Introduction to Mathematical Analysis I - Second Edition

44 2.3 MONOTONE SEQUENCES Monotone Convergence Theorem (Theorem 2.3.1 ) , there exist a , b ∈ R such that lim n → ∞ a n = a and lim n → ∞ b n = b . Since a n ≤ b n for all n , by Theorem 2.1.5 , we get a ≤ b . Now, we also have a n ≤ a and b ≤ b n for all n ∈ N (since { a n } is increasing and { b n } is decreasing). This shows that if a ≤ x ≤ b , then x ∈ I n for all n ∈ N . Thus, [ a , b ] ⊂ T ∞ n = 1 I n . It follows that T ∞ n = 1 I n 6 = /0. This proves part (a) . Now note also that T ∞ n = 1 I n ⊂ [ a , b ] . Indeed, if x ∈ T ∞ n = 1 I n , then x ∈ I n for all n . Therefore, a n ≤ x ≤ b n for all n . Using Theorem 2.1.5 , we conclude a ≤ x ≤ b . Thus, x ∈ [ a , b ] . This proves the desired inclusion and, hence, T ∞ n = 1 I n = [ a , b ] . We now prove part (b) . Suppose the lengths of the intervals I n converge to zero. This means b n − a n → 0 as n → ∞ . Then b = lim n → ∞ b n = lim n → ∞ [( b n − a n )+ a n ] = a . It follows that T ∞ n = 1 I n = { a } as desired. When a monotone sequence is not bounded, it does not converge. However, the behavior follows a clear pattern. To make this precise we provide the following definition. Definition 2.3.2 A sequence { a n } is said to diverge to ∞ if for every M ∈ R , there exists N ∈ N such that a n > M for all n ≥ N . In this case, we write lim n → ∞ a n = ∞ . Similarly, we say that { a n } diverges to − ∞ and write lim n → ∞ a n = − ∞ if for every M ∈ R , there exists N ∈ N such that a n < M for all n ≥ N . Remark 2.3.4 We should not confuse a sequence that diverges to ∞ (that is, one that satisfies the previous definition), with a divergent sequence (that is, one that does not converge). Example 2.3.4 Consider the sequence a n = n 2 + 1 5 n . We will show, using Definition 2.3.2 , that lim n → ∞ a n = ∞ . Let M ∈ R . Note that n 2 + 1 5 n = n 5 + 1 5 n ≥ n 5 . Choose N > 5 M . Then, if n ≥ N , we have a n ≥ n 5 ≥ N 5 > M . Th following result completes the description of the behavior of monotone sequences. Theorem 2.3.5 If a sequence { a n } is increasing and not bounded above, then lim n → ∞ a n = ∞ . Similarly, if { a n } is decreasing and not bounded below, then lim n → ∞ a n = − ∞ . Proof: Fix any real number M . Since { a n } is not bounded above, there exists N ∈ N such that a N ≥ M . Then a n ≥ a N ≥ M for all n ≥ N because { a n } is increasing. Therefore, lim n → ∞ a n = ∞ . The proof for the second case is similar.