{"id":2761,"date":"2024-08-15T23:55:33","date_gmt":"2024-08-15T23:55:33","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=2761"},"modified":"2024-11-21T22:45:18","modified_gmt":"2024-11-21T22:45:18","slug":"series-and-their-notations-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/series-and-their-notations-learn-it-5\/","title":{"raw":"Series and Their Notations: Learn It 5","rendered":"Series and Their Notations: Learn It 5"},"content":{"raw":"

Finding Sums of Infinite Series<\/h2>\r\nWhen the sum of an infinite geometric series exists, we can calculate the sum.\r\n

The formula for the sum of an infinite series is related to the formula for the sum of the first [latex]n[\/latex]\u00a0terms of a geometric series.<\/p>\r\n

[latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex]<\/p>\r\nAs\u00a0[latex]n[\/latex] gets large, [latex]r^n[\/latex] gets very small. We say that as\u00a0[latex]n[\/latex] increases without bound,\u00a0[latex]r^n[\/latex] approaches [latex]0[\/latex]. As\u00a0[latex]r^n[\/latex] approaches [latex]0[\/latex],\u00a0[latex]1-r^n[\/latex] approaches [latex]1[\/latex]. When this happens the numerator approaches [latex]a_1[\/latex]. This gives us the formula for the sum of an infinite geometric series.\r\n\r\n

\r\n

formula for the sum of an infinite geometric series<\/h3>\r\nThe formula for the sum of an infinite geometric series with [latex]-1 < r < 1[\/latex] is\r\n

[latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex]<\/p>\r\n\r\n<\/section>

We will examine an infinite series with [latex]r=\\frac{1}{2}[\/latex]. What happens to [latex]r^n[\/latex] as\u00a0[latex]n[\/latex] increases?\r\n

[latex]\\begin{align} &{\\left(\\frac{1}{2}\\right)}^{2} = \\frac{1}{4} \\\\&{\\left(\\frac{1}{2}\\right)}^{3} = \\frac{1}{8} \\\\&{\\left(\\frac{1}{2}\\right)}^{4} = \\frac{1}{16} \\end{align}[\/latex]<\/p>\r\nThe value of [latex]r^n[\/latex] decreases rapidly. What happens for greater values of\u00a0[latex]n[\/latex]?\r\n

[latex]\\begin{align} &{\\left(\\frac{1}{2}\\right)}^{10} = \\frac{1}{1\\text{,}024} \\\\&{\\left(\\frac{1}{2}\\right)}^{20} = \\frac{1}{1\\text{,}048\\text{,}576} \\\\&{\\left(\\frac{1}{2}\\right)}^{30} = \\frac{1}{1\\text{,}073\\text{,}741\\text{,}824} \\end{align}[\/latex]<\/p>\r\n\r\n<\/section>

How To: Given an infinite geometric series, find its sum.<\/strong>\r\n
    \r\n \t
  1. Identify [latex]a_1[\/latex] and\u00a0[latex]r[\/latex].<\/li>\r\n \t
  2. Confirm that [latex]-1 < r < 1[\/latex].<\/li>\r\n \t
  3. Substitute values for\u00a0[latex]a_1[\/latex] and\u00a0r<\/em> into the formula,\u00a0[latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex].<\/li>\r\n \t
  4. Simplify to find\u00a0[latex]S[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
    Find the sum, if it exists, for the following:\r\n
      \r\n \t
    1. [latex]10+9+8+7+\\dots[\/latex]<\/li>\r\n \t
    2. [latex]248.6+99.44+39.776+\\dots[\/latex]<\/span><\/li>\r\n \t
    3. [latex]\\sum\\limits _{k=1}^{\\infty}4\\text{,}374\\cdot\\left(-\\dfrac{1}{3}\\right)^{k-1}[\/latex]<\/li>\r\n \t
    4. [latex]\\sum\\limits _{k=1}^{\\infty}\\dfrac{1}{9}\\cdot\\left(\\dfrac{4}{3}\\right)^{k}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"18513\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"18513\"]\r\n
        \r\n \t
      1. There is not a constant ratio; the series is not geometric.<\/li>\r\n \t
      2. There is a constant ratio; the series is geometric. [latex]a_1=248.6[\/latex] and [latex]r=\\dfrac{99.44}{248.6}=0.4[\/latex], so the sum exists. Substitute\u00a0[latex]a_1=248.6[\/latex] and [latex]r=0.4[\/latex] into the formula and simplify to find the sum.\r\n
        [latex]\\begin{align} \\\\ &S=\\frac{a_1}{1-r} \\\\[1.5mm] &S=\\frac{248.6}{1-0.4}=\\frac{1243}{3} \\\\ \\text{ }\\end{align}[\/latex]<\/center><\/li>\r\n \t
      3. The formula is exponential, so the series is geometric with [latex]r=-\\frac{1}{3}[\/latex]. Find [latex]a_1[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.\r\n
        [latex]\\begin{align} \\\\ a_1=4\\text{,}374\\cdot\\left(-\\frac{1}{3}\\right)^{1-1}=4\\text{,}374 \\\\ \\text{ }\\end{align}[\/latex]<\/center>Substitute [latex]4\\text{,}374[\/latex] and [latex]r=-\\frac{1}{3}[\/latex] into the formula, and simplify to find the sum.\r\n
        [latex]\\begin{align}\\\\&S=\\frac{a_1}{1-r} \\\\[1.5mm] &S=\\frac{4\\text{,}374}{1-\\left(-\\frac{1}{3}\\right)}=3\\text{,}280.5 \\\\ \\text{ }\\end{align}[\/latex]<\/center><\/li>\r\n \t
      4. The formula is exponential, so the series is geometric, but [latex]r>1[\/latex].\u00a0The sum does not exist.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        Find the equivalent fraction for the repeating decimal [latex]0.\\overline{3}[\/latex].[reveal-answer q=\"624358\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624358\"]We notice the repeating decimal [latex]0.\\overline{3}=0.333\\dots[\/latex].\u00a0so we can rewrite the repeating decimal as a sum of terms.\r\n

        [latex]0.\\overline{3}=0.3+0.03+0.003+\\dots[\/latex]<\/p>\r\nLooking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to [latex]0.1[\/latex] in the second term, and the second term multiplied to [latex]0.1[\/latex] in the third term.\r\n

        [latex]\\begin{align}0.\\overline{3}&=0.3+0.3\\cdot(0.1)+0.3\\cdot(0.01)+0.3\\cdot(0.001)+\\dots \\\\ &=0.3+0.3\\cdot(0.1)+0.3\\cdot(0.1)^2+0.3\\cdot(0.1)^3+\\dots\\end{align}[\/latex]<\/p>\r\nNotice the pattern; we multiply each consecutive term by a common ratio of [latex]0.1 [\/latex] starting with the first term of [latex]0.3[\/latex]. So, substituting into our formula for an infinite geometric sum, we have\r\n

        [latex]S=\\dfrac{a_1}{1-r} =\\dfrac{0.3}{1-0.1} =\\dfrac{0.3}{0.9} =\\dfrac{1}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

        [ohm2_question hide_question_numbers=1]24954[\/ohm2_question]<\/section>
        [ohm2_question hide_question_numbers=1]24955[\/ohm2_question]<\/section>","rendered":"

        Finding Sums of Infinite Series<\/h2>\n

        When the sum of an infinite geometric series exists, we can calculate the sum.<\/p>\n

        The formula for the sum of an infinite series is related to the formula for the sum of the first [latex]n[\/latex]\u00a0terms of a geometric series.<\/p>\n

        [latex]{S}_{n}=\\dfrac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex]<\/p>\n

        As\u00a0[latex]n[\/latex] gets large, [latex]r^n[\/latex] gets very small. We say that as\u00a0[latex]n[\/latex] increases without bound,\u00a0[latex]r^n[\/latex] approaches [latex]0[\/latex]. As\u00a0[latex]r^n[\/latex] approaches [latex]0[\/latex],\u00a0[latex]1-r^n[\/latex] approaches [latex]1[\/latex]. When this happens the numerator approaches [latex]a_1[\/latex]. This gives us the formula for the sum of an infinite geometric series.<\/p>\n

        \n

        formula for the sum of an infinite geometric series<\/h3>\n

        The formula for the sum of an infinite geometric series with [latex]-1 < r < 1[\/latex] is\n\n\n

        [latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex]<\/p>\n<\/section>\n

        We will examine an infinite series with [latex]r=\\frac{1}{2}[\/latex]. What happens to [latex]r^n[\/latex] as\u00a0[latex]n[\/latex] increases?<\/p>\n

        [latex]\\begin{align} &{\\left(\\frac{1}{2}\\right)}^{2} = \\frac{1}{4} \\\\&{\\left(\\frac{1}{2}\\right)}^{3} = \\frac{1}{8} \\\\&{\\left(\\frac{1}{2}\\right)}^{4} = \\frac{1}{16} \\end{align}[\/latex]<\/p>\n

        The value of [latex]r^n[\/latex] decreases rapidly. What happens for greater values of\u00a0[latex]n[\/latex]?<\/p>\n

        [latex]\\begin{align} &{\\left(\\frac{1}{2}\\right)}^{10} = \\frac{1}{1\\text{,}024} \\\\&{\\left(\\frac{1}{2}\\right)}^{20} = \\frac{1}{1\\text{,}048\\text{,}576} \\\\&{\\left(\\frac{1}{2}\\right)}^{30} = \\frac{1}{1\\text{,}073\\text{,}741\\text{,}824} \\end{align}[\/latex]<\/p>\n<\/section>\n

        How To: Given an infinite geometric series, find its sum.<\/strong><\/p>\n
          \n
        1. Identify [latex]a_1[\/latex] and\u00a0[latex]r[\/latex].<\/li>\n
        2. Confirm that [latex]-1 < r < 1[\/latex].<\/li>\n
        3. Substitute values for\u00a0[latex]a_1[\/latex] and\u00a0r<\/em> into the formula,\u00a0[latex]S=\\dfrac{{a}_{1}}{1-r}[\/latex].<\/li>\n
        4. Simplify to find\u00a0[latex]S[\/latex].<\/li>\n<\/ol>\n<\/section>\n
          Find the sum, if it exists, for the following:<\/p>\n
            \n
          1. [latex]10+9+8+7+\\dots[\/latex]<\/li>\n
          2. [latex]248.6+99.44+39.776+\\dots[\/latex]<\/span><\/li>\n
          3. [latex]\\sum\\limits _{k=1}^{\\infty}4\\text{,}374\\cdot\\left(-\\dfrac{1}{3}\\right)^{k-1}[\/latex]<\/li>\n
          4. [latex]\\sum\\limits _{k=1}^{\\infty}\\dfrac{1}{9}\\cdot\\left(\\dfrac{4}{3}\\right)^{k}[\/latex]<\/li>\n<\/ol>\n