{"id":1538,"date":"2025-07-29T15:46:59","date_gmt":"2025-07-29T15:46:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=1538"},"modified":"2025-09-11T16:00:22","modified_gmt":"2025-09-11T16:00:22","slug":"sequences-and-their-properties-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/sequences-and-their-properties-learn-it-5\/","title":{"raw":"Sequences and Their Properties: Learn It 5","rendered":"Sequences and Their Properties: Learn It 5"},"content":{"raw":"

Monotone Convergence Theorem<\/h2>\r\nWe now turn to one of the most important theorems about sequences: the Monotone Convergence Theorem. This powerful result gives us a way to prove that certain sequences converge even when we can't find their exact limits. Before we can state the main theorem, we need to understand what it means for a sequence to be bounded.\r\n\r\n
\r\n

bounded sequences<\/h3>\r\n
    \r\n \t
  • A sequence [latex]\\{{a}_{n}\\}[\/latex] is a bounded sequence<\/strong> if it is bounded above and bounded below.\r\n
      \r\n \t
    • If a sequence is not bounded, it is an unbounded sequence<\/strong>.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
    • A sequence [latex]\\{{a}_{n}\\}[\/latex] is bounded above<\/strong> if there exists a real number [latex]M[\/latex] such that\r\n
      [latex]{a}_{n}\\le M[\/latex]<\/center>\r\nfor all positive integers [latex]n[\/latex].<\/li>\r\n \t
    • A sequence [latex]\\{{a}_{n}\\}[\/latex] is bounded below<\/strong> if there exists a real number [latex]M[\/latex] such that\r\n
      [latex]M\\le {a}_{n}[\/latex]<\/center>\r\nfor all positive integers [latex]n[\/latex].<\/li>\r\n<\/ul>\r\n<\/section>
      \r\n
        \r\n \t
      • The sequence [latex]{\\frac{1}{n}} = {1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\ldots}[\/latex] is bounded above by [latex]1[\/latex] and bounded below by [latex]0[\/latex], so it's bounded.<\/li>\r\n \t
      • The sequence [latex]{2^n} = {2, 4, 8, 16, \\ldots}[\/latex] is bounded below by [latex] 2 [\/latex] but has no upper bound, so it's unbounded.<\/li>\r\n<\/ul>\r\n<\/section>\r\n

        Here's a crucial insight: if a sequence is unbounded, it cannot converge. Why? Because unbounded sequences have terms that become arbitrarily large in magnitude, preventing them from settling down to any finite limit.<\/p>\r\n\r\n

        \r\n

        convergent sequences are bounded<\/h3>\r\n

        If a sequence [latex]{a_n}[\/latex] converges, then it is bounded.<\/p>\r\n[latex]\\\\[\/latex]\r\n

        Important:<\/strong> The converse is NOT true. A bounded sequence doesn't have to converge.<\/p>\r\n\r\n<\/section>

        \u00a0The sequence [latex]\\left\\{{\\left(-1\\right)}^{n}\\right\\}[\/latex] is bounded, but the sequence diverges because the sequence oscillates between [latex]1[\/latex] and [latex]-1[\/latex] and never approaches a finite number.<\/section>So boundedness is necessary for convergence, but not sufficient. We need something more. The missing ingredient is monotonicity\u2014sequences that consistently move in one direction.\r\n\r\n
        \r\n

        monotone sequences<\/h3>\r\n

        A sequence [latex]{a_n}[\/latex] is:<\/p>\r\n\r\n

          \r\n \t
        • Increasing<\/strong> (for [latex]n \\geq n_0[\/latex]) if [latex]a_n \\leq a_{n+1}[\/latex] for all [latex]n \\geq n_0[\/latex]<\/li>\r\n \t
        • Decreasing<\/strong> (for [latex]n \\geq n_0[\/latex]) if [latex]a_n \\geq a_{n+1}[\/latex] for all [latex]n \\geq n_0[\/latex]<\/li>\r\n \t
        • Monotone<\/strong> if it's either increasing or decreasing (at least eventually)<\/li>\r\n<\/ul>\r\n

