we need [latex]N>22.36[\/latex]. To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is [latex]N=23[\/latex].<\/li>\r\n<\/ul>\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n<\/section>","rendered":"

            Estimating the Value of a Series<\/h2>\n

            When we know a series [latex]\\displaystyle\\sum_{n=1}^{\\infty} a_n[\/latex] converges, we often want to estimate its sum. We can approximate this sum using any finite partial sum [latex]\\displaystyle\\sum_{n=1}^{N} a_n[\/latex] where [latex]N[\/latex] is a positive integer.<\/p>\n

            The key question is: How good is this approximation?<\/strong><\/p>\n

            When we approximate an infinite series with its [latex]N[\/latex]th partial sum, we define the remainder<\/strong> as:<\/p>\n

            [latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex]<\/div>\n

            This remainder [latex]R_N[\/latex] represents the “error” in our approximation\u2014how much we’re missing by stopping at the [latex]N[\/latex]th term instead of continuing infinitely.<\/p>\n

            For series that satisfy the conditions of the integral test, we can estimate how large this remainder is.<\/p>\n

            \n

            remainder estimate from the integral test<\/h3>\n

            Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a convergent series with positive terms. Suppose there exists a function [latex]f[\/latex] satisfying the following three conditions:<\/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 1[\/latex].<\/li>\n<\/ol>\n

              Let [latex]S_N[\/latex] be the [latex]N[\/latex]th partial sum. Then:<\/p>\n

              [latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}<{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n

               <\/p>\n

              Remainder Estimate<\/strong>: The remainder [latex]R_N = \\displaystyle\\sum_{n=1}^{\\infty} a_n - S_N[\/latex] satisfies:<\/p>\n

              [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n

               <\/p>\n

              This gives us upper and lower bounds on how much error we make when approximating the series with [latex]S_N[\/latex].<\/p>\n<\/section>\n

              Visualizing the Remainder Estimate<\/h3>\n

              Figure 4 illustrates how the remainder estimate works.<\/p>\n

              \"This
              Figure 4. Given a continuous, positive, decreasing function [latex]f[\/latex] and a sequence of positive terms [latex]{a}_{n}[\/latex] such that [latex]{a}_{n}=f\\left(n\\right)[\/latex] for all positive integers [latex]n[\/latex], (a) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots <{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex], or (b) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots >{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]. Therefore, the integral is either an overestimate or an underestimate of the error.<\/figcaption><\/figure>\n

              We can represent the remainder [latex]R_N = a_{N+1} + a_{N+2} + a_{N+3} + \\cdots[\/latex] as the sum of areas of rectangles.<\/p>\n

              From the visual comparison, we see that the total area of these rectangles is:<\/p>\n

                \n
              • Bounded above<\/strong> by [latex]\\int_N^{\\infty} f(x)dx[\/latex]<\/li>\n
              • Bounded below<\/strong> by [latex]\\int_{N+1}^{\\infty} f(x)dx[\/latex]<\/li>\n<\/ul>\n

                This gives us the inequalities:<\/p>\n

                [latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots >{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]<\/div>\n

                and<\/p>\n

                [latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots <{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n

                Therefore:<\/p>\n

                [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n

                 <\/p>\n

                Since the infinite series equals the partial sum plus the remainder:<\/p>\n

                [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={S}_{N}+{R}_{N}[\/latex],<\/div>\n

                 <\/p>\n

                We can substitute our bounds for [latex]R_N[\/latex] to get bounds for the entire series:<\/p>\n

                [latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}<{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n
                \n
                This result gives you a “sandwich” estimate for your series. Calculate [latex]S_N[\/latex] (which you can compute exactly), then add the lower and upper integral bounds to get a range where the true sum must lie. The integrals tell you either an overestimate or underestimate of your approximation error.<\/section>\n<\/div>\n
                \n

                Consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex].<\/p>\n

                  \n
                1. Calculate [latex]{S}_{10}=\\displaystyle\\sum _{n=1}^{10}\\frac{1}{{n}^{3}}[\/latex] and estimate the error.<\/li>\n
                2. Determine the least value of [latex]N[\/latex] necessary such that [latex]{S}_{N}[\/latex] will estimate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] to within [latex]0.001[\/latex].<\/li>\n<\/ol>\n