{"id":1113,"date":"2025-06-30T16:34:48","date_gmt":"2025-06-30T16:34:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1113"},"modified":"2025-08-18T15:25:56","modified_gmt":"2025-08-18T15:25:56","slug":"alternating-series-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/alternating-series-learn-it-3\/","title":{"raw":"Alternating Series: Learn It 3","rendered":"Alternating Series: Learn It 3"},"content":{"raw":"

Absolute and Conditional Convergence<\/h2>\r\nWhen studying series with both positive and negative terms, we need to consider two different types of convergence behavior.\r\n

Consider any series [latex]\\displaystyle\\sum_{n=1}^{\\infty}a_n[\/latex] and its related series [latex]\\displaystyle\\sum_{n=1}^{\\infty}|a_n|[\/latex] (formed by taking the absolute value of each term). The relationship between these two series reveals important information about convergence.<\/p>\r\n

Let's examine two key examples:<\/p>\r\n

Example 1: The Alternating Harmonic Series<\/strong><\/p>\r\n

Consider [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n}[\/latex]. The series of absolute values is: [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\left|\\frac{(-1)^{n+1}}{n}\\right| = \\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex]<\/p>\r\n

We know the alternating harmonic series converges, but the harmonic series diverges. This creates an interesting situation. Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.<\/p>\r\n

Example 2: An Alternating [latex]p[\/latex]-Series<\/strong><\/p>\r\n

Consider [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n^2}[\/latex]. The series of absolute values is: [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\left|\\frac{(-1)^{n+1}}{n^2}\\right| = \\sum_{n=1}^{\\infty}\\frac{1}{n^2}[\/latex]<\/p>\r\n

Both series converge in this case. Since both of these series converge, we say the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex] exhibits absolute convergence.<\/p>\r\n\r\n

\r\n

absolute and conditional convergence<\/h3>\r\nA series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits absolute convergence<\/strong> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges.\r\n\r\n[latex]\\\\[\/latex]\r\n\r\nA series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits conditional convergence<\/span> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges but [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] diverges.\r\n\r\n<\/section>The alternating harmonic series demonstrates that a series can converge even when its absolute value series diverges. However, the following theorem shows that absolute convergence guarantees regular convergence.\r\n\r\n
\r\n

theorem: absolute convergence implies convergence<\/h3>\r\nIf [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.\r\n\r\n<\/section>
\r\n

Proof<\/strong><\/p>\r\n\r\n\r\n

\r\n

Suppose that [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges. We show this by using the fact that [latex]{a}_{n}=|{a}_{n}|[\/latex] or [latex]{a}_{n}=\\text{-}|{a}_{n}|[\/latex] and therefore [latex]|{a}_{n}|+{a}_{n}=2|{a}_{n}|[\/latex] or [latex]|{a}_{n}|+{a}_{n}=0[\/latex]. Therefore, [latex]0\\le |{a}_{n}|+{a}_{n}\\le 2|{a}_{n}|[\/latex]. Consequently, by the comparison test, since [latex]2\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, the series<\/p>\r\n\r\n

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

converges. By using the algebraic properties for convergent series, we conclude that<\/p>\r\n\r\n

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

converges.<\/p>\r\n

[latex]_\\blacksquare[\/latex]<\/p>\r\n\r\n<\/section>

\r\n
\r\n

For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.<\/p>\r\n\r\n

    \r\n \t
  1. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{\\left(3n+1\\right)}[\/latex]<\/li>\r\n \t
  2. [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos\\left(n\\right)}{{n}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n
    \r\n
      \r\n \t
    1. We can see that\r\n<\/span>\r\n
      [latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{{\\left(-1\\right)}^{n+1}}{3n+1}|=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{3n+1}[\/latex]<\/div>\r\n\r\n<\/span>\r\ndiverges by using the limit comparison test with the harmonic series. In fact,\r\n<\/span>\r\n
      [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left(3n+1\\right)}}{\\frac{1}{n}}=\\frac{1}{3}[\/latex].<\/div>\r\n\r\n<\/span>\r\nTherefore, the series does not converge absolutely. However, since\r\n<\/span>\r\n
      [latex]\\frac{1}{3\\left(n+1\\right)+1}<\\frac{1}{3n+1}\\text{and}\\frac{1}{3n+1}\\to 0[\/latex],<\/div>\r\n\r\n<\/span>\r\nthe series converges. We can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{\\left(3n+1\\right)}[\/latex] converges conditionally.<\/li>\r\n \t
    2. Noting that [latex]|\\cos{n}|\\le 1[\/latex], to determine whether the series converges absolutely, compare\r\n<\/span>\r\n
      [latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{\\cos{n}}{{n}^{2}}|[\/latex]<\/div>\r\n\r\n<\/span>\r\nwith the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, by the comparison test, [latex]\\displaystyle\\sum _{n=1}^{\\infty }|\\frac{\\cos{n}}{{n}^{2}}|[\/latex] converges, and therefore [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\cos{n}}{{n}^{2}}[\/latex] converges absolutely.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
      \r\n
      \r\n