          Note:<\/strong> \"Eventually\" means the pattern holds from some point [latex]n_0[\/latex] onward\u2014early terms can behave differently.<\/p>\r\n\r\n<\/section>Consider a bounded sequence [latex]{a_n}[\/latex] that is also increasing: [latex]a_1 \\leq a_2 \\leq a_3 \\leq \\ldots[\/latex]. Since the sequence is increasing, the terms aren't oscillating. This leaves only two possibilities: the sequence could diverge to infinity, or it could converge. But here's the crucial point\u2014since the sequence is bounded above, it cannot<\/em> diverge to infinity.\u00a0 We conclude that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges.\r\n

          The good news is that we don't need a sequence to be monotone from the very beginning\u2014it just needs to be monotone eventually<\/em>. This makes the theorem much more practical.The monotone behavior doesn't need to start immediately.<\/p>\r\n

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

          [latex]\\left\\{2,0,3,0,4,0,1,-\\frac{1}{2},-\\frac{1}{3},-\\frac{1}{4}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\r\n

          Even though the sequence is not increasing for all values of [latex]n[\/latex], we see that [latex]-\\frac{1}{2}<\\text{-}\\frac{1}{3}<\\text{-}\\frac{1}{4}<\\cdots [\/latex]. Therefore, starting with the eighth term, [latex]{a}_{8}=-\\frac{1}{2}[\/latex], the sequence is increasing. In this case, we say the sequence is eventually<\/em> increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.<\/p>\r\n\r\n

          Why this works:<\/strong> Remember, convergence depends only on what happens as [latex]n \\to \\infty[\/latex]. The first few terms don't affect the ultimate behavior of the sequence.<\/section>We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.\r\n\r\n
          \r\n

          Monotone Convergence Theorem<\/h3>\r\nIf [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a bounded sequence and there exists a positive integer [latex]{n}_{0}[\/latex] such that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is monotone for all [latex]n\\ge {n}_{0}[\/latex], then [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges.\r\n\r\n<\/section>\r\n

          The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 6).<\/p>\r\n\r\n

          <\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"456\"]\"A Figure 6. Since the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing and bounded above, it must converge.[\/caption]<\/figure>\r\n

