By Jon Rogawski

Calculus: Early Transcendentals, Multivariable: Chapters 10-17 through Rogawski, Jon [W. H. Freeman, 2011] ( Hardcover ) 2d version [Hardcover]

**Read Online or Download Calculus: Early Transcendentals, Multivariable, 2nd Edition: Chapters 10-17 PDF**

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**Extra resources for Calculus: Early Transcendentals, Multivariable, 2nd Edition: Chapters 10-17**

**Sample text**

I) cos πn n! (ii) n 2 (iii) (−1)n+1 n (iv) n+1 (b) −1, 1, −1, 1, . . (c) 1, −1, 1, −1, . . (d) 12 , 24 , 68 , 24 16 . . 1 for n = 1, 2, 3, . . Write out the first three terms 2. Let an = 2n − 1 of the following sequences. (a) bn = an+1 (b) cn = an+3 (c) dn = an2 (d) en = 2an − an+1 In Exercises 3–12, calculate the first four terms of the sequence, starting with n = 1. 3. cn = 3n n! 4. bn = (2n − 1)! n! 5. a1 = 2, an+1 = 2an2 − 3 6. b1 = 1, bn = bn−1 + 8. cn = (−1)2n+1 In Exercises 27–30, use Theorem 4 to determine the limit of the sequence.

Figure 10 and Table 2 suggest that an increases initially and then tends to zero. In the next example, we verify that an = R n /n! converges to zero for all R. 7. 542 C H A P T E R 10 INFINITE SERIES Rn = 0 for all R. n→∞ n! E X A M P L E 9 Prove that lim Solution Assume first that R > 0 and let M be the positive integer such that M ≤R

An = n + (n + 1) + (n + 2) + · · · + (2n) n→∞ 12n + 2 −9 + 4n 25. an = ln n n3 + 1 33. Use the limit definition to prove that lim n−2 = 0. 1 1 1 1 + + ··· + 2 3 n 11. b1 = 2, b2 = 3, 24. an = n2 + 1 n→∞ 7. bn = 5 + cos πn 9. cn = 1 + n 23. an = 1. Match each sequence with its general term: 42. an = 47. dn = ln 5n − ln n! 48. dn = ln(n2 + 4) − ln(n2 − 1) 4 1/3 49. an = 2 + 2 n 51. cn = ln en 53. yn = n 2 2n + 1 3n + 4 50. bn = tan−1 1 − 52. cn = n n + n1/n n 54. 1 55. yn = en + (−3)n 5n 56. bn = (−1)n n3 + 2−n 3n3 + 4−n π n 58.