      To understand the fundamental difference between absolute and conditional convergence, we need to explore what happens when we rearrange the terms of a conditionally convergent series.<\/p>\r\nConsider the alternating harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n}[\/latex], which we know converges conditionally. While rearranging terms in a finite sum doesn't change the result, infinite series behave very differently.\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      We can demonstrate this by rearranging the terms to make the series diverge to infinity.<\/p>\r\n

      Begin by adding enough positive terms to produce a sum larger than some target value [latex]M > 0[\/latex]. For example, let [latex]M = 10[\/latex] and find an integer [latex]k[\/latex] such that:<\/p>\r\n

      [latex]1 + \\frac{1}{3} + \\frac{1}{5} + \\cdots + \\frac{1}{2k-1} > 10[\/latex]<\/p>\r\n

      This is possible because [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{2n-1}[\/latex] diverges to infinity. Then subtract [latex]\\frac{1}{2}[\/latex].<\/p>\r\n

      Next, add more positive terms until the sum reaches [latex]100[\/latex]. Find another integer [latex]j > k[\/latex] such that:<\/p>\r\n

      [latex]1 + \\frac{1}{3} + \\cdots + \\frac{1}{2k-1} - \\frac{1}{2} + \\frac{1}{2k+1} + \\cdots + \\frac{1}{2j+1} > 100[\/latex]<\/p>\r\n

      Then subtract [latex]\\frac{1}{4}[\/latex]. By continuing this process\u2014alternating between adding enough positive terms to reach increasingly large values and subtracting single negative terms\u2014we create a rearranged series whose partial sums grow without bound and therefore diverges.<\/p>\r\nThe alternating harmonic series can also be rearranged to converge to entirely different values. In fact, it can be rearranged to converge to any real number [latex]r[\/latex] we choose! We'll see in the next example how to rearrange the terms to create a series that converges to [latex]\\frac{3\\ln(2)}{2}[\/latex].\r\n\r\nIn general, any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] that converges conditionally can be rearranged so that the new series diverges or converges to a different real number. A series that converges absolutely does not have this property. For any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] that converges absolutely, the value of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is the same for any rearrangement of the terms. This result is known as the Riemann Rearrangement Theorem, which is beyond the scope of this book.\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      Use the fact that<\/p>\r\n\r\n

      [latex]1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\cdots =\\text{ln}2[\/latex]<\/div>\r\n \r\n

      to rearrange the terms in the alternating harmonic series so the sum of the rearranged series is [latex]\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n

      \r\n

      Let<\/p>\r\n\r\n

      [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{5}-\\frac{1}{6}+\\frac{1}{7}-\\frac{1}{8}+\\cdots [\/latex].<\/div>\r\n \r\n

      Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\text{ln}\\left(2\\right)[\/latex], by the algebraic properties of convergent series,<\/p>\r\n\r\n

      [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{2}{a}_{n}=\\frac{1}{2}-\\frac{1}{4}+\\frac{1}{6}-\\frac{1}{8}+\\cdots =\\frac{1}{2}\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}=\\frac{\\text{ln}2}{2}[\/latex].<\/div>\r\n \r\n

      Now introduce the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] such that for all [latex]n\\ge 1[\/latex], [latex]{b}_{2n - 1}=0[\/latex] and [latex]{b}_{2n}=\\frac{{a}_{n}}{2}[\/latex]. Then<\/p>\r\n\r\n

      [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}=0+\\frac{1}{2}+0-\\frac{1}{4}+0+\\frac{1}{6}+0-\\frac{1}{8}+\\cdots =\\frac{\\text{ln}2}{2}[\/latex].<\/div>\r\n \r\n

      Then using the algebraic limit properties of convergent series, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converge, the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)[\/latex] converges and<\/p>\r\n\r\n

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

      Now adding the corresponding terms, [latex]{a}_{n}[\/latex] and [latex]{b}_{n}[\/latex], we see that<\/p>\r\n\r\n

