{"id":874,"date":"2025-06-20T17:19:19","date_gmt":"2025-06-20T17:19:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=874"},"modified":"2025-09-11T16:09:16","modified_gmt":"2025-09-11T16:09:16","slug":"direction-fields-and-eulers-method-learn-it-4-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-learn-it-4-2\/","title":{"raw":"Introduction to Series: Learn It 4","rendered":"Introduction to Series: Learn It 4"},"content":{"raw":"

Telescoping Series<\/h2>\r\nLet's examine a special type of series where terms cancel out in a predictable way. Consider the series:\r\n\r\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}[\/latex].\r\n

We can analyze this series using a technique called partial fractions<\/strong>. Notice that:<\/p>\r\n\r\n

[latex]\\frac{1}{n\\left(n+1\\right)}=\\frac{1}{n}-\\frac{1}{n+1}[\/latex].<\/div>\r\n
Partial Fractions Review<\/strong>: To verify this decomposition, find a common denominator:\r\n

[latex]\\frac{1}{n}-\\frac{1}{n+1}=\\frac{n+1}{n(n+1)}-\\frac{n}{n(n+1)}=\\frac{1}{n(n+1)}[\/latex] \u2713<\/p>\r\n\r\n<\/section>\r\n

Using this decomposition, we can rewrite the series as:<\/p>\r\n\r\n

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

Let's calculate the first few partial sums [latex]{{S}_{k}}[\/latex]:<\/p>\r\n\r\n

[latex]\\begin{array}{l}{S}_{1}=1-\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)=1-\\frac{1}{3}\\hfill \\\\ {S}_{3}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)=1-\\frac{1}{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n

Notice the pattern? The middle terms cancel each other out! In general:<\/p>\r\n\r\n

[latex]{S}_{k}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots +\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right)=1-\\frac{1}{k+1}[\/latex].<\/div>\r\n \r\n

The series collapses like a telescope, where most sections disappear into each other, leaving only the first and last terms visible. For this reason, we call a series that has this property a telescoping series<\/strong>.<\/p>\r\nFor this series, since [latex]{S}_{k}=1 - \\frac{1}{k+1}[\/latex] and [latex]\\frac{1}{k+1}\\to 0[\/latex] as [latex]k\\to \\infty [\/latex], the sequence of partial sums converges to [latex]1[\/latex]. Therefore, [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}=1[\/latex].\r\n\r\n

\r\n

telescoping series<\/h3>\r\nA telescoping series<\/strong> is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.\r\n\r\n[latex]\\\\[\/latex]\r\n\r\nThe key characteristic is that consecutive terms have parts that subtract to zero.\r\n\r\n<\/section>\r\n

Any series of the form:<\/p>\r\n

[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{b}_{n}-{b}_{n+1}\\right]=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)+\\cdots [\/latex]<\/p>\r\n

is a telescoping series. Let's see why by examining the partial sums:<\/p>\r\n

[latex]\\begin{array}{l}{S}_{1}={b}_{1}-{b}_{2}\\hfill \\\\ {S}_{2}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)={b}_{1}-{b}_{3}\\hfill \\\\ {S}_{3}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)={b}_{1}-{b}_{4}.\\hfill \\end{array}[\/latex]<\/p>\r\n

The pattern is clear: the [latex]k[\/latex]th partial sum simplifies to:<\/p>\r\n

[latex]{S}_{k}={b}_{1}-{b}_{k+1}[\/latex].<\/p>\r\nSince we can express each partial sum as [latex]{S}{k}={b}<\/em>{1}-{b}_{k+1}[\/latex], the convergence behavior becomes straightforward to analyze.\r\n\r\n

\r\n

convergence of general telescoping series<\/h3>\r\nA telescoping series [latex]\\displaystyle\\sum {n=1}^{\\infty }\\left[{b}{n}-{b}{n+1}\\right][\/latex] converges if and only if the sequence [latex]{{b}{k+1}}[\/latex] converges.\r\n\r\n[latex]\\\\[\/latex]\r\n\r\nIf [latex]{b}{k+1} \\to B[\/latex] as [latex]k \\to \\infty[\/latex], then:\r\n