          In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.<\/p>\r\n\r\n

          \r\n
          \r\n

          For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.<\/p>\r\n\r\n

            \r\n \t
          1. [latex]\\left\\{\\frac{{4}^{n}}{n\\text{!}}\\right\\}[\/latex]<\/li>\r\n \t
          2. [latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that\r\n<\/span>\r\n
            [latex]{a}_{1}=2\\text{ and }{a}_{n+1}=\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\text{ for all }n\\ge 2[\/latex].<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n
            \r\n
              \r\n \t
            1. Writing out the first few terms, we see that\r\n<\/span>\r\n
              [latex]\\left\\{\\frac{{4}^{n}}{n\\text{!}}\\right\\}=\\left\\{4,8,\\frac{32}{3},\\frac{32}{3},\\frac{128}{15}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\r\n\r\n<\/span>\r\nAt first, the terms increase. However, after the third term, the terms decrease. In fact, the terms decrease for all [latex]n\\ge 3[\/latex]. We can show this as follows.\r\n<\/span>\r\n
              [latex]{a}_{n+1}=\\frac{{4}^{n+1}}{\\left(n+1\\right)\\text{!}}=\\frac{4}{n+1}\\cdot \\frac{{4}^{n}}{n\\text{!}}=\\frac{4}{n+1}\\cdot {a}_{n}\\le {a}_{n}\\text{ if }n\\ge 3[\/latex].<\/div>\r\n\r\n<\/span>\r\nTherefore, the sequence is decreasing for all [latex]n\\ge 3[\/latex]. Further, the sequence is bounded below by [latex]0[\/latex] because [latex]\\frac{{4}^{n}}{n\\text{!}}\\ge 0[\/latex] for all positive integers [latex]n[\/latex]. Therefore, by the Monotone Convergence Theorem, the sequence converges.\r\n<\/span>\r\nTo find the limit, we use the fact that the sequence converges and let [latex]L=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. Now note this important observation. Consider [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}[\/latex]. Since\r\n<\/span>\r\n
              [latex]\\left\\{{a}_{n+1}\\right\\}=\\left\\{{a}_{2,}{a}_{3},{a}_{4}\\text{,}\\ldots\\right\\}[\/latex],<\/div>\r\n \r\n\r\nthe only difference between the sequences [latex]\\left\\{{a}_{n+1}\\right\\}[\/latex] and [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is that [latex]\\left\\{{a}_{n+1}\\right\\}[\/latex] omits the first term. Since a finite number of terms does not affect the convergence of a sequence,\r\n\r\n\u00a0<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=L[\/latex].<\/div>\r\n\r\n<\/span>\r\nCombining this fact with the equation\r\n<\/span>\r\n
              [latex]{a}_{n+1}=\\frac{4}{n+1}{a}_{n}[\/latex]<\/div>\r\n\r\n<\/span>\r\nand taking the limit of both sides of the equation\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{4}{n+1}{a}_{n}[\/latex],<\/div>\r\n\r\n<\/span>\r\nwe can conclude that\r\n<\/span>\r\n
              [latex]L=0\\cdot L=0[\/latex].<\/div>\r\n <\/li>\r\n \t
            2. Writing out the first several terms,\r\n<\/span>\r\n
              [latex]\\left\\{2,\\frac{5}{4},\\frac{41}{40},\\frac{3281}{3280}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\r\n\r\n<\/span>\r\nwe can conjecture that the sequence is decreasing and bounded below by [latex]1[\/latex]. To show that the sequence is bounded below by [latex]1[\/latex], we can show that\r\n<\/span>\r\n
              [latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\ge 1[\/latex].<\/div>\r\n\r\n<\/span>\r\nTo show this, first rewrite\r\n<\/span>\r\n
              [latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}=\\frac{{a}_{n}^{2}+1}{2{a}_{n}}[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince [latex]{a}_{1}>0[\/latex] and [latex]{a}_{2}[\/latex] is defined as a sum of positive terms, [latex]{a}_{2}>0[\/latex]. Similarly, all terms [latex]{a}_{n}>0[\/latex]. Therefore,\r\n<\/span>\r\n
              [latex]\\frac{{a}_{n}^{2}+1}{2{a}_{n}}\\ge 1[\/latex]<\/div>\r\n\r\n<\/span>\r\nif and only if\r\n<\/span>\r\n
              [latex]{a}_{n}^{2}+1\\ge 2{a}_{n}[\/latex].<\/div>\r\n\r\n<\/span>\r\nRewriting the inequality [latex]{a}_{n}^{2}+1\\ge 2{a}_{n}[\/latex] as [latex]{a}_{n}^{2}-2{a}_{n}+1\\ge 0[\/latex], and using the fact that\r\n<\/span>\r\n
              [latex]{a}_{n}^{2}-2{a}_{n}+1={\\left({a}_{n}-1\\right)}^{2}\\ge 0[\/latex]<\/div>\r\n\r\n<\/span>\r\nbecause the square of any real number is nonnegative, we can conclude that\r\n<\/span>\r\n
              [latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\ge 1[\/latex].<\/div>\r\n \r\n

              \r\n<\/span>\r\nTo show that the sequence is decreasing, we must show that [latex]{a}_{n+1}\\le {a}_{n}[\/latex] for all [latex]n\\ge 1[\/latex]. Since [latex]1\\le {a}_{n}^{2}[\/latex], it follows that\r\n<\/span><\/p>\r\n\r\n

