Ans-11 - Answers Assignment#11 Exercises 2.2 1(a False sn = 0(c False The example in(b will work here cfw_sn tn = cfw_1 1 1 1 1 > 0 for all n sn 0 n(b

# Ans-11 - Answers Assignment#11 Exercises 2.2 1(a False sn =...

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Answers, Assignment #11 Exercises 2.2 1. (a) False. s n = 1 n > 0 for all n ; s n 0. (b) False. { s n } = { 1 , - 1 , 1 , - 1 , · ··} ; { t n } = {- 1 , 1 , - 1 , 1 , ·· ·} ; { s n + t n } = { 0 , 0 , 0 , . .. } → 0. (c) False. The example in (b) will work here; { s n t n } = {- 1 , - 1 , - 1 , . . . } → - 1. (d) True. Suppose { s n } → A and { s n + t n } → B . Then t n = ( s n + t n ) - s n B - A . (e) False. Let s n = 1 n , t n = n . Then { s n t n } = { 1 , 1 , 1 , . .. } → 1, but { t n } diverges. (f) False. Let s n = ( - 1) n n = {- 1 , 2 , - 3 , 4 , - 5 ,. .. } . { s n } is not bounded above, { s n } does not diverge to + . 2. (a) s n → - 2 (b) s n 0 (c) { s n } diverges; for large n, s n ≈ ± 1 2 (d) s n = 2 3 n 3 2 n = 8 9 n 0 (e) s n diverges to + (f) s n diverges to + 3. (a) n 2 + 1 - n = n 2 + 1 - n n 2 + 1 + n n 2 + 1 + n = 1 n 2 + 1 + n 0. (b) n 2 + n - n = n 2 + n - n n 2 + n + n n 2 + n + n = n n 2 + n + n 1 2 . 4. (1) Suppose s n s, t n t . Let > 0. Then there exists a positive integer N 1 such that | s n - s | < 2 for all n > N 1 . There exists a positive integer N 2 such that | t n - t | < 2 for all n > N 2 . Let N = max { N 1 , N 2 } . Then | ( s n + t n ) - ( s + t ) | = | ( s n - s ) + ( t n - t ) | ≤ | s n - s | + | t n - t | < 2 + 2 = for all n > N . (3) Let > 0. Since s n s , { s n } is bounded, say | s n | ≤ M for all n . There exists a positive integer N 1 such that | s n - s | < 2 | t | for all n > N 1 (assuming t = 0) There exists a positive integer N 2 such that | t n - t | < 2 M for all n > N 2 .  #### You've reached the end of your free preview.

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• Fall '08
• Staff
• Tn, Natural number, Limit of a sequence, Sn
• • • 