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6. Using only Definition 4.1.: (a) For any real number k *(b) For any real number k 4n +1 (c) lim n+3 n+3 (e) lim- n -13 7. Using any of the results in

6. Using only Definition 4.1.: (a) For any real number k *(b) For any real number k 4n +1 (c) lim n+3 n+3 (e) lim- n -13 7. Using any of the results in

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section: Chapter Questions
Problem 63RE
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#6 part c
8:50 Aa Section 4.1. Convergence 169 4.1 EXERCISES Exercises marked with * are used in later sections, and exercises marked with * have hints or solutions in the back of the book. 1. Mark each statement True or False. Justify cach answer. (a) If (s.) is a sequence and s, S, then /-. (b) Ifs, s, then for every e> 0 there exists Ne N such that n2 N implies (c) If s, k and t, → k, then s, -t, for all n e N. (d) Every convergent sequence is bounded. 2. Mark each statement True or False. Justify each answer. (a) Ifs, 0, then for every e> 0 there exists Ne N such that n2 N implies (b) If for every e>0 there exists Ne N such thatn2 N implies s, < 6, then S.0. (c) Given sequences (s,) and (a.), if for some s e R, k> 0 and m eN we have s,-sS kla. for all n> m, then lim s, =s. (d) Ifs, →s and s, 1, then s = t. 3. Write out the first seven terms of each sequence. (a) a, -n * (-1)" (b) b, = 2n+1 (c) c,- cos с, соs (d) d, 3n-1 4. Find k>0 and meN so that 6n' + 17n s kn' for all integers n2 m. 5. Find k>0 and me N so that n- 7n 2 ku for all integers n2m. 6. Using only Definition 4.1.2, prove the following. (a) For any real number k, lim, (k/n) - 0. *(b) For any real number k > 0, lim,(1/n) = 0. () lim An +1 4 n+3 (d) lim Sin -0 (d) lim n+3 (e) lim- n-13 n+2 (f) lim =0 n +n-3 7. Using any of the results in this section, prove the following. Sn -6 (a) lim =0 2+ 3n (b) lim 2n - 7n 6n +3n = 3 2n -5 (c) lim (d) lim V <> 169 Reader Contents Notebook Bookmarks Flashcards IK
Transcribed Image Text:8:50 Aa Section 4.1. Convergence 169 4.1 EXERCISES Exercises marked with * are used in later sections, and exercises marked with * have hints or solutions in the back of the book. 1. Mark each statement True or False. Justify cach answer. (a) If (s.) is a sequence and s, S, then /-. (b) Ifs, s, then for every e> 0 there exists Ne N such that n2 N implies (c) If s, k and t, → k, then s, -t, for all n e N. (d) Every convergent sequence is bounded. 2. Mark each statement True or False. Justify each answer. (a) Ifs, 0, then for every e> 0 there exists Ne N such that n2 N implies (b) If for every e>0 there exists Ne N such thatn2 N implies s, < 6, then S.0. (c) Given sequences (s,) and (a.), if for some s e R, k> 0 and m eN we have s,-sS kla. for all n> m, then lim s, =s. (d) Ifs, →s and s, 1, then s = t. 3. Write out the first seven terms of each sequence. (a) a, -n * (-1)" (b) b, = 2n+1 (c) c,- cos с, соs (d) d, 3n-1 4. Find k>0 and meN so that 6n' + 17n s kn' for all integers n2 m. 5. Find k>0 and me N so that n- 7n 2 ku for all integers n2m. 6. Using only Definition 4.1.2, prove the following. (a) For any real number k, lim, (k/n) - 0. *(b) For any real number k > 0, lim,(1/n) = 0. () lim An +1 4 n+3 (d) lim Sin -0 (d) lim n+3 (e) lim- n-13 n+2 (f) lim =0 n +n-3 7. Using any of the results in this section, prove the following. Sn -6 (a) lim =0 2+ 3n (b) lim 2n - 7n 6n +3n = 3 2n -5 (c) lim (d) lim V <> 169 Reader Contents Notebook Bookmarks Flashcards IK
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