TEST 1 SOLUTIONS MATH 1002

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1 October 17, TEST 1 SOLUTIONS MATH Indicate whether each it below exists or does not exist. If the it exists then write what it is. No proofs are required. For example, 1 n exists and is equal to 0. (a) (b) (c) 3n exists and is equal to 73 + ( 1)n does not exist 21/n exists and is equal to 1 (Theorem 9.7 (d)) 2. Indicate whether each sequence below is monotone or not. If it is monotone then indicate whether it is increasing or decreasing. No proofs are required. For example, ( 1 n ) (a) (3 n ) n=1 monotone and increasing (b) (73 + ( 1) n ) n=1 not monotone is monotone and decreasing n=1 (c) (2 1/n ) n=1 monotone and decreasing 3. Suppose (a k ) k=1 is a sequence. Provide the definition for the infinite series k=1 a k to converge. Solution: k=1 a k converges if converges. n a k k=1

2 October 17, Suppose (s n ) is a sequence and a R. Provide the definition of s n = a. /10 (Hint: For all ɛ... ) Solution: For all ɛ > 0 there exists M N such that n > M implies s n a < ɛ. 5. Using only the definition above, prove that 3n + 2 = 4 3. Show your scratch work and indicate your logic with, or. Solution: Scratch: If n N then 3n < ɛ 3() 4(3n + 2) 3(3n + 2) < ɛ 11 9n + 6 < ɛ 11 < ɛ since 9, 6 and n are positive 9n + 6 9n + 6 > 1 11 ɛ n > 1 9 (11 ɛ 6) Formal proof: Suppose ɛ > 0. Choose M N such that M > 1 9 ( 11 ɛ 6). If n > M then n > 1 9 ( 11 ɛ 6) so that 9n > 1 ɛ 11 9n + 6 < ɛ 11 9n + 6 < ɛ since 9, 6 and n are positive 3() 4(3n + 2) 3(3n + 2) < ɛ 3n < ɛ Grading: 3 for correct style in formal proof. 3 for correct choice of M. 4 for logically arguing the final inequality.

3 October 17, /6 6. Using only theorems from the text, prove that n n + 1 =. No theorem numbers are required, but try to make clear which theorems 1 you are using and why you may use them. You may also use n = 0 without proof. Solution: n n + 1 = n2 (1 + 6/n 2 ( ) (1 + 6/n 2 ) ) = n n(1 + 1/n) 1 + 1/n By Thm /n = 1+ 1/n = 1+0 (as both its converge.) By Thm 9.4 1/n 2 = ( 1/n)( 1/n) = (0)(0) (again both its converge). By Thms 9.3 and /n 2 = /n 2 = 1 + 6(0) = 1. By Thm 9.6 (1+6/n2 ) 1+1/n 1 0). = 1/1 = 1 (it in denominator converges to By Thm 9.10 n = 1 1/n =, since n > 0 and 1/n = 0. Finally, by Thm 9.9 n2 + 6 n + 1 = n ( (1 + 6/n 2 ) ) = 1 + 1/n since the (1+6/n2 ) 1+1/n = 1 > 0. Grading: 1 for putting the quotient into the right form. 2 for giving it of numerator or denominator. 1 for giving it of (1+6/n2 ) 1+1/n. 2 for final it. /6 7. Let s n = cos(nπ) for all n N. Define a subsequence (t k ) of (s n ) which is a constant sequence. Justify that your definition is a constant subsequence. (Hint: Replace the question mark in t k = cos(?π) with the right thing.) Solution: Let t k = cos(2kπ) = 1 for all k N. In this way, t k = s 2k for all k N. Moreover, 2k < 2k + 2 = 2(k + 1) for all k N. Clearly t k is the constant sequence (1) k=1. Grading: 2 for definition of t k. 1 for identifying constant. 3 for proof of strictly increasing indices.

4 October 17, Suppose a is a positive real number and n N. /4 (a) Prove that a a n a using rules about inequalities. You don t have to label the rules you are using, but try to make clear what they are. Solution: 0 < n 0 < 1/n 1 n < n < 1 a(1 1 ) < a since a is positive n a a n < a a a n a (b) Provide an upper bound for the sequence ( 1 n) 1 and explain n=1 why it is an upper bound. Solution: Take a = 1 in part (a) to obtain ( 1 n) 1 1 for all n N. This means that 1 is an upper bound by definition. 9. (Not easy! Do last.) Suppose (s n ) is a bounded sequence and for all ɛ > 0 there exists M N such that if n, m > M then s n s m < ɛ. Prove that (s n ) converges by using a subsequence and the definition of convergence. Solution: By the Bolzano-Weierstrass Theorem there is a subsequence (t k ) and a R such that t k = a. Suppose ɛ > 0. There exists M 1 N such that k > M 1 implies t k a < ɛ. By hypothesis, there is also an M 2 N so that n, m > M 2 implies s n s m < ɛ. Now, t k = s nk for strictly increasing n k < n k+1. Therefore there exists k > M 1 such that n k > M 2. If n > M 2 then s n a = s n t k + t k a s n s nk + t k a < ɛ + ɛ = ɛ. Grading: 1 for use of Bolzano-Weierstrass Theorem. 2 for everything else.

5 October 17, Indicate whether each of the following statements is true or false. Provide some justification for each statement. (a) Suppose S is a bounded nonempty subset of R and a = sup S. Then there exists b < a which is an upper bound for S. Solution: False. If c is an upper bound then a c (part of definition of sup). The contrapositive of this true statement is that if c < a then c is not an upper bound. Since the contrapositive is true then b < a implies that it is never an upper bound. (b) Suppose (s n ) n=1 is a sequence, a > 0, and s n = a. Then s n 0 for all n N. Solution: False. The sequence ( 1, 1, 1, 1, 1,...) converges to 1 > 0 but its first entry is 1 < 0 Total /55