{"id":873,"date":"2025-06-20T17:19:17","date_gmt":"2025-06-20T17:19:17","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=873"},"modified":"2025-08-15T18:18:42","modified_gmt":"2025-08-15T18:18:42","slug":"direction-fields-and-eulers-method-learn-it-3-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-learn-it-3-2\/","title":{"raw":"Introduction to Series: Learn It 3","rendered":"Introduction to Series: Learn It 3"},"content":{"raw":"
\r\n

Geometric Series<\/h2>\r\n

A geometric series<\/strong> is any series that we can write in the form:<\/p>\r\n\r\n

[latex]a+ar+a{r}^{2}+a{r}^{3}+\\cdots =\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex].<\/div>\r\n

Because the ratio of each term in this series to the previous term is [latex]r[\/latex], the number [latex]r[\/latex] is called the common ratio<\/strong>. We refer to [latex]a[\/latex] as the initial term<\/strong> because it is the first term in the series.<\/p>\r\n\r\n

\r\n

components of a geometric series<\/h3>\r\nThe number [latex]r[\/latex] is called the common ratio<\/strong> because it's the ratio between consecutive terms. The value [latex]a[\/latex] is the initial term<\/strong> since it's the first term in the series.\r\n\r\n<\/section>
\r\n

Let's look at a concrete example. The series:<\/p>\r\n

[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\cdots [\/latex]<\/p>\r\n

is a geometric series with initial term [latex]a=1[\/latex] and common ratio [latex]r=\\frac{1}{2}[\/latex].<\/p>\r\n\r\n<\/section>\r\n

When Does a Geometric Series Converge?<\/h3>\r\n

The key question is: when does a geometric series actually converge to a finite value? To answer this, we need to examine the partial sums. Consider the geometric series [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex]\u00a0when [latex]a>0[\/latex].<\/p>\r\nThe sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is:\r\n

[latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}a{r}^{n - 1}=a+ar+a{r}^{2}+\\cdots +a{r}^{k - 1}[\/latex].<\/div>\r\n \r\n

Case 1: When [latex]r=1[\/latex] <\/strong><\/p>\r\nWhen [latex]r=1[\/latex], our partial sum becomes:\r\n

[latex]{S}_{k}=a+a\\left(1\\right)+a{\\left(1\\right)}^{2}+\\cdots +a{\\left(1\\right)}^{k - 1}=ak[\/latex].<\/div>\r\n \r\n

Since [latex]a>0[\/latex], we have [latex]ak\\to \\infty [\/latex] as [latex]k\\to \\infty [\/latex]. The sequence of partial sums is unbounded, so the series diverges when [latex]r=1[\/latex].<\/p>\r\n

Case 2: When [latex]r\\ne 1[\/latex]<\/strong><\/p>\r\nFor [latex]r\\ne 1[\/latex], to find the limit of [latex]\\left\\{{S}_{k}\\right\\}[\/latex], multiply the geometric series general equation by [latex]1-r[\/latex].\r\n

[latex]\\begin{array}{cc}\\hfill \\left(1-r\\right){S}_{k}& =a\\left(1-r\\right)\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)\\hfill \\\\ & =a\\left[\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)-\\left(r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k}\\right)\\right]\\hfill \\\\ & =a\\left(1-{r}^{k}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Notice how most terms cancel out\u2014this is the key insight! Therefore:<\/p>\r\n\r\n

[latex]{S}_{k}=\\frac{a\\left(1-{r}^{k}\\right)}{1-r}\\text{for }r\\ne 1[\/latex].<\/div>\r\n

From our discussion in the previous section, we know that the geometric sequence [latex]{r}^{k}\\to 0[\/latex] if [latex]|r|<1[\/latex] and that [latex]{r}^{k}[\/latex] diverges if [latex]|r|>1[\/latex] or [latex]r= \\pm{1}[\/latex].<\/p>\r\n

Using this knowledge:<\/p>\r\n\r\n

    \r\n \t
  • For [latex]|r|<1[\/latex]: [latex]{S}_{k}\\to \\frac{a}{1-r}[\/latex], so [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}[\/latex]<\/li>\r\n \t
  • For [latex]|r|\\ge 1[\/latex]: [latex]{S}_{k}[\/latex] diverges, so [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex] diverges.<\/li>\r\n<\/ul>\r\n \r\n\r\n<\/section>
    \r\n

    convergence of geometric series<\/h3>\r\n

    A geometric series is a series of the form<\/p>\r\n\r\n

    [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=a+ar+a{r}^{2}+a{r}^{3}+\\cdots [\/latex].<\/div>\r\n \r\n

