Hw_3_Solutions

# Hw_3_Solutions - MATH 413 HW 3 Solution to selected...

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MATH 413 HW 3: Solution to selected problems. September 19, 2007 3.1.3.5. Prove that lim sup n →∞ ( x n + y n ) lim sup n →∞ x n +lim sup n →∞ y n if both limsups are finite, and give an example where equality does not hold. Proof: Let l > k , where k is a number large enough such that sup j>k x j and sup j>k y j are finite (the existence of this k is guaranteed by hypoth- esis and Definition 3.1.2), clearly x l + y l sup j>k x j + sup j>k y j (where the sum is well defined because the sups are finite). Since the supremum of a set is the least upper bound and l > k was arbitrary, we conclude that sup j>k ( x j + y j ) sup j>k x j + sup j>k y j for k large enough. By taking limit when k goes to infinity in the last equation and by Definition 3.1.2 we obtain the desired inequality. For an example where we have strict inequality, define x n = ( - 1) n and y n = ( - 1) n +1 , then lim sup n →∞ x n = lim sup n →∞ y n = 1, but since x n + y n = 0 for all n , lim sup n →∞ ( x n + y n ) = 0 < 2 = lim sup n →∞ x n + lim sup n →∞ y n . 3.1.3.9. Can there exist a sequence whose set of limit points is exactly 1 , 1 2 , 1 3 , . . . ? Solution: The answer is no. Suppose n x n is such a sequence. Then by Theorem 3.1.4 (which is also true with liminf and inf replaced by limsup and sup, respectively) inf n N { 1 n } = 0 would be a limit-point of n x n ,

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