Introduction to Mathematical Analysis I - Second Edition

150 Solutions and Hints for Selected Exercises b n + 1 = a n + b n 2 ≤ b n + b n 2 = b n for all n ∈ N . It follows that { a n } is monotone increasing and bounded above by b 1 , and { b n } is decreasing and bounded below by a 1 . Let x = lim n → ∞ a n and y = lim n → ∞ b n . Then x = √ xy and y = x + y 2 . Therefore, x = y . SECTION 2.4 Exercise 2.4.1 . Here we use the fact that in R a sequence is a Cauchy sequence if and only if it is convergent. (a) Not a Cauchy sequence. See Example 2.1.7 . (b) A Cauchy sequence. This sequence converges to 0. (c) A Cauchy sequence. This sequence converges to 1. (d) A Cauchy sequence. This sequence converges to 0 (see Exercise 2.1.5 ) . SECTION 2.5 Exercise 2.5.4 . (a) Define α n = sup k ≥ n ( a n + b n ) , β n = sup k ≥ n a k , γ n = sup k ≥ n b k . By the definition, limsup n → ∞ ( a n + b n ) = lim n → ∞ α n , limsup n → ∞ a n = lim n → ∞ β n , limsup n → ∞ b n = lim n → ∞ γ n . By Exercise 2.5.3 , α n ≤ β n + γ n for all n ∈ N . This implies lim n → ∞ α n ≤ lim n → ∞ β n + lim n → ∞ γ n for all n ∈ N . Therefore, limsup n → ∞ ( a n + b n ) ≤ limsup n → ∞ a n + limsup n → ∞ b n . This conclusion remains valid for unbounded sequences provided that the right-hand side is well- defined. Note that the right-hand side is not well-defined, for example, when limsup n → ∞ a n = ∞ and limsup n → ∞ b n = − ∞ . (b) Define α n = inf k ≥ n ( a n + b n ) , β n = inf k ≥ n a k , γ n = inf k ≥ n b k . Proceed as in part (a), but use part (b) of Exercise 2.5.3 . (c) Consider a n = ( − 1 ) n and b n = ( − 1 ) n + 1 .