{"id":923,"date":"2025-06-20T17:22:56","date_gmt":"2025-06-20T17:22:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=923"},"modified":"2025-09-11T16:48:47","modified_gmt":"2025-09-11T16:48:47","slug":"introduction-to-power-series-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/introduction-to-power-series-learn-it-3\/","title":{"raw":"Introduction to Power Series: Learn It 3","rendered":"Introduction to Power Series: Learn It 3"},"content":{"raw":"

Interval and Radius of Convergence<\/h2>\r\n

The interval of convergence<\/strong> is the set of all [latex]x[\/latex]-values where the power series converges. When we have the third case behavior, this interval has length [latex]2R[\/latex] and is centered at [latex]x = a[\/latex]. The value [latex]R[\/latex] represents the distance from the center to either endpoint, so we call it the radius of convergence<\/strong>.<\/p>\r\n\r\n

Consider the familiar geometric series [latex]\\displaystyle\\sum_{n=0}^{\\infty} x^n[\/latex]. This series converges for all [latex]x[\/latex] in the interval [latex](-1,1)[\/latex] and diverges when [latex]|x| \\geq 1[\/latex]. The interval of convergence is [latex](-1,1)[\/latex] , which has length 2, giving us a radius of convergence [latex]R = 1[\/latex].<\/section>
\r\n
\r\n

interval and radius of convergence<\/h3>\r\n

For the power series [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex]:<\/p>\r\n\r\n

    \r\n \t
  • The interval of convergence<\/strong> is the set of all real numbers [latex]x[\/latex] where the series converges<\/li>\r\n \t
  • The radius of convergence<\/strong> [latex]R[\/latex] is defined as:\r\n
      \r\n \t
    • [latex]R = 0[\/latex] if the series converges only at [latex]x = a[\/latex]<\/li>\r\n \t
    • [latex]R = \\infty[\/latex] if the series converges for all real numbers [latex]x[\/latex]<\/li>\r\n \t
    • [latex]R > 0[\/latex] if the series converges for [latex]|x-a| < R[\/latex] and diverges for [latex]|x-a| > R[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section>\r\n

      Figure 1 illustrates these three cases visually. In case (c), notice that the series behavior at the endpoints [latex]x = a + R[\/latex] and [latex]x = a - R[\/latex] requires separate investigation.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 1. For a series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}[\/latex] graph (a) shows a radius of convergence at [latex]R=0[\/latex], graph (b) shows a radius of convergence at [latex]R=\\infty [\/latex], and graph (c) shows a radius of convergence at R. For graph (c) we note that the series may or may not converge at the endpoints [latex]x=a+R[\/latex] and [latex]x=a-R[\/latex].[\/caption]\r\n

      To determine the interval of convergence for a power series, we typically apply the ratio test.<\/p>\r\n\r\n

      \r\n

      Rules for Solving Absolute Value Inequalities<\/strong><\/p>\r\n

      When working with convergence conditions like [latex]|x-a| < R[\/latex], remember these key rules:<\/p>\r\n\r\n

        \r\n \t
      • [latex]|x - a| \\leq b[\/latex] is equivalent to [latex]-b \\leq x - a \\leq b[\/latex]<\/li>\r\n \t
      • Adding or subtracting the same number to both sides preserves the inequality<\/li>\r\n \t
      • Multiplying or dividing by a positive number preserves the inequality direction<\/li>\r\n \t
      • Multiplying or dividing by a negative number reverses the inequality direction<\/li>\r\n \t
      • If [latex]|x^n| \\leq a[\/latex], then [latex]-\\sqrt[n]{a} \\leq x \\leq \\sqrt[n]{a}[\/latex]<\/li>\r\n<\/ul>\r\n

        Express your final answer using interval notation.<\/p>\r\n\r\n<\/section>\r\n

        In the next example, we show the three different possibilities illustrated in Figure 1.<\/p>\r\n\r\n

        \r\n
        \r\n

        For each of the following series, find the interval and radius of convergence.<\/p>\r\n\r\n

