Pre-Calc Homework Solutions 366

Pre-Calc Homework Solutions 366 - 366 Section 9.1 1 1 2n...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 366 Section 9.1 1 1 2n for n 2n 1 ,a 3 2 1 , and so on. Thus 9 8. (a) We graph the points n, 1, 2, 3, ... . 2. (a) Note that a0 an 1 n . 3 1, a1 (b) Note that a1 an [0, 23.5] by [ 2, 2] 1, a2 1 ( 1)n n 1 ,a 2 3 1 , and so on. Thus 3 . 5, a1 5 . 10n 1 n 1 alternate 2 (b) lim an n 1 n 1 lim 2n 2n 2 2 1 1 for n n (c) Note that a0 an 1, 2, 3, ... . 5(0.1)n 0.5, a2 0.05, and so on. Thus 9. (a) We graph the points n, 2 3. Different, since the terms of n 1 between positive and negative, while the terms of n 1 1 n 1 are all negative. 2 [0, 23.5] by [ 1, 3] 4. The same, since both series can be represented as 1 n (b) lim an n lim 2 n 2 ln (n 1) for n n 1 1, 2, 3, ... . 1 2 1 4 1 8 .... 10. (a) We graph the points n, 5. The same, since both series can be represented as 1 1 2 1 4 1 8 .... 1 n 1 1 1 2 2 1 1 1 .... 2 4 8 1 1 2 3 5 4 6. Different, since [0, 23.5] by [ 1, 1] 1 n 1 ( 1) 2n 1 n n 1 1 4 1 8 ... but 1 (b) lim an n ln (n 1) lim n n lim n (n 1) 1 0 7. Converges; n 0 2 n 3 3 Section 9.1 Exercises 1. (a) Let un represent the value of * in the n th term, starting with n and u1 un 1 u4 8. Diverges, because the terms do not approach zero. 9. Converges; 5 4 2 n 3 2 3 3 n 0 1 5 4 15 4 1. Then 1 , so 16 1 u1 1, 1 u2 1 1 , 4 u3 1 , 9 10. Diverges, because the common ratio is do not approach zero. 1 and the terms 1 1, u2 n2, or * 4, u3 n2. 9, and u4 16. We may write 11. Diverges, because the terms alternate between 1 and and do not approach zero. 12. Converges; 3( 0.1)n n 0 1 3 ( 0.1) 30 11 (b) Let un represent the value of * in the nth term, starting with n 1 u3 0. Then 1 , so u0 16 1 u0 1, 1, u1 1 u1 1 1 , 4 u2 1 , and 9 13. Converges; sin n n 0 4 1 2 2 n 3 4, u2 (n 9, and 1)2, or * (n 1)2. 1 1 2 1 u3 (c) If * 16. We may write un 3, the series is 1 1 4 2 1 2 ... ( 1)3 ( 1)6 ( 1)4 1 2 2 ( 1)5 1 3 2 n 0 1 2 2( 2 n 1 1 1) 1) 1 2 2 2 2 1 1 2 ... , which is the same as the desired 3. 2 2 1 ( 2 1)( 2 2 2 series. Thus let * 14. Diverges, because the terms do not approach zero. ...
View Full Document

This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online