    If [latex]|r|<1[\/latex], the series converges, and<\/p>\r\n\r\n

    [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}\\text{ for }|r|<1[\/latex].<\/div>\r\n \r\n

    If [latex]|r|\\ge 1[\/latex], the series diverges.<\/p>\r\n\r\n<\/section>\r\n

    Recognizing Geometric Series in Different Forms<\/h3>\r\n

    Geometric series don't always appear in the standard form. You might encounter series where the index starts at a different value or the exponent has a different linear expression. The key is recognizing when you can rewrite the series in geometric form.<\/p>\r\n\r\n

    Identifying Geometric Series<\/strong>: Look for a constant ratio between consecutive terms. If you find one, you likely have a geometric series that can be rewritten in standard form.<\/section>Let's work through an example.\r\n\r\n
    \r\n

    Consider the series:<\/p>\r\n

    [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}[\/latex]<\/p>\r\n

    To see if this is geometric, write out the first several terms:<\/p>\r\n

    [latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}& ={\\left(\\frac{2}{3}\\right)}^{2}+{\\left(\\frac{2}{3}\\right)}^{3}+{\\left(\\frac{2}{3}\\right)}^{4}+\\cdots \\hfill \\\\ & =\\frac{4}{9}+\\frac{4}{9}\\cdot \\left(\\frac{2}{3}\\right)+\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/p>\r\nWe see that the initial term is [latex]a=\\frac{4}{9}[\/latex] and the ratio is [latex]r=\\frac{2}{3}[\/latex].\r\n

    This means we can rewrite the series in standard form:<\/p>\r\n

    [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}[\/latex].<\/p>\r\n

    Since [latex]|r|=\\frac{2}{3}<1[\/latex], the series converges. Using our convergence formula:<\/p>\r\n

    [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}=\\frac{\\frac{4}{9}}{1 - \\frac{2}{3}}=\\frac{\\frac{4}{9}}{\\frac{1}{3}}=\\frac{4}{9}\\cdot\\frac{3}{1}=\\frac{4}{3}[\/latex]<\/p>\r\n\r\n<\/section>

    \r\n
    \r\n

    Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.<\/p>\r\n\r\n

      \r\n \t
    1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-3\\right)}^{n+1}}{{4}^{n - 1}}[\/latex]<\/li>\r\n \t
    2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{2n}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n
      \r\n
        \r\n \t
      1. Writing out the first several terms in the series, we have\r\n<\/span>\r\n
        [latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-3\\right)}^{n+1}}{{4}^{n - 1}}& =\\frac{{\\left(-3\\right)}^{2}}{{4}^{0}}+\\frac{{\\left(-3\\right)}^{3}}{4}+\\frac{{\\left(-3\\right)}^{4}}{{4}^{2}}+\\cdots \\hfill \\\\ & ={\\left(-3\\right)}^{2}+{\\left(-3\\right)}^{2}\\cdot \\left(\\frac{-3}{4}\\right)+{\\left(-3\\right)}^{2}\\cdot {\\left(\\frac{-3}{4}\\right)}^{2}+\\cdots \\hfill \\\\ & =9+9\\cdot \\left(\\frac{-3}{4}\\right)+9\\cdot {\\left(\\frac{-3}{4}\\right)}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThe initial term [latex]a=-3[\/latex] and the ratio [latex]r=-\\frac{3}{4}[\/latex]. Since [latex]|r|=\\frac{3}{4}<1[\/latex], the series converges to\r\n<\/span>\r\n
        [latex]\\frac{9}{1-\\left(-\\frac{3}{4}\\right)}=\\frac{9}{\\frac{7}{4}}=\\frac{36}{7}[\/latex].<\/div>\r\n <\/li>\r\n \t
      2. Writing this series as\r\n<\/span>\r\n
        [latex]{e}^{2}\\displaystyle\\sum _{n=1}^{\\infty }{\\left({e}^{2}\\right)}^{n - 1}[\/latex]<\/div>\r\n\r\n<\/span>\r\nwe can see that this is a geometric series where [latex]r={e}^{2}>1[\/latex]. Therefore, the series diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>We now turn our attention to a nice application of geometric series. We show how they can be used to write repeating decimals as fractions of integers.\r\n\r\n
        \r\n
        \r\n