      [latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)& =\\left(1+0\\right)+\\left(-\\frac{1}{2}+\\frac{1}{2}\\right)+\\left(\\frac{1}{3}+0\\right)+\\left(-\\frac{1}{4}-\\frac{1}{4}\\right)+\\left(\\frac{1}{5}+0\\right)+\\left(-\\frac{1}{6}+\\frac{1}{6}\\right)\\hfill \\\\ \\\\ & +\\left(\\frac{1}{7}+0\\right)+\\left(\\frac{1}{8}-\\frac{1}{8}\\right)+\\cdots \\hfill \\\\ & =1+\\frac{1}{3}-\\frac{1}{2}+\\frac{1}{5}+\\frac{1}{7}-\\frac{1}{4}+\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

      We notice that the series on the right side of the equal sign is a rearrangement of the alternating harmonic series. Since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left({a}_{n}+{b}_{n}\\right)=\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex], we conclude that<\/p>\r\n\r\n

      [latex]1+\\frac{1}{3}-\\frac{1}{2}+\\frac{1}{5}+\\frac{1}{7}-\\frac{1}{4}+\\cdots =\\frac{3\\text{ln}\\left(2\\right)}{2}[\/latex].<\/div>\r\n \r\n

      Therefore, we have found a rearrangement of the alternating harmonic series having the desired property.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

      [ohm_question hide_question_numbers=1]230805[\/ohm_question]<\/section>","rendered":"

      Absolute and Conditional Convergence<\/h2>\n

      When studying series with both positive and negative terms, we need to consider two different types of convergence behavior.<\/p>\n

      Consider any series [latex]\\displaystyle\\sum_{n=1}^{\\infty}a_n[\/latex] and its related series [latex]\\displaystyle\\sum_{n=1}^{\\infty}|a_n|[\/latex] (formed by taking the absolute value of each term). The relationship between these two series reveals important information about convergence.<\/p>\n

      Let’s examine two key examples:<\/p>\n

      Example 1: The Alternating Harmonic Series<\/strong><\/p>\n

      Consider [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n}[\/latex]. The series of absolute values is: [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\left|\\frac{(-1)^{n+1}}{n}\\right| = \\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex]<\/p>\n

      We know the alternating harmonic series converges, but the harmonic series diverges. This creates an interesting situation. Since the alternating harmonic series converges, but the harmonic series diverges, we say the alternating harmonic series exhibits conditional convergence.<\/p>\n

      Example 2: An Alternating [latex]p[\/latex]-Series<\/strong><\/p>\n

      Consider [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{(-1)^{n+1}}{n^2}[\/latex]. The series of absolute values is: [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\left|\\frac{(-1)^{n+1}}{n^2}\\right| = \\sum_{n=1}^{\\infty}\\frac{1}{n^2}[\/latex]<\/p>\n

      Both series converge in this case. Since both of these series converge, we say the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{{n}^{2}}[\/latex] exhibits absolute convergence.<\/p>\n

      \n

      absolute and conditional convergence<\/h3>\n

      A series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits absolute convergence<\/strong> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges.<\/p>\n

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

      A series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] exhibits conditional convergence<\/span> if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges but [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] diverges.<\/p>\n<\/section>\n

      The alternating harmonic series demonstrates that a series can converge even when its absolute value series diverges. However, the following theorem shows that absolute convergence guarantees regular convergence.<\/p>\n

      \n

      theorem: absolute convergence implies convergence<\/h3>\n

      If [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/p>\n<\/section>\n

      \n

      Proof<\/strong><\/p>\n

      \n

      Suppose that [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges. We show this by using the fact that [latex]{a}_{n}=|{a}_{n}|[\/latex] or [latex]{a}_{n}=\\text{-}|{a}_{n}|[\/latex] and therefore [latex]|{a}_{n}|+{a}_{n}=2|{a}_{n}|[\/latex] or [latex]|{a}_{n}|+{a}_{n}=0[\/latex]. Therefore, [latex]0\\le |{a}_{n}|+{a}_{n}\\le 2|{a}_{n}|[\/latex]. Consequently, by the comparison test, since [latex]2\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] converges, the series<\/p>\n

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

       <\/p>\n

      converges. By using the algebraic properties for convergent series, we conclude that<\/p>\n

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

       <\/p>\n

      converges.<\/p>\n

      [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

      \n
      \n

      For each of the following series, determine whether the series converges absolutely, converges conditionally, or diverges.<\/p>\n

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