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Unformatted text preview: PRACTICE PROBLEMS FOR SECOND MATH 3100 MIDTERM 1) Let { a n } n =1 be a sequence of real numbers. a) Say what it means for the infinite series n =1 a n to converge. Solution: For n N , put S n = a 1 + ... + a n . Then the infinite series converges if the sequence of partial sums S n converges, and the limit S of the sequence of partial sums is called the sum of the series. b) Say what it means for n =1 a n to absolutely converge. Solution: This means that the series n =1  a n  converges. c) Say what it means for n =1 a n to nonabsolutely converge. Solution: This means that the series n =1 a n converges but is not absolutely con vergent. (In other words, it is a synonym for conditional convergence .) 2) Let { S n } n =1 be any real sequence. Show that there exists a real sequence { a n } such that for all n N , S n = a 1 + ... + a n . Solution: Put a 1 = S 1 , and for all n > 1 put a n = S n S n 1 . Then a 1 + ... + a n = S 1 + ( S 2 S 1 ) + ... + ( S n S n 1 ) = S n . 3) Let { a n } n =1 and { b n } n =1 be two positive sequences such that lim n a n b n = . a) Show that if n =1 a n converges, then n =1 b n converges. Solution: Since the ratio a n b n approaches infinity, there exists N N such that for all n N , a n b n 1, i.e., a n b n . Thus n = N b n n = N a n < , so the series n b n converges by comparison. (As usual, if we are interested only in the convergence / absolute convergence / divergence of a series and not its sum, we may feel free to neglect any finite number of terms.) b) show that if n =1 b n diverges, then n =1 a n diverges. Solution: This is just the contrapositive of part a). Thus, since part a) holds, so does part b)....
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 Spring '08
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 Real Numbers

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