Introduction to Mathematical Analysis I - Second Edition

42 2.3 MONOTONE SEQUENCES Then { b n } is increasing and bounded above (if M is a lower bound for { a n } , then − M is an upper bound for { b n } ). Let ` = lim n → ∞ b n = lim n → ∞ ( − a n ) . Then { a n } converges to − ` by Theorem 2.2.1 . Remark 2.3.2 It follows from the proof of Theorem 2.3.1 that if { a n } is increasing and bounded above, then lim n → ∞ a n = sup { a n : n ∈ N } . Similarly, if { a n } is decreasing and bounded below, then lim n → ∞ a n = inf { a n : n ∈ N } . Example 2.3.1 Given r ∈ R with | r | < 1, define a n = r n for n ∈ N . Then lim n → ∞ a n = 0 . This is clear if r = 0. Let us first consider the case where 0 < r < 1. Then 0 ≤ a n + 1 = ra n ≤ a n for all n . Therefore, { a n } is decreasing and bounded below. By Theorem 2.3.1 , the sequence converges. Let ` = lim n → ∞ a n . Since a n + 1 = ra n for all n , taking limits on both sides gives ` = r ` . Thus, ( 1 − r ) ` = 0 and, hence, ` = 0. In the general case, we only need to consider the sequence defined by b n = | a n | for n ∈ N ; see Exercise 2.1.3 . Example 2.3.2 Consider the sequence { a n } defined as follows: a 1 = 2 (2.6) a n + 1 = a n + 5 3 for n ≥ 1 . (2.7) First we will show that the sequence is increasing. We prove by induction that for all n ∈ N , a n < a n + 1 . Since a 2 = a 1 + 5 3 = 7 3 > 2 = a 1 , the statement is true for n = 1. Next, suppose a k < a k + 1 for some k ∈ N . Then a k + 5 < a k + 1 + 5 and ( a k + 5 ) / 3 < ( a k + 1 + 5 ) / 3. Therefore, a k + 1 = a k + 5 3 < a k + 1 + 5 3 = a k + 2 . It follows by induction that the sequence is increasing. Next we prove that the sequence is bounded by 3. Again, we proceed by induction. The statement is clearly true for n = 1. Suppose that a k ≤ 3 for some k ∈ N . Then a k + 1 = a k + 5 3 ≤ 3 + 5 3 = 8 3 ≤ 3 . It follows that a n ≤ 3 for all n ∈ N .