1 Let N max N 1 N 2 Then s n t n st s n t n s n t s n t st s n t n t t s n s M

# 1 let n max n 1 n 2 then s n t n st s n t n s n t s n

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1 Let N = max { N 1 , N 2 } . Then | s n t n - st | = | s n t n - s n t + s n t - st | ≤ | s n || t n - t | + | t || s n - s | ≤ M | t n - t | + | t || s n - s | < 2 M + 2 | t | = for all n > N . Note: if t = 0, then | s n t n - st | < / 2 < for all n > N . 5. (1) Let M be a real number. Since s n + there is a positive integer N such that s n > M for all n > N . Since t n s n for all n , t n > M for all n > M so t n diverges to + . (2) Modify the proof of (1). 6. Suppose s n L, u n L and s n t n u n for all n . Let > 0. There exists a positive integer N 1 such that | s n - L | < for all n > N 1 ; that is L - < s n < L + when n > N 1 . There exists a positive integer N 2 such that | t n - L | < for all n > N 2 ; that is L - < t n < L + when n > N 2 . Let N = max { N 1 , N 2 } . Then, for all n > N , L - < s n t n u n < L + which implies | t n - L | < . Therefore t n L . Exercises 2.3 1. (a) True. Theorem 8. (b) True. Theorem 8. (c) True. A convergent sequence is bounded, monotone or not!! (d) False. s n = ( - 1) n n is convergent ( 0) but not monotone. 2. (a) ( - 1) n n Cauchy but not monotone. (b) { 1 , 2 , 3 , 4 , · ··} monotone but not Cauchy. (c) { 1 , 0 , 1 , 0 , · ··} bounded but not Cauchy. 2 5. Bounded: by induction. Claim 0 < s n < 3 for all n . Let S be the set of positive integers for which the claim is true.  #### You've reached the end of your free preview.

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