Introduction to Mathematical Analysis I - Second Edition

149 Solving this quadratic equation yields ` = − 1 or ` = 2. Therefore, lim n → ∞ a n = 2. Define a more general sequence as follows: a 1 = c > 0 , a n + 1 = √ c + a n for n ∈ N . We can prove that { a n } is monotone increasing and bounded above by 1 + √ 1 + 4 c 2 . In fact, { a n } converges to this limit. The number 1 + √ 1 + 4 c 2 is obtained by solving the equation ` = √ c + ` , where ` > 0. Exercise 2.3.2 . (a) The limit is 3. (b) The limit is 3. (c) We use the well-known inequality a + b + c 3 ≥ 3 √ abc for a , b , c ≥ 0 . By induction, we see that a n > 0 for all n ∈ N . Moreover, a n + 1 = 1 3 ( 2 a n + 1 a 2 n ) = 1 3 ( a n + a n + 1 a 2 n ) ≥ 1 3 3 s a n · a n · 1 a 2 n = 1 . We also have, for n ≥ 2, a n + 1 − a n = 1 3 2 a n + 1 a 2 n − a n = − a 3 n + 1 3 a 2 n = − ( a n − 1 )( a 2 n + a n + 1 ) 3 a 2 n < 0 . Thus, { a n } is monotone deceasing (for n ≥ 2) and bounded below. We can show that lim n → ∞ a n = 1. (d) Use the inequality a + b 2 ≥ √ ab for a , b ≥ 0 to show that a n + 1 ≥ √ b for all n ∈ N . Then follow part 3 to show that { a n } is monotone decreasing. The limit is √ b . Exercise 2.3.3 . (a) Let { a n } be the given sequence. Observe that a n + 1 = √ 2 a n . Then show that { a n } is monotone increasing and bounded above. The limit is 2. (b) Let { a n } be the given sequence. Then a n + 1 = 1 2 + a n . Show that { a 2 n + 1 } is monotone decreasing and bounded below; { a 2 n } is monotone increasing and bounded above. Thus, { a n } converges by Exercise 2.1.12 . The limit is √ 2 − 1 . Exercise 2.3.5 . Observe that b n + 1 = a n + b n 2 ≥ p a n b n = a n + 1 for all n ∈ N . Thus, a n + 1 = p a n b n ≥ √ a n a n = a n for all n ∈ N ,