          \r\n \t
        1. [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n \t
        2. [latex]\\displaystyle\\sum _{n=0}^{\\infty }n\\text{!}{x}^{n}[\/latex]<\/li>\r\n \t
        3. [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(x - 2\\right)}^{n}}{\\left(n+1\\right){3}^{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n
          \r\n
            \r\n \t
          1. To check for convergence, apply the ratio test. We have\r\n<\/span>\r\n
            [latex]\\begin{array}{cc}\\hfill \\rho & =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{\\frac{{x}^{n+1}}{\\left(n+1\\right)\\text{!}}}{\\frac{{x}^{n}}{n\\text{!}}}|\\hfill \\\\ & =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{x}^{n+1}}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{{x}^{n}}|\\hfill \\\\ & =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{x}^{n+1}}{\\left(n+1\\right)\\cdot n\\text{!}}\\cdot \\frac{n\\text{!}}{{x}^{n}}|\\hfill \\\\ & =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{x}{n+1}|\\hfill \\\\ & =|x|\\underset{n\\to \\infty }{\\text{lim}}\\frac{1}{n+1}\\hfill \\\\ & =0<1\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

            \r\n<\/span>\r\nfor all values of x<\/em>. Therefore, the series converges for all real numbers x<\/em>. The interval of convergence is [latex]\\left(\\text{-}\\infty ,\\infty \\right)[\/latex] and the radius of convergence is [latex]R=\\infty [\/latex].<\/p>\r\n<\/li>\r\n \t

          2. Apply the ratio test. For [latex]x\\ne 0[\/latex], we see that\r\n<\/span>\r\n
            [latex]\\begin{array}{cc}\\hfill \\rho & \\hfill =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{\\left(n+1\\right)\\text{!}{x}^{n+1}}{n\\text{!}{x}^{n}}|\\\\ & =\\underset{n\\to \\infty }{\\text{lim}}|\\left(n+1\\right)x|\\hfill \\\\ & =|x|\\underset{n\\to \\infty }{\\text{lim}}\\left(n+1\\right)\\hfill \\\\ & =\\infty .\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nTherefore, the series diverges for all [latex]x\\ne 0[\/latex]. Since the series is centered at [latex]x=0[\/latex], it must converge there, so the series converges only for [latex]x\\ne 0[\/latex]. The interval of convergence is the single value [latex]x=0[\/latex] and the radius of convergence is [latex]R=0[\/latex].<\/li>\r\n \t
          3. In order to apply the ratio test, consider\r\n<\/span>\r\n
            [latex]\\begin{array}{cc}\\hfill \\rho & =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{\\frac{{\\left(x - 2\\right)}^{n+1}}{\\left(n+2\\right){3}^{n+1}}}{\\frac{{\\left(x - 2\\right)}^{n}}{\\left(n+1\\right){3}^{n}}}|\\hfill \\\\ & =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{\\left(x - 2\\right)}^{n+1}}{\\left(n+2\\right){3}^{n+1}}\\cdot \\frac{\\left(n+1\\right){3}^{n}}{{\\left(x - 2\\right)}^{n}}|\\hfill \\\\ & =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{\\left(x - 2\\right)\\left(n+1\\right)}{3\\left(n+2\\right)}|\\hfill \\\\ & =\\frac{|x - 2|}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThe ratio [latex]\\rho <1[\/latex] if [latex]|x - 2|<3[\/latex]. Since [latex]|x - 2|<3[\/latex] implies that [latex]-3<x - 2<3[\/latex], the series converges absolutely if [latex]-1<x<5[\/latex]. The ratio [latex]\\rho >1[\/latex] if [latex]|x - 2|>3[\/latex]. Therefore, the series diverges if [latex]x<-1[\/latex] or [latex]x>5[\/latex]. The ratio test is inconclusive if [latex]\\rho =1[\/latex]. The ratio [latex]\\rho =1[\/latex] if and only if [latex]x=-1[\/latex] or [latex]x=5[\/latex]. We need to test these values of x<\/em> separately. For [latex]x=-1[\/latex], the series is given by\r\n<\/span>\r\n
            [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}}{n+1}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\cdots [\/latex].<\/div>\r\n\r\n<\/span>\r\nSince this is the alternating harmonic series, it converges. Thus, the series converges at [latex]x=-1[\/latex]. For [latex]x=5[\/latex], the series is given by\r\n<\/span>\r\n
            [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{n+1}=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots [\/latex].<\/div>\r\n\r\n<\/span>\r\nThis is the harmonic series, which is divergent. Therefore, the power series diverges at [latex]x=5[\/latex]. We conclude that the interval of convergence is [latex]\\left[-1,5\\right)[\/latex] and the radius of convergence is [latex]R=3[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
            [ohm_question hide_question_numbers=1]311404[\/ohm_question]<\/section>","rendered":"

