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Hw3 - due Oct 28, 10am A. Using the definition of limit, show that limn→∞ 1/n5 = 0 and that lim→∞ 1/n¹/5 = 0.

Hw3 - due Oct 28, 10am A. Using the definition of limit, show that limn→∞ 1/n5 = 0 and that lim→∞ 1/n¹/5 = 0.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.5: The Binomial Theorem
Problem 21E
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20:27 < hw3 ...I LTE Hw3 - due Oct 28, 10am A. Using the definition of limit, show that limn→∞ 1/ n5 = 0 and that lim→∞ 1/ n¹/5 = 0. B. Using the definition of limit (so, without using Arithmetic of Limits), show that i. limɲ→∞ (4 + n) | 2n = 1/2 ii. limn→∞ 2/n + 3/(n+1) = 0 C. Suppose that (sn) and (t) are sequences so that sn= tn except for finitely many values of n. Using the definition of limit, explain why if limn → ∞ Sn = s, then also limn → ∞ tn = s. D. Suppose real sequences (sn) and (tn) are bounded. (That is, that their ranges are bounded sets.) i. Show the sequence given by (Sn + tn) is bounded. ii. For any real number a, show that the sequence (a sn) is bounded. E. i. For a convergent real sequence s, and a real number a, show that if sn ≥ a for all but finitely many values of n, then limn→∞ Sn ≥a. ii. For each value of a € R, give an example of a convergent sequence sn with sn > a for all n, but where limn→∞ Sn = a. Add submission
Transcribed Image Text:20:27 < hw3 ...I LTE Hw3 - due Oct 28, 10am A. Using the definition of limit, show that limn→∞ 1/ n5 = 0 and that lim→∞ 1/ n¹/5 = 0. B. Using the definition of limit (so, without using Arithmetic of Limits), show that i. limɲ→∞ (4 + n) | 2n = 1/2 ii. limn→∞ 2/n + 3/(n+1) = 0 C. Suppose that (sn) and (t) are sequences so that sn= tn except for finitely many values of n. Using the definition of limit, explain why if limn → ∞ Sn = s, then also limn → ∞ tn = s. D. Suppose real sequences (sn) and (tn) are bounded. (That is, that their ranges are bounded sets.) i. Show the sequence given by (Sn + tn) is bounded. ii. For any real number a, show that the sequence (a sn) is bounded. E. i. For a convergent real sequence s, and a real number a, show that if sn ≥ a for all but finitely many values of n, then limn→∞ Sn ≥a. ii. For each value of a € R, give an example of a convergent sequence sn with sn > a for all n, but where limn→∞ Sn = a. Add submission
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