Solution suppose a n converges to l let b n a n 1 and

This preview shows page 10 - 12 out of 95 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
A First Course in Differential Equations with Modeling Applications
The document you are viewing contains questions related to this textbook.
Chapter 6 / Exercise 3
A First Course in Differential Equations with Modeling Applications
Zill
Expert Verified
solution Suppose { a n } converges to L . Let b n = a n + 1 , and let > 0. Because { a n } converges to L , there exists an M such that | a n L | < for n > M . Now, let M = M 1. Then, whenever n > M , n + 1 > M + 1 = M . Thus, for n > M , | b n L | = | a n + 1 L | < . Hence, { b n } converges to L . Let { a n } be a sequence such that lim n →∞ | a n | exists and is nonzero. Show that lim n →∞ a n exists if and only if there exists an integer M such that the sign of a n does not change for n > M . 83. Proceed as in Example 12 to show that the sequence 3, 3 3, 3 3 3 , . . . is increasing and bounded above by M = 3. Then prove that the limit exists and find its value. solution This sequence is defined recursively by the formula: a n + 1 = 3 a n , a 1 = 3 . Consider the following inequalities: a 2 = 3 a 1 = 3 3 > 3 = a 1 a 2 > a 1 ; a 3 = 3 a 2 > 3 a 1 = a 2 a 3 > a 2 ; a 4 = 3 a 3 > 3 a 2 = a 3 a 4 > a 3 . In general, if we assume that a k > a k 1 , then a k + 1 = 3 a k > 3 a k 1 = a k . Hence, by mathematical induction, a n + 1 > a n for all n ; that is, the sequence { a n } is increasing. Because a n + 1 = 3 a n , it follows that a n 0 for all n . Now, a 1 = 3 < 3. If a k 3, then a k + 1 = 3 a k 3 · 3 = 3 . Thus, by mathematical induction, a n 3 for all n . Since { a n } is increasing and bounded, it follows by the Theorem on Bounded Monotonic Sequences that this sequence is converging. Denote the limit by L = lim n →∞ a n . Using Exercise 81, it follows that L = lim n →∞ a n + 1 = lim n →∞ 3 a n = 3 lim n →∞ a n = 3 L. Thus, L 2 = 3 L , so L = 0 or L = 3. Because the sequence is increasing, we have a n a 1 = 3 for all n . Hence, the limit also satisfies L 3. We conclude that the appropriate solution is L = 3; that is, lim n →∞ a n = 3. Let { a n } be the sequence defined recursively by a 0 = 0 , a n + 1 = 2 + a n Thus, a 1 = 2 , a 2 = 2 + 2 , a 3 = 2 + 2 + 2 , . . . . (a) Show that if a n < 2, then a n + 1 < 2. Conclude by induction that a n < 2 for all n . (b) Show that if a n < 2, then a n a n + 1 . Conclude by induction that { a n } is increasing. (c) Use (a) and (b) to conclude that L = lim n →∞ a n exists. Then compute L by showing that L = 2 + L . Further Insights and Challenges 85. Show that lim n →∞ n n ! = ∞ . Hint: Verify that n ! ≥ (n/ 2 ) n/ 2 by observing that half of the factors of n ! are greater than or equal to n/ 2. solution We show that n ! ≥ ( n 2 ) n/ 2 . For n 4 even, we have: n ! = 1 · · · · · n 2 n 2 factors · n 2 + 1 · · · · · n n 2 factors n 2 + 1 · · · · · n n 2 factors . Since each one of the n 2 factors is greater than n 2 , we have: n ! ≥ n 2 + 1 · · · · · n n 2 factors n 2 · · · · · n 2 n 2 factors = n 2 n/ 2 . For n 3 odd, we have: n ! = 1 · · · · · n 1 2 n 1 2 factors · n + 1 2 · · · · · n n + 1 2 factors n + 1 2 · · · · · n n + 1 2 factors .
We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
A First Course in Differential Equations with Modeling Applications
The document you are viewing contains questions related to this textbook.
Chapter 6 / Exercise 3
A First Course in Differential Equations with Modeling Applications
Zill
Expert Verified
May 23, 2011 S E C T I O N 10.1 Sequences 643 Since each one of the n + 1 2 factors is greater than n 2 , we have: n ! ≥ n + 1 2 · · · · · n n + 1 2 factors n 2 · · · · · n 2 n + 1 2 factors = n 2 (n + 1 )/ 2 = n 2 n/ 2 n 2 n 2 n/ 2 .

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture