If a n then s n Proof Note first that a n for all n N Since a n 0 there exists

If a n then s n proof note first that a n for all n n

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If a n 0 , then s n 0 . Proof: Note first that a n 0 for all n > N . Since a n 0, there exists a positive integer N 1 such that | a n | < /k . Without loss of generality, assume that N 1 N . Then, for all n > N 1 , | s n - 0 | = | s n | ≤ k a n < k k = . 14
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Therefore, s n 0. Corollary Let { s n } and { a n } be sequences and let s R . Suppose that there is a positive number k and a positive integer N such that | s n - s | ≤ k a n for all n > N. If a n 0, then s n s . Exercises 2.1 1. True – False. Justify your answer by citing a theorem, giving a proof, or giving a counter- example. (a) If s n s , then s n +1 s . (b) If s n s and t n s , then there is a positive integer N such that s n = t n for all n > N . (c) Every bounded sequence converges (d) If to each > 0 there is a positive integer N such that n > N implies s n < , then s n 0. (e) If s n s , then s is an accumulation point of the set S = { s 1 , s 2 , · · ·} . 2. Prove that lim 3 n + 1 n + 2 = 3. 3. Prove that lim sin n n = 0. 4. Prove or give a counterexample: (a) If { s n } converges, then {| s n |} converges. (b) If {| s n |} converges, then { s n } converges. 5. Give an example of: (a) A convergent sequence of rational numbers having an irrational limit. (b) A convergent sequence of irrational numbers having a rational limit. 6. Give the first six terms of the sequence and then give the n th term (a) s 1 = 1 , s n +1 = 1 2 ( s n + 1) (b) s 1 = 1 , s n +1 = 1 2 s n + 1 (c) s 1 = 1 , s n +1 = 2 s n + 1 7. use induction to prove the following assertions: (a) If s 1 = 1 and s n +1 = n + 1 2 n s n , then s n = n 2 n - 1 . (b) If s 1 = 1 and s n +1 = s n - 1 n ( n + 1) , then s n = 1 n . 15
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8. Letrbe a real number,r= 0. Define a sequence{Sn}by 9. Set a n = 1 n ( n + 1) , n = 1 , 2 , 3 , . . . , and form the sequence S 1 = a 1 S 2 = a 1 + a 2 S 3 = a 1 + a 2 + a 3 . . . S n = a 1 + a 2 + a 3 + · · · + a n . . . Find a formula for S n . II.2. LIMIT THEOREMS THEOREM 4. Suppose s n s and t n t . Then: 1. s n + t n s + t . 2. s n - t n s - t . 3. s n t n st . Special case: ks n ks for any number k . 4. s n /t n s/t provided t = 0 and t n = 0 for all n . THEOREM 5. Suppose s n s and t n t . If s n t n for all n , then s t . Proof: Suppose s > t . Let = s - t 2 . Since s n s , there exists a positive integer N 1 such that | s n - s | < for all n > N 1 . This implies that s - < s n < s + for all n > N 1 . Similarly, there exists a positive integer N 2 such that t - < t n < t + for all n > N 2 . Let N = max { N 1 , N 2 } . Then, for all n > N , we have t n < t + = t + s - t 2 = s + t 2 = s - < s n 16
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which contradicts the assumption s n t n for all n . Corollary Suppose t n t . If t n 0 for all n , then t 0. Infinite Limits Definition 3. A sequence { s n } diverges to + ( s n + ) if to each real number M there is a positive integer N such that s n > M for all n > N . { s n } diverges to -∞ ( s n → -∞ ) if to each real number M there is a positive integer N such that s n < M for all n > N .
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