prelim2_311_sol - Mathematics 311 Instructor: 1. Let an =...

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Mathematics 311 Solutions to Preliminary Exam 2 Introduction to Analysis Instructor: Marius Ionescu 1. Let a n = n . (a) (5pts) Prove that for every ε > 0 there is N 1 such that | a n +1 - a n | < ε if n N . Solution: Let ε > 0. Then | a n +1 - a n | < ε is equivalent with | n + 1 - n | < ε . Simplifying, this inequality is equivalent with 1 n + 1 + n < ε. Note that the previous expression is positive so we do not need the absolute value bars. Since 1 n + 1 + n < 1 2 n if we choose N = (1 / (2 ε )) 2 , it follows that | a n +1 - a n | < ε if n N . (b) (5pts) Is a n a Cauchy sequence? Why? Solution: a n is not a Cauchy sequence. To prove this, suppose by contradiction that a n is Cauchy. Then a n must be convergent by Theorem 6.4. The sequence a n is, however, divergent. Thus it can not be Cauchy.
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Mathematics 311 Solutions to Preliminary Exam 2 Introduction to Analysis 2. (a) (7pts) Show that if a n converges absolutely, then so does a 2 n . Is this true without the hypothesis of absolute convergence (prove or give a counterexample)? Solution: If the series a n converges absolutely it converges. Then lim n →∞ a n = 0. Then there is N 1 such that | a n | < 1 for all n N . It follows that a 2 n ≤ | a n | for all n N . The comparison test for positive series implies that n = N a 2 n converges. From the tail theorem we conclude that a 2 n converges. The conclusion fails without the hypothesis of absolute convergence. Consider ( - 1) n n . This series is convergent by the Cauchy test. The series 1 n , however, diverges (it is a
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prelim2_311_sol - Mathematics 311 Instructor: 1. Let an =...

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