            Interval and Radius of Convergence<\/h2>\n

            The interval of convergence<\/strong> is the set of all [latex]x[\/latex]-values where the power series converges. When we have the third case behavior, this interval has length [latex]2R[\/latex] and is centered at [latex]x = a[\/latex]. The value [latex]R[\/latex] represents the distance from the center to either endpoint, so we call it the radius of convergence<\/strong>.<\/p>\n

            Consider the familiar geometric series [latex]\\displaystyle\\sum_{n=0}^{\\infty} x^n[\/latex]. This series converges for all [latex]x[\/latex] in the interval [latex](-1,1)[\/latex] and diverges when [latex]|x| \\geq 1[\/latex]. The interval of convergence is [latex](-1,1)[\/latex] , which has length 2, giving us a radius of convergence [latex]R = 1[\/latex].<\/section>\n
            \n
            \n

            interval and radius of convergence<\/h3>\n

            For the power series [latex]\\displaystyle\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex]:<\/p>\n

              \n
            • The interval of convergence<\/strong> is the set of all real numbers [latex]x[\/latex] where the series converges<\/li>\n
            • The radius of convergence<\/strong> [latex]R[\/latex] is defined as:\n
                \n
              • [latex]R = 0[\/latex] if the series converges only at [latex]x = a[\/latex]<\/li>\n
              • [latex]R = \\infty[\/latex] if the series converges for all real numbers [latex]x[\/latex]<\/li>\n
              • [latex]R > 0[\/latex] if the series converges for [latex]|x-a| < R[\/latex] and diverges for [latex]|x-a| > R[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<\/section>\n

                Figure 1 illustrates these three cases visually. In case (c), notice that the series behavior at the endpoints [latex]x = a + R[\/latex] and [latex]x = a - R[\/latex] requires separate investigation.<\/p>\n

                \"This
                Figure 1. For a series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}[\/latex] graph (a) shows a radius of convergence at [latex]R=0[\/latex], graph (b) shows a radius of convergence at [latex]R=\\infty [\/latex], and graph (c) shows a radius of convergence at R. For graph (c) we note that the series may or may not converge at the endpoints [latex]x=a+R[\/latex] and [latex]x=a-R[\/latex].<\/figcaption><\/figure>\n

                To determine the interval of convergence for a power series, we typically apply the ratio test.<\/p>\n

                \n

                Rules for Solving Absolute Value Inequalities<\/strong><\/p>\n

                When working with convergence conditions like [latex]|x-a| < R[\/latex], remember these key rules:<\/p>\n

                  \n
                • [latex]|x - a| \\leq b[\/latex] is equivalent to [latex]-b \\leq x - a \\leq b[\/latex]<\/li>\n
                • Adding or subtracting the same number to both sides preserves the inequality<\/li>\n
                • Multiplying or dividing by a positive number preserves the inequality direction<\/li>\n
                • Multiplying or dividing by a negative number reverses the inequality direction<\/li>\n
                • If [latex]|x^n| \\leq a[\/latex], then [latex]-\\sqrt[n]{a} \\leq x \\leq \\sqrt[n]{a}[\/latex]<\/li>\n<\/ul>\n

                  Express your final answer using interval notation.<\/p>\n<\/section>\n

                  In the next example, we show the three different possibilities illustrated in Figure 1.<\/p>\n

                  \n
                  \n

                  For each of the following series, find the interval and radius of convergence.<\/p>\n

                    \n
                  1. [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/li>\n
                  2. [latex]\\displaystyle\\sum _{n=0}^{\\infty }n\\text{!}{x}^{n}[\/latex]<\/li>\n
                  3. [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(x - 2\\right)}^{n}}{\\left(n+1\\right){3}^{n}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n