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# Chapter 9 - INSTRUCTOR'S SOLUTIONS MANUAL SECTION 9.1(PAGE...

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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 9.1 (PAGE 478) CHAPTER 9. SEQUENCES, SERIES, AND POWER SERIES Section 9.1 Sequences and Convergence (page 478) 1. 2 n 2 n 2 + 1 = 2 2 n 2 + 1 = 1 , 8 5 , 9 5 , . . . is bounded, positive, increasing, and converges to 2. 2. 2 n n 2 + 1 = 1 , 4 5 , 3 5 , 8 17 , . . . is bounded, positive, decreasing, and converges to 0. 3. 4 ( 1 ) n n = 5 , 7 2 , 13 3 , . . . is bounded, positive, and converges to 4. 4. sin 1 n = sin 1 , sin 1 2 , sin 1 3 , . . . is bounded, positive, decreasing, and converges to 0. 5. n 2 1 n = n 1 n = 0 , 3 2 , 8 3 , 15 4 , . . . is bounded below, positive, increasing, and diverges to infinity. 6. e n π n = e π , e π 2 , e π 3 , . . . is bounded, positive, decreasing, and converges to 0, since e < π . 7. e n π n / 2 = e π n . Since e / π > 1, the sequence is bounded below, positive, increasing, and diverges to infinity. 8. ( 1 ) n n e n = 1 e , 2 e 2 , 3 e 3 , . . . is bounded, alternat- ing, and converges to 0. 9. { 2 n / n n } is bounded, positive, decreasing, and converges to 0. 10. ( n ! ) 2 ( 2 n ) ! = 1 n + 1 2 n + 2 3 n + 3 · · · n 2 n 1 2 n . Also, a n + 1 a n = ( n + 1 ) 2 ( 2 n + 2 )( 2 n + 1 ) < 1 2 . Thus the sequence ( n ! ) 2 ( 2 n ) ! is positive, decreasing, bounded, and convergent to 0. 11. { n cos ( n π/ 2 ) } = { 0 , 2 , 0 , 4 , 0 , 6 , . . . } is divergent. 12. sin n n = sin 1 , sin 2 2 , sin 3 3 , . . . is bounded and con- verges to 0. 13. { 1 , 1 , 2 , 3 , 3 , 4 , 5 , 5 , 6 , . . . } is divergent. 14. lim 5 2 n 3 n 7 = lim 5 n 2 3 7 n = − 2 3 . 15. lim n 2 4 n + 5 = lim n 4 n 1 + 5 n = ∞ . 16. lim n 2 n 3 + 1 = lim 1 n 1 + 1 n 3 = 0 . 17. lim ( 1 ) n n n 3 + 1 = 0 . 18. lim n 2 2 n + 1 1 n 3 n 2 = lim 1 2 n n + 1 n 2 1 n 2 1 n 3 = − 1 3 . 19. lim e n e n e n + e n = lim 1 e 2 n 1 + e 2 n = 1 . 20. lim n sin 1 n = lim x 0 + sin x x = lim x 0 + cos x 1 = 1 . 21. lim n 3 n n = lim 1 + 3 n n = e 3 by l’Hˆopital’s Rule. 22. lim n ln ( n + 1 ) = lim x →∞ x ln ( x + 1 ) = lim x →∞ 1 1 x + 1 = lim x →∞ x + 1 = ∞ . 23. lim ( n + 1 n ) = lim n + 1 n n + 1 + n = 0 . 24. lim n n 2 4 n = lim n 2 ( n 2 4 n ) n + n 2 4 n = lim 4 n n + n 2 4 n = lim 4 1 + 1 4 n = 2 . 25. lim ( n 2 + n n 2 1 ) = lim n 2 + n ( n 2 1 ) n 2 + n + n 2 1 = lim n + 1 n 1 + 1 n + 1 1 n 2 = lim 1 + 1 n 1 + 1 n + 1 1 n 2 = 1 2 . 347

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SECTION 9.1 (PAGE 478) R. A. ADAMS: CALCULUS 26. If a n = n 1 n + 1 n , then lim a n = lim n 1 n n n n + 1 n = lim 1 1 n n lim 1 + 1 n n = e 1 e = e 2 (by Theorem 6 of Section 3.4). 27. a n = ( n ! ) 2 ( 2 n ) ! = ( 1 · 2 · 3 · · · n )( 1 · 2 · 3 · · · n ) 1 · 2 · 3 · · · n · ( n + 1 ) · ( n + 2 ) · · · 2 n = 1 n + 1 · 2 n + 2 · 3 n + 3 · · · n n + n 1 2 n . Thus lim a n = 0. 28. We have lim n 2 2 n = 0 since 2 n grows much faster than n 2 and lim 4 n n ! = 0 by Theorem 3(b). Hence, lim n 2 2 n n ! = lim n 2 2 n · 2 2 n n ! = lim n 2 2 n lim 4 n n ! = 0 . 29. a n = π n 1 + 2 2 n 0 < a n < (π/ 4 ) n . Since π/ 4 < 1, therefore (π/ 4 ) n 0 as n → ∞ . Thus lim a n = 0. 30. Let a 1 = 1 and a n + 1 = 1 + 2 a n for n = 1 , 2 , 3 , . . . . Then we have a 2 = 3 > a 1 . If a k + 1 > a k for some k , then a k + 2 = 1 + 2 a k + 1 > 1 + 2 a k = a k + 1 .
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Chapter 9 - INSTRUCTOR'S SOLUTIONS MANUAL SECTION 9.1(PAGE...

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