Introduction to Mathematical Analysis I - Second Edition

148 Solutions and Hints for Selected Exercises (b) The limit is calculated as follows: lim n → ∞ 3 p n 3 + 3 n 2 − n = lim n → ∞ 3 √ n 3 + 3 n 2 − n 3 p ( n 3 + 3 n 2 ) 2 + n 3 √ n 3 + 3 n 2 + n 2 3 p ( n 3 + 3 n 2 ) 2 + n 3 √ n 3 + 3 n 2 + n 2 ) = lim n → ∞ 3 n 2 3 p ( n 3 + 3 n 2 ) 2 + n 3 √ n 3 + 3 n 2 + n 2 = lim n → ∞ 3 n 2 3 p n 6 ( 1 + 3 / n ) 2 + n 3 p n 3 ( 1 + 3 / n )+ n 2 = lim n → ∞ 3 n 2 n 2 3 p ( 1 + 3 / n ) 2 + 3 p ( 1 + 3 / n )+ 1 = lim n → ∞ 3 3 p ( 1 + 3 / n ) 2 + 3 p ( 1 + 3 / n )+ 1 = 1 . (c) We use the result in par (a) and part (b) to obtain lim n → ∞ ( 3 p n 3 + 3 n 2 − p n 2 + 1 ) = lim n → ∞ 3 p n 3 + 3 n 2 − n + n − p n 2 + 1 = lim n → ∞ 3 p n 3 + 3 n 2 − n + lim n → ∞ n − p n 2 + 1 = 1 − 1 / 2 = 1 / 2 . Using a similar technique, we can find the following limit: lim n → ∞ 3 p an 3 + bn 2 + cn + d − p α n 2 + β n + γ , where a > 0 and α > 0. SECTION 2.3 Exercise 2.3.1 . (a) Clearly, a 1 < 2. Suppose that a k < 2 for k ∈ N . Then a k + 1 = p 2 + a k < √ 2 + 2 = 2 . By induction, a n < 2 for all n ∈ N . (b) Clearly, a 1 = √ 2 < p 2 + √ 2 = a 2 . Suppose that a k < a k + 1 for k ∈ N . Then a k + 2 < a k + 1 + 2 , which implies p a k + 2 < p a k + 1 + 2 . Thus, a k + 1 < a k + 2 . By induction, a n < a n + 1 for all n ∈ N . Therefore, { a n } is an increasing sequence. (c) By the monotone convergence theorem, lim n → ∞ a n exists. Let ` = lim n → ∞ a n . Since a n + 1 = √ 2 + a n and lim n → ∞ a n + 1 = ` , we have ` = √ 2 + ` or ` 2 = 2 + `.