14 Let vextendsingle vextendsingle vextendsingle vextendsingle 1 n 1 n 1

14 let vextendsingle vextendsingle vextendsingle

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14. Let vextendsingle vextendsingle vextendsingle vextendsingle 1 n - 1 n + 1 - 0 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 1 n - 1 n + 1 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle n + 1 - n n ( n + 1) vextendsingle vextendsingle vextendsingle vextendsingle = 1 n 2 + n < 1 n 2 < 1 n . Since 1 n 0 as n → ∞ , the result follows by Theorem 16.8. 1
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17. Let s n = k n /n !. (a)Consider the sequence of ratios: 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 17.4. 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 17.4. Section 18. 3. (a) s 1 = 1 and s n +1 = 1 4 ( s n + 5) for all n > 1. Note: s 2 = 3 2 , s 3 = 13 8 , s 4 = 53 32 . . . . ( s n ) is increasing: By induction – 1 S : s 2 = 1 4 ( s 1 + 5) = 1 4 (6) = 6 4 = 3 2 > 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 .
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