              [latex]{a}_{n}^{2}+1\\le 2{a}_{n}^{2}[\/latex].<\/div>\r\n\r\n<\/span>\r\nDividing both sides by [latex]2{a}_{n}[\/latex], we obtain\r\n<\/span>\r\n
              [latex]\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\le {a}_{n}[\/latex].<\/div>\r\n\r\n<\/span>\r\nUsing the definition of [latex]{a}_{n+1}[\/latex], we conclude that\r\n<\/span>\r\n
              [latex]{a}_{n+1}=\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\le {a}_{n}[\/latex].<\/div>\r\n\r\n<\/span>\r\nSince [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is bounded below and decreasing, by the Monotone Convergence Theorem, it converges.\r\n<\/span>\r\nTo find the limit, let [latex]L=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex]. Then using the recurrence relation and the fact that [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}[\/latex], we have\r\n<\/span>\r\n
              [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\left(\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\right)[\/latex],<\/div>\r\n\r\n<\/span>\r\nand therefore\r\n<\/span>\r\n
              [latex]L=\\frac{L}{2}+\\frac{1}{2L}[\/latex].<\/div>\r\n\r\n<\/span>\r\nMultiplying both sides of this equation by [latex]2L[\/latex], we arrive at the equation\r\n<\/span>\r\n
              [latex]2{L}^{2}={L}^{2}+1[\/latex].<\/div>\r\n\r\n<\/span>\r\nSolving this equation for [latex]L[\/latex], we conclude that [latex]{L}^{2}=1[\/latex], which implies [latex]L=\\pm 1[\/latex]. Since all the terms are positive, the limit [latex]L=1[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
              [ohm_question hide_question_numbers=1]311394[\/ohm_question]<\/section>","rendered":"

              Monotone Convergence Theorem<\/h2>\n

              We now turn to one of the most important theorems about sequences: the Monotone Convergence Theorem. This powerful result gives us a way to prove that certain sequences converge even when we can’t find their exact limits. Before we can state the main theorem, we need to understand what it means for a sequence to be bounded.<\/p>\n

              \n

              bounded sequences<\/h3>\n
                \n
              • A sequence [latex]\\{{a}_{n}\\}[\/latex] is a bounded sequence<\/strong> if it is bounded above and bounded below.\n
                  \n
                • If a sequence is not bounded, it is an unbounded sequence<\/strong>.<\/li>\n<\/ul>\n<\/li>\n
                • A sequence [latex]\\{{a}_{n}\\}[\/latex] is bounded above<\/strong> if there exists a real number [latex]M[\/latex] such that\n
                  [latex]{a}_{n}\\le M[\/latex]<\/div>\n

                  for all positive integers [latex]n[\/latex].<\/li>\n

                • A sequence [latex]\\{{a}_{n}\\}[\/latex] is bounded below<\/strong> if there exists a real number [latex]M[\/latex] such that\n
                  [latex]M\\le {a}_{n}[\/latex]<\/div>\n

                  for all positive integers [latex]n[\/latex].<\/li>\n<\/ul>\n<\/section>\n

                  \n
                    \n
                  • The sequence [latex]{\\frac{1}{n}} = {1, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{4}, \\ldots}[\/latex] is bounded above by [latex]1[\/latex] and bounded below by [latex]0[\/latex], so it’s bounded.<\/li>\n
                  • The sequence [latex]{2^n} = {2, 4, 8, 16, \\ldots}[\/latex] is bounded below by [latex]2[\/latex] but has no upper bound, so it’s unbounded.<\/li>\n<\/ul>\n<\/section>\n

                    Here’s a crucial insight: if a sequence is unbounded, it cannot converge. Why? Because unbounded sequences have terms that become arbitrarily large in magnitude, preventing them from settling down to any finite limit.<\/p>\n

                    \n

                    convergent sequences are bounded<\/h3>\n

                    If a sequence [latex]{a_n}[\/latex] converges, then it is bounded.<\/p>\n

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

                    Important:<\/strong> The converse is NOT true. A bounded sequence doesn’t have to converge.<\/p>\n<\/section>\n

                    \u00a0The sequence [latex]\\left\\{{\\left(-1\\right)}^{n}\\right\\}[\/latex] is bounded, but the sequence diverges because the sequence oscillates between [latex]1[\/latex] and [latex]-1[\/latex] and never approaches a finite number.<\/section>\n

                    So boundedness is necessary for convergence, but not sufficient. We need something more. The missing ingredient is monotonicity\u2014sequences that consistently move in one direction.<\/p>\n