        Use a geometric series to write [latex]3.\\overline{26}[\/latex] as a fraction of integers.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n

        \r\n

        Since [latex]3.\\overline{26}=3.262626\\ldots [\/latex], first we write<\/p>\r\n\r\n

        [latex]\\begin{array}{cc}\\hfill 3.262626\\ldots& =3+\\frac{26}{100}+\\frac{26}{1000}+\\frac{26}{100,000}+\\cdots \\hfill \\\\ & =3+\\frac{26}{{10}^{2}}+\\frac{26}{{10}^{4}}+\\frac{26}{{10}^{6}}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

        Ignoring the term 3, the rest of this expression is a geometric series with initial term [latex]a=\\frac{26}{{10}^{2}}[\/latex] and ratio [latex]r=\\frac{1}{{10}^{2}}[\/latex]. Therefore, the sum of this series is<\/p>\r\n\r\n

        [latex]\\frac{\\frac{26}{{10}^{2}}}{1-\\left(\\frac{1}{{10}^{2}}\\right)}=\\frac{\\frac{26}{{10}^{2}}}{\\frac{99}{{10}^{2}}}=\\frac{26}{99}[\/latex].<\/div>\r\n \r\n

        Thus,<\/p>\r\n\r\n

        [latex]3.262626\\ldots =3+\\frac{26}{99}=\\frac{323}{99}[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"
        \n

        Geometric Series<\/h2>\n

        A geometric series<\/strong> is any series that we can write in the form:<\/p>\n

        [latex]a+ar+a{r}^{2}+a{r}^{3}+\\cdots =\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex].<\/div>\n

        Because the ratio of each term in this series to the previous term is [latex]r[\/latex], the number [latex]r[\/latex] is called the common ratio<\/strong>. We refer to [latex]a[\/latex] as the initial term<\/strong> because it is the first term in the series.<\/p>\n

        \n

        components of a geometric series<\/h3>\n

        The number [latex]r[\/latex] is called the common ratio<\/strong> because it’s the ratio between consecutive terms. The value [latex]a[\/latex] is the initial term<\/strong> since it’s the first term in the series.<\/p>\n<\/section>\n

        \n

        Let’s look at a concrete example. The series:<\/p>\n

        [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{1}{2}\\right)}^{n - 1}=1+\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\cdots[\/latex]<\/p>\n

        is a geometric series with initial term [latex]a=1[\/latex] and common ratio [latex]r=\\frac{1}{2}[\/latex].<\/p>\n<\/section>\n

        When Does a Geometric Series Converge?<\/h3>\n

        The key question is: when does a geometric series actually converge to a finite value? To answer this, we need to examine the partial sums. Consider the geometric series [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex]\u00a0when [latex]a>0[\/latex].<\/p>\n

        The sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] is:<\/p>\n

        [latex]{S}_{k}=\\displaystyle\\sum _{n=1}^{k}a{r}^{n - 1}=a+ar+a{r}^{2}+\\cdots +a{r}^{k - 1}[\/latex].<\/div>\n

         <\/p>\n

        Case 1: When [latex]r=1[\/latex] <\/strong><\/p>\n

        When [latex]r=1[\/latex], our partial sum becomes:<\/p>\n

        [latex]{S}_{k}=a+a\\left(1\\right)+a{\\left(1\\right)}^{2}+\\cdots +a{\\left(1\\right)}^{k - 1}=ak[\/latex].<\/div>\n

         <\/p>\n

        Since [latex]a>0[\/latex], we have [latex]ak\\to \\infty[\/latex] as [latex]k\\to \\infty[\/latex]. The sequence of partial sums is unbounded, so the series diverges when [latex]r=1[\/latex].<\/p>\n

        Case 2: When [latex]r\\ne 1[\/latex]<\/strong><\/p>\n

        For [latex]r\\ne 1[\/latex], to find the limit of [latex]\\left\\{{S}_{k}\\right\\}[\/latex], multiply the geometric series general equation by [latex]1-r[\/latex].<\/p>\n

