Math 3333 Homework 5 Solutions Name: Peoplesoft ID: Show your work. If a problem requires a proof, explain and justify your steps carefully. Homework papers should be legible and neat, and the pages should be stapled together in the correct order. Illegible work may not be graded. Homework should be submitted in class on the indicated due date. Submissions by email or to the math office will not be accepted. 1. TRUE/FALSE: If the statement is true, prove it. If the statement is false, give a counter-example (a) If s n → 0, then given any positive number there corresponds a positive integer N such that s n < for all n > N . True: s n → 0 means that for any > 0, there is a positive integers N , such that, for all n > N , | s n - 0 | = | s n | < ⇒ - < s n < . In particular, s n < for all n > N .

(b) If for each positive number there is a positive integer N such that s n < for all n > N , then s n → 0. False: Let s n = - n. Note that s n 6→ 0. ( s n diverges to -∞ .) However, for each > 0, we have s n = - n < 0 < for all n > N = 1. (c) If s n → s and s n > 0 for all n , then s > 0. False: Let s n = 1 n . Then, s n > 0 for all n , but s n → s = 0.

(d) If ( s n ) and ( t n ) are divergent sequences, then ( s n + t n ) is divergent. False: Let s n = ( - 1) n and t n = ( - 1) n +1 . That is, ( s n ) = ( - 1 , 1 , - 1 , 1 , . . . ) ( t n ) = (1 , - 1 , 1 , - 1 , . . . ) Then, s n + t n = 0 for all n . Therefore, s n diverges, t n diverges, and s n + t n → 0. (e) If ( s n ) and ( s n + t n ) are convergent sequences, then ( t n ) is convergent. True: Suppose s n → s and s n + t n → r . Then, by Theorem 1.2, Section 17, t n = [ s n + t n ] - s n → r - s .