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m25-hw7-solutions - Math 25 Advanced Calculus Fall 2010...

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Math 25: Advanced Calculus Fall 2010 Homework # 7 – Solutions 7.1: Prove that lim n →∞ ( n + 1 - n ) = 0. [Hint: What would you have done in calculus?] Solution: Multiplying by the conjugate of n + 1 - n gives lim n →∞ n + 1 - n = lim n →∞ ( n + 1 - n )( n + 1 + n ) n + 1 + n = 1 n + 1 + n . Now we use the squeeze theorem. Since n + 1+ n > n , it follows that 1 n +1+ n < 1 n . Thus 0 < 1 n + 1 + n < 1 n for all n N . We know that lim n →∞ 1 n = 0, and hence by the squeeze theorem, lim n →∞ n + 1 - n = 0 , as well. 1
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7.2: Decide whether or not each of the following statements is true or false. In either case, you must justify your answer with a proof. For all of these scenarios, assume that { a n } and { b n } are sequences of real numbers. (a). If { a n } is unbounded, then {| a n |} diverges to infinity. FALSE: Let { a n } be the sequence defined by a n = ( 0 if n is odd , n 2 if n is even . This is the sequence: 0 , 1 , 0 , 2 , 0 , 3 , 0 , 4 , 0 , 5 , 0 , 6 , . . . . For any M > 0, there is a natural number n > M by the Archimedean principle, and hence a 2 n > M . Thus { a n } is unbounded. The sequence does not diverge to infinity because for any n , there is a term a k = 0 for some k > n .
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