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chap10-et-instructor-solutions

# chap10-et-instructor-solutions - 10 INFINITE SERIES 10.1...

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10 INFINITE SERIES 10.1 Sequences Preliminary Questions 1. What is a 4 for the sequence a n = n 2 n ? SOLUTION Substituting n = 4 in the expression for a n gives a 4 = 4 2 4 = 12 . 2. Which of the following sequences converge to zero? (a) n 2 n 2 + 1 (b) 2 n (c) 1 2 n SOLUTION (a) This sequence does not converge to zero: lim n →∞ n 2 n 2 + 1 = lim x →∞ x 2 x 2 + 1 = lim x →∞ 1 1 + 1 x 2 = 1 1 + 0 = 1 . (b) This sequence does not converge to zero: this is a geometric sequence with r = 2 > 1; hence, the sequence diverges to . (c) Recall that if | a n | converges to 0, then a n must also converge to zero. Here, 1 2 n = 1 2 n , which is a geometric sequence with 0 < r < 1; hence, ( 1 2 ) n converges to zero. It therefore follows that ( 1 2 ) n converges to zero. 3. Let a n be the n th decimal approximation to 2. That is, a 1 = 1, a 2 = 1 . 4, a 3 = 1 . 41, etc. What is lim n →∞ a n ? SOLUTION lim n →∞ a n = 2. 4. Which sequence is defined recursively? (a) a n = 2 + n 1 (b) b n = 4 + b n 1 SOLUTION (a) a n can be computed directly, since it depends on n only and not on preceding terms. Therefore a n is defined explicitly and not recursively. (b) b n is computed in terms of the preceding term b n 1 , hence the sequence { b n } is defined recursively. 5. Theorem 5 says that every convergent sequence is bounded. Which of the following statements follow from Theorem 5 and which are false? If false, give a counterexample. (a) If { a n } is bounded, then it converges. (b) If { a n } is not bounded, then it diverges. (c) If { a n } diverges, then it is not bounded. SOLUTION (a) This statement is false. The sequence a n = cos π n is bounded since 1 cos π n 1 for all n , but it does not converge: since a n = cos n π = ( 1 ) n , the terms assume the two values 1 and 1 alternately, hence they do not approach one value. (b) By Theorem 5, a converging sequence must be bounded. Therefore, if a sequence is not bounded, it certainly does not converge. (c) The statement is false. The sequence a n = ( 1 ) n is bounded, but it does not approach one limit.

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S E C T I O N 10.1 Sequences 1133 Exercises 1. Match the sequence with the general term: a 1 , a 2 , a 3 , a 4 , . . . General term (a) 1 2 , 2 3 , 3 4 , 4 5 , . . . (i) cos π n (b) 1 , 1 , 1 , 1 , . . . (ii) n ! 2 n (c) 1 , 1 , 1 , 1 , . . . (iii) ( 1 ) n + 1 (d) 1 2 , 2 4 , 6 8 , 24 16 . . . (iv) n n + 1 SOLUTION (a) The numerator of each term is the same as the index of the term, and the denominator is one more than the numerator; hence a n = n n + 1 , n = 1 , 2 , 3 , . . . . (b) The terms of this sequence are alternating between 1 and 1 so that the positive terms are in the even places. Since cos π n = 1 for even n and cos π n = − 1 for odd n , we have a n = cos π n , n = 1 , 2 , . . . . (c) The terms a n are 1 for odd n and 1 for even n . Hence, a n = ( 1 ) n + 1 , n = 1 , 2 , . . . (d) The numerator of each term is n ! , and the denominator is 2 n ; hence, a n = n ! 2 n , n = 1 , 2 , 3 , . . . . 2. Let a n = 1 2 n 1 for n = 1 , 2 , 3 , . . . . Write out the first three terms of the following sequences.
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