        [latex]\\begin{array}{cc}\\hfill \\left(1-r\\right){S}_{k}& =a\\left(1-r\\right)\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)\\hfill \\\\ & =a\\left[\\left(1+r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k - 1}\\right)-\\left(r+{r}^{2}+{r}^{3}+\\cdots +{r}^{k}\\right)\\right]\\hfill \\\\ & =a\\left(1-{r}^{k}\\right).\\hfill \\end{array}[\/latex]<\/div>\n

         <\/p>\n

        Notice how most terms cancel out\u2014this is the key insight! Therefore:<\/p>\n

        [latex]{S}_{k}=\\frac{a\\left(1-{r}^{k}\\right)}{1-r}\\text{for }r\\ne 1[\/latex].<\/div>\n

        From our discussion in the previous section, we know that the geometric sequence [latex]{r}^{k}\\to 0[\/latex] if [latex]|r|<1[\/latex] and that [latex]{r}^{k}[\/latex] diverges if [latex]|r|>1[\/latex] or [latex]r= \\pm{1}[\/latex].<\/p>\n

        Using this knowledge:<\/p>\n

          \n
        • For [latex]|r|<1[\/latex]: [latex]{S}_{k}\\to \\frac{a}{1-r}[\/latex], so [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}[\/latex]<\/li>\n
        • For [latex]|r|\\ge 1[\/latex]: [latex]{S}_{k}[\/latex] diverges, so [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex] diverges.<\/li>\n<\/ul>\n

           <\/p>\n<\/section>\n

          \n

          convergence of geometric series<\/h3>\n

          A geometric series is a series of the form<\/p>\n

          [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=a+ar+a{r}^{2}+a{r}^{3}+\\cdots[\/latex].<\/div>\n

           <\/p>\n

          If [latex]|r|<1[\/latex], the series converges, and<\/p>\n

          [latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}=\\frac{a}{1-r}\\text{ for }|r|<1[\/latex].<\/div>\n

           <\/p>\n

          If [latex]|r|\\ge 1[\/latex], the series diverges.<\/p>\n<\/section>\n

          Recognizing Geometric Series in Different Forms<\/h3>\n

          Geometric series don’t always appear in the standard form. You might encounter series where the index starts at a different value or the exponent has a different linear expression. The key is recognizing when you can rewrite the series in geometric form.<\/p>\n

          Identifying Geometric Series<\/strong>: Look for a constant ratio between consecutive terms. If you find one, you likely have a geometric series that can be rewritten in standard form.<\/section>\n

          Let’s work through an example.<\/p>\n

          \n

          Consider the series:<\/p>\n

          [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}[\/latex]<\/p>\n

          To see if this is geometric, write out the first several terms:<\/p>\n

          [latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n+2}& ={\\left(\\frac{2}{3}\\right)}^{2}+{\\left(\\frac{2}{3}\\right)}^{3}+{\\left(\\frac{2}{3}\\right)}^{4}+\\cdots \\hfill \\\\ & =\\frac{4}{9}+\\frac{4}{9}\\cdot \\left(\\frac{2}{3}\\right)+\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{2}+\\cdots .\\hfill \\end{array}[\/latex]<\/p>\n

          We see that the initial term is [latex]a=\\frac{4}{9}[\/latex] and the ratio is [latex]r=\\frac{2}{3}[\/latex].<\/p>\n

          This means we can rewrite the series in standard form:<\/p>\n

          [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}[\/latex].<\/p>\n

          Since [latex]|r|=\\frac{2}{3}<1[\/latex], the series converges. Using our convergence formula:<\/p>\n

          [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{4}{9}\\cdot {\\left(\\frac{2}{3}\\right)}^{n - 1}=\\frac{\\frac{4}{9}}{1 - \\frac{2}{3}}=\\frac{\\frac{4}{9}}{\\frac{1}{3}}=\\frac{4}{9}\\cdot\\frac{3}{1}=\\frac{4}{3}[\/latex]<\/p>\n<\/section>\n

          \n
          \n

          Determine whether each of the following geometric series converges or diverges, and if it converges, find its sum.<\/p>\n

            \n
          1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-3\\right)}^{n+1}}{{4}^{n - 1}}[\/latex]<\/li>\n
          2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }{e}^{2n}[\/latex]<\/li>\n<\/ol>\n<\/div>\n