                    \n

                    monotone sequences<\/h3>\n

                    A sequence [latex]{a_n}[\/latex] is:<\/p>\n

                      \n
                    • Increasing<\/strong> (for [latex]n \\geq n_0[\/latex]) if [latex]a_n \\leq a_{n+1}[\/latex] for all [latex]n \\geq n_0[\/latex]<\/li>\n
                    • Decreasing<\/strong> (for [latex]n \\geq n_0[\/latex]) if [latex]a_n \\geq a_{n+1}[\/latex] for all [latex]n \\geq n_0[\/latex]<\/li>\n
                    • Monotone<\/strong> if it’s either increasing or decreasing (at least eventually)<\/li>\n<\/ul>\n

                      Note:<\/strong> “Eventually” means the pattern holds from some point [latex]n_0[\/latex] onward\u2014early terms can behave differently.<\/p>\n<\/section>\n

                      Consider a bounded sequence [latex]{a_n}[\/latex] that is also increasing: [latex]a_1 \\leq a_2 \\leq a_3 \\leq \\ldots[\/latex]. Since the sequence is increasing, the terms aren’t oscillating. This leaves only two possibilities: the sequence could diverge to infinity, or it could converge. But here’s the crucial point\u2014since the sequence is bounded above, it cannot<\/em> diverge to infinity.\u00a0 We conclude that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges.<\/p>\n

                      The good news is that we don’t need a sequence to be monotone from the very beginning\u2014it just needs to be monotone eventually<\/em>. This makes the theorem much more practical.The monotone behavior doesn’t need to start immediately.<\/p>\n

                      Consider the sequence:<\/p>\n

                      [latex]\\left\\{2,0,3,0,4,0,1,-\\frac{1}{2},-\\frac{1}{3},-\\frac{1}{4}\\text{,}\\ldots\\right\\}[\/latex].<\/div>\n

                      Even though the sequence is not increasing for all values of [latex]n[\/latex], we see that [latex]-\\frac{1}{2}<\\text{-}\\frac{1}{3}<\\text{-}\\frac{1}{4}<\\cdots[\/latex]. Therefore, starting with the eighth term, [latex]{a}_{8}=-\\frac{1}{2}[\/latex], the sequence is increasing. In this case, we say the sequence is eventually<\/em> increasing. Since the sequence is bounded above, it converges. It is also true that if a sequence is decreasing (or eventually decreasing) and bounded below, it also converges.<\/p>\n

                      Why this works:<\/strong> Remember, convergence depends only on what happens as [latex]n \\to \\infty[\/latex]. The first few terms don’t affect the ultimate behavior of the sequence.<\/section>\n

                      We now have the necessary definitions to state the Monotone Convergence Theorem, which gives a sufficient condition for convergence of a sequence.<\/p>\n

                      \n

                      Monotone Convergence Theorem<\/h3>\n

                      If [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is a bounded sequence and there exists a positive integer [latex]{n}_{0}[\/latex] such that [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is monotone for all [latex]n\\ge {n}_{0}[\/latex], then [latex]\\left\\{{a}_{n}\\right\\}[\/latex] converges.<\/p>\n<\/section>\n

                      The proof of this theorem is beyond the scope of this text. Instead, we provide a graph to show intuitively why this theorem makes sense (Figure 6).<\/p>\n

                      <\/figcaption>
                      \"A
                      Figure 6. Since the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] is increasing and bounded above, it must converge.<\/figcaption><\/figure>\n<\/figure>\n

                      In the following example, we show how the Monotone Convergence Theorem can be used to prove convergence of a sequence.<\/p>\n

                      \n
                      \n

                      For each of the following sequences, use the Monotone Convergence Theorem to show the sequence converges and find its limit.<\/p>\n

                        \n
                      1. [latex]\\left\\{\\frac{{4}^{n}}{n\\text{!}}\\right\\}[\/latex]<\/li>\n
                      2. [latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that
                        \n<\/span><\/p>\n
                        [latex]{a}_{1}=2\\text{ and }{a}_{n+1}=\\frac{{a}_{n}}{2}+\\frac{1}{2{a}_{n}}\\text{ for all }n\\ge 2[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n