[latex]\\displaystyle\\sum {n=1}^{\\infty }\\left[{b}{n}-{b}{n+1}\\right]={b}_{1}-B[\/latex]<\/p>\r\n\r\n<\/section>

\r\n

This gives us a powerful method for evaluating certain infinite series. Instead of struggling with complex partial sum formulas, we just need to:<\/p>\r\n\r\n

    \r\n \t
  1. Identify the sequence [latex]{b_n}[\/latex]<\/li>\r\n \t
  2. Check if [latex]\\underset{n\\to \\infty }{\\text{lim}} b_n[\/latex] exists<\/li>\r\n \t
  3. Apply the formula [latex]{b}_{1}-\\underset{n\\to \\infty }{\\text{lim}} b_n[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>\r\n

    In the next example, we show how to use these ideas to analyze a telescoping series of this form.<\/p>\r\n\r\n

    \r\n
    \r\n

    Determine whether the telescoping series<\/p>\r\n\r\n

    [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\cos\\left(\\frac{1}{n}\\right)-\\cos\\left(\\frac{1}{n+1}\\right)\\right][\/latex]<\/div>\r\n \r\n

    converges or diverges. If it converges, find its sum.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n

    \r\n

    By writing out terms in the sequence of partial sums, we can see that<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill {S}_{1}& =\\hfill & \\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\hfill \\\\ \\hfill {S}_{2}& =\\hfill & \\left(\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{2}\\right)-\\cos\\left(\\frac{1}{3}\\right)\\right)=\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{3}\\right)\\hfill \\\\ \\hfill {S}_{3}& =\\hfill & \\left(\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{2}\\right)-\\cos\\left(\\frac{1}{3}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{3}\\right)-\\cos\\left(\\frac{1}{4}\\right)\\right)\\hfill \\\\ & =\\hfill & \\cos\\left(1\\right)-\\cos\\left(\\frac{1}{4}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

    In general,<\/p>\r\n\r\n

    [latex]{S}_{k}=\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{k+1}\\right)[\/latex].<\/div>\r\n \r\n

    Since [latex]\\frac{1}{\\left(k+1\\right)}\\to 0[\/latex] as [latex]k\\to \\infty [\/latex] and [latex]\\cos{x}[\/latex] is a continuous function, [latex]\\cos\\left(\\frac{1}{\\left(k+1\\right)}\\right)\\to \\cos\\left(0\\right)=1[\/latex]. Therefore, we conclude that [latex]{S}_{k}\\to \\cos\\left(1\\right)-1[\/latex]. The telescoping series converges and the sum is given by<\/p>\r\n\r\n

    [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\cos\\left(\\frac{1}{n}\\right)-\\cos\\left(\\frac{1}{n+1}\\right)\\right]=\\cos\\left(1\\right)-1[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
    [ohm_question hide_question_numbers=1]311396[\/ohm_question]<\/section>","rendered":"

    Telescoping Series<\/h2>\n

    Let’s examine a special type of series where terms cancel out in a predictable way. Consider the series:<\/p>\n

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

    We can analyze this series using a technique called partial fractions<\/strong>. Notice that:<\/p>\n

    [latex]\\frac{1}{n\\left(n+1\\right)}=\\frac{1}{n}-\\frac{1}{n+1}[\/latex].<\/div>\n
    Partial Fractions Review<\/strong>: To verify this decomposition, find a common denominator:<\/p>\n

    [latex]\\frac{1}{n}-\\frac{1}{n+1}=\\frac{n+1}{n(n+1)}-\\frac{n}{n(n+1)}=\\frac{1}{n(n+1)}[\/latex] \u2713<\/p>\n<\/section>\n

