Pre-Calc Homework Solutions 394

Pre-Calc Homework Solutions 394 - 394 Chapter 9 Review an...

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 394 Chapter 9 Review an an x 1 2n 2n 3 3 12. lim n 1 lim n 2n x 1 1 2n 1 x 12 12 1, 16. This is a geometric series with r absolutely when x2 2 1 x2 2 1 , so it converges x 3. It The series converges absolutely when x or 0 x 2. 0: 2: n 0 1, or 3 diverges for all other values of x. ( 1) ( 1) 2n 1 n 2n 1 Check x n 0 ( 1) converges 2n 1 n (a) (b) ( (c) ( 3 3, 3, 3) 3) conditionally by the Alternating Series Test. Check x n 0 ( 1)n converges conditionally by the 2n 1 (d) None 17. f (x) 1 1 x Alternating Series Test. (a) 1 (b) [0, 2] (c) (0, 2) (d) At x 13. lim n 1 x 1 . Sum 4 x2 1 x2 2 ... 1 1 4 ( 1)nx n 4 . 5 ..., evaluated at x 0 and x lim n 2 1)! x 2n 2n 1 2 18. f (x) 2n n! x 2n ln (1 x) x 2 . Sum 3 x3 3 ... 2 3 ( 1)n ln 5 . 3 1x n n , an an 1 (n evaluated at x 19. f (x) sin x x ln 1 x5 5! lim n (n 1)x 2 2 0, x , x 0 0 0. x3 3! ... sin ( 1)n 0. x 2n 1 (2n 1)! ..., evaluated at x The series converges only at x (a) 0 (b) x (c) x 0 only 0 21. f (x) 10x n 1 ln (n 1) n . Sum 1 3 x2 2! x4 4! 20. f (x) cos x ... cos 3 xn n! ( 1)n 1 . 2 x 2n (2n)! ..., evaluated at x ex 1 . Sum x2 2! (d) None 14. lim n x e ln 2 x 1 3 ... 2. ..., evaluated at an an 1 lim ln n 10x n 10x 1, x 22. f (x) ln 2. Sum tan 1 The series converges absolutely for 10x or 1 10 x x 1 . 10 1 : 10 n x3 3 x5 5 ... tan 1 ( 1)n 1 3 x 2n 1 2n 1 ..., evaluated at x 2 . Sum Check n Series Test. Check n ( 1) converges by the Alternating ln n n 6 . (Note that 1 when n is replaced by n becomes ( 1) n 1 x 2n 1 1, the general term of tan x 2n 1 , which matches the general term 1 : 10 n 1 Test, since ln n 1 10 1 1 , 10 10 1 1 , 10 10 2 1 diverges by the Direct Comparison ln n 1 1 for n 2 and diverges. n n 2 n given in the exercise.) 23. Replace x by 6x in the Maclaurin series for the end of Section 9.2. 1 1 6x 1 1 x given at (a) (b) (c) 1 1 (6x) 6x (6x)2 36x 2 ... ... (6x)n (6x)n ... ... 1 1 x 24. Replace x by x 3 in the Maclaurin series for 1 10 given at (d) At x 15. lim n the end of Section 9.2. 1 an an 1 (n 2)! x n 1 (n 1)! x n n lim lim (n n 2) x (x 0) 1 x3 1 1 (x 3) x3 (x 3) 2 x6 ... ... ( x 3)n ... ... The series converges only at x (a) 0 (b) x (c) x 0 only 0 0. ( 1)nx 3n 25. The Maclaurin series for a polynomial is the polynomial itself: 1 2x 2 x 9. (d) None ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online