a Consider the sequence of ratios s n 1 s n k n 1 n 1 k n n k n 1 n 1 n k n k n

A consider the sequence of ratios s n 1 s n k n 1 n 1

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(a)Consider the sequence of ratios: 0. Since s n 0 as n → ∞ , 1 /s n = n ! k n → ∞ as n → ∞ by Theorem 4.2.13. If k < 0, then k n is alternately negative and positive, and so the limit does not exist (odd numbered terms → -∞ , even numbered terms → ∞ ). 18. Given that ( s n ) is a convergent sequence. Suppose s n s . Let ( t n ) be the sequence ( t n ) = ( b, b, b, b, . . . ). Then t n b . Since s n b for all n , it follows that s b by Theorem 4.2.4. Now let ( u n ) be the sequence ( u n ) = ( a, a, a, a, . . . ). Then u n a . Since a s n for all n , it follows that a s , again by Theorem 4.2.4. 2
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Section 4.3 3. (a) s 1 = 1 and s n +1 = 1 5 ( s n + 7) for all n > 1. Note: s 2 = 8 5 , s 3 = 43 25 , s 4 = 218 125 . . . . ( s n ) appears to be increasing: Proof by induction – 1 S : s 2 = 1 5 ( s 1 + 7) = 1 5 (8) = 8 5 > 1 = s 1 . Therefore 1 S . Assume that k S . That is, assume s k +1 s k . Prove that k + 1 S . That is, prove that s k +2 s k +1 . s k +2 = 1 5 ( s k +1 + 7) 1 5 ( s k + 7) = s k +1 . Therefore, k + 1 S . Thus, S = N — the sequence is increasing. ( s n ) is bounded: Claim s n < 2 for all n ; by induction. Let S be the set of positive integers for which s n < 2. 1 S : s 1 = 1 < 2. Thus 1 S . Assume that k S . That is, assume s k < 2. Prove that k + 1 S . That is, prove s k +1 < 2. s k +1 = 1 5 ( s k + 7) 1 5 (2 + 7) = 9 5 < 2. Therefore, k + 1 S and S = N . Now, since ( s n ) is increasing and bounded above, s n s 2. For large n , s n s and s n +1 s . Therefore, for large n s = 1 5 ( s + 7) 5 s = s + 7 s = 7 4 . (c) s 1 = 2 and s n +1 = 1 4 (2 s n + 7) for all n > 1. ( s n ) is increasing: By induction – 1 S : s 2 = 1 4 (2 s 1 + 7) = 1 4 (11) = 11 4 > 2 = s 1 . Therefore 1 S . Assume that k S . That is, assume s k +1 s k . 3
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Prove that k + 1 S . That is, prove that s k +2 s k +1 .
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  • Math, lim, Natural number, Sn

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