    Using this decomposition, we can rewrite the series as:<\/p>\n

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

    Let’s calculate the first few partial sums [latex]{{S}_{k}}[\/latex]:<\/p>\n

    [latex]\\begin{array}{l}{S}_{1}=1-\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)=1-\\frac{1}{3}\\hfill \\\\ {S}_{3}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)=1-\\frac{1}{4}.\\hfill \\end{array}[\/latex]<\/div>\n

    Notice the pattern? The middle terms cancel each other out! In general:<\/p>\n

    [latex]{S}_{k}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots +\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right)=1-\\frac{1}{k+1}[\/latex].<\/div>\n

     <\/p>\n

    The series collapses like a telescope, where most sections disappear into each other, leaving only the first and last terms visible. For this reason, we call a series that has this property a telescoping series<\/strong>.<\/p>\n

    For this series, since [latex]{S}_{k}=1 - \\frac{1}{k+1}[\/latex] and [latex]\\frac{1}{k+1}\\to 0[\/latex] as [latex]k\\to \\infty[\/latex], the sequence of partial sums converges to [latex]1[\/latex]. Therefore, [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}=1[\/latex].<\/p>\n

    \n

    telescoping series<\/h3>\n

    A telescoping series<\/strong> is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.<\/p>\n

    [latex]\\\\[\/latex]<\/p>\n

    The key characteristic is that consecutive terms have parts that subtract to zero.<\/p>\n<\/section>\n

    Any series of the form:<\/p>\n

    [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{b}_{n}-{b}_{n+1}\\right]=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)+\\cdots[\/latex]<\/p>\n

    is a telescoping series. Let’s see why by examining the partial sums:<\/p>\n

    [latex]\\begin{array}{l}{S}_{1}={b}_{1}-{b}_{2}\\hfill \\\\ {S}_{2}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)={b}_{1}-{b}_{3}\\hfill \\\\ {S}_{3}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)={b}_{1}-{b}_{4}.\\hfill \\end{array}[\/latex]<\/p>\n

    The pattern is clear: the [latex]k[\/latex]th partial sum simplifies to:<\/p>\n

    [latex]{S}_{k}={b}_{1}-{b}_{k+1}[\/latex].<\/p>\n

    Since we can express each partial sum as [latex]{S}{k}={b}<\/em>{1}-{b}_{k+1}[\/latex], the convergence behavior becomes straightforward to analyze.<\/p>\n

    \n

    convergence of general telescoping series<\/h3>\n

    A telescoping series [latex]\\displaystyle\\sum {n=1}^{\\infty }\\left[{b}{n}-{b}{n+1}\\right][\/latex] converges if and only if the sequence [latex]{{b}{k+1}}[\/latex] converges.<\/p>\n

    [latex]\\\\[\/latex]<\/p>\n

    If [latex]{b}{k+1} \\to B[\/latex] as [latex]k \\to \\infty[\/latex], then:<\/p>\n

    [latex]\\displaystyle\\sum {n=1}^{\\infty }\\left[{b}{n}-{b}{n+1}\\right]={b}_{1}-B[\/latex]<\/p>\n<\/section>\n

    \n

    This gives us a powerful method for evaluating certain infinite series. Instead of struggling with complex partial sum formulas, we just need to:<\/p>\n

      \n
    1. Identify the sequence [latex]{b_n}[\/latex]<\/li>\n
    2. Check if [latex]\\underset{n\\to \\infty }{\\text{lim}} b_n[\/latex] exists<\/li>\n
    3. Apply the formula [latex]{b}_{1}-\\underset{n\\to \\infty }{\\text{lim}} b_n[\/latex]<\/li>\n<\/ol>\n<\/section>\n

      In the next example, we show how to use these ideas to analyze a telescoping series of this form.<\/p>\n

      \n
      \n

      Determine whether the telescoping series<\/p>\n

      [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\cos\\left(\\frac{1}{n}\\right)-\\cos\\left(\\frac{1}{n+1}\\right)\\right][\/latex]<\/div>\n

       <\/p>\n

      converges or diverges. If it converges, find its sum.<\/p>\n<\/div>\n