# 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 2 n 2 + 1 ² = ± 1 , 8 5 , 9 5 ,... ² is bounded, positive, increasing, and converges to 2. 2. ± 2 n n 2 + 1 ² = ± 1 , 4 5 , 3 5 , 8 17 ² is bounded, positive, decreasing, and converges to 0. 3. ± 4 ( 1 ) n n ² = ± 5 , 7 2 , 13 3 ² is bounded, positive, and converges to 4. 4. ± sin 1 n ² = ± sin 1 , sin ³ 1 2 ´ , sin ³ 1 3 ´ ² is bounded, positive, decreasing, and converges to 0. 5. ± n 2 1 n ² = ± n 1 n ² = ± 0 , 3 2 , 8 3 , 15 4 ² is bounded below, positive, increasing, and diverges to inﬁnity. 6. ± e n π n ² = ± e π , µ e π 2 , µ e π 3 ² is bounded, positive, decreasing, and converges to 0, since e . 7. ± e n π n / 2 ² = ±³ e π ´ n ² . Since e / π> 1, the sequence is bounded below, positive, increasing, and diverges to inﬁnity. 8. ± ( 1 ) n n e n ² = ± 1 e , 2 e 2 , 3 e 3 ² 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 ³ 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 ) ! ² 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 ² 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 ³ n 3 n ´ n = lim ³ 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 ³ 1 x + 1 ´ = lim x →∞ x + 1 . 23. lim ( n + 1 n ) = lim n + 1 n n + 1 + n = 0 . 24. lim µ n ¹ 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 + º 1 4 n = 2 . 25. lim ( ¹ n 2 + n ¹ n 2 1 ) = lim n 2 + n ( n 2 1 ) n 2 + n + n 2 1 = lim n + 1 n » º 1 + 1 n + º 1 1 n 2 ¼ = lim 1 + 1 n º 1 + 1 n + º 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 ² n , then lim a n = lim ± n 1 n ² n ± n n + 1 ² n = lim ± 1 1 n ² n ³ lim ± 1 + 1 n ² 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 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 !
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## This note was uploaded on 12/16/2009 for the course FEW, FEWEB 400567 taught by Professor Moerdersen during the Fall '09 term at Vrije Universiteit Amsterdam.

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

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