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111 - uous Exercise 66 Determine whether the sequence is...

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MATH 153 Selected Solutions for § 11.1 Exercise 10 : Find a formula for the general term a n of the following sequence, assuming that the pattern of the first few terms continues. 1 , 1 3 , 1 9 , 1 27 , 1 81 , . . . Solution : I like to write all of the symbols that appear in every term, then figure out formulas for the parts that change. Noticing that 1 = 1 1 , we start with a n = 1 . We only have the denominator left to solve; the pattern there is 1 , 3 , 9 , 27 , 81 , . . . . These are powers of 3: 3 0 = 1 , 3 1 = 3 , 3 2 = 9 , 3 3 = 27 , 3 4 = 81 , . . . . Thus, we can say that a n = 1 3 n . Exercise 26 : Determine whether the sequence converges or diverges. If it converges, find the limit. a n = ( - 1) n n 3 n 3 + 2 n 2 + 1 Solution : lim n →∞ a n = lim n →∞ ( - 1) n n 3 n 3 + 2 n 2 + 1 = lim n →∞ ( - 1) n 1 + 2 /n + 1 /n 3 = lim n →∞ ( - 1) n · 1 . This last limit does not exist, as the function oscillates between - 1 and 1. Thus, the original sequence diverges. 1
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2 Exercise 36 : Determine whether the sequence converges or diverges. If it converges, find the limit. a n = ln( n + 1) - ln n Solution : lim n →∞ a n = lim n →∞ ln( n + 1) - ln n = lim n →∞ ln n + 1 n = ln lim n →∞ n + 1 n = ln 1 = 0 . Note that we can bring the limit inside the natural log because ln is contin-
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Unformatted text preview: uous. Exercise 66 : Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? a n = n + 1 n Solution : The sequence is increasing. To see this, we need to show that a n +1 > a n for all n . Plugging into the formula gives that n + 1 + 1 n + 1 > n + 1 n . This is true if and only if 1 + 1 n + 1 > 1 n , or 1 > 1 n-1 n + 1 . But 1 n-1 n +1 = 1 n ( n +1) , so we can rewrite the last line as 1 > 1 n ( n + 1) . This is clearly true if n ≥ 1, so the original inequality a n +1 > a n must also be true. Thus, the sequence a n is increasing. The sequence is not bounded; let M be any real number, and let n be an integer greater than or equal to M . Then a n = n + 1 n > n ≥ M, so a n > M . As M was arbitrary, there cannot be any number that bounds the sequence above....
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