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ch_11_review

# ch_11_review - (2 n e n(3 n Ratio L = 0&amp;lt 1 17 X 2...

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1. Show that: lim n →∞ n sin 1 n = 1 lim n →∞ n tan 1 n = 1 lim n →∞ n cos 1 n = 2. Show that lim n →∞ ± 1 + 3 n + 5 n 2 ² n = e 3 . 3. Show that lim n →∞ n n = 1 and lim n →∞ n c = 1 if c > 0 . 4. Use #3 and the squeeze theorem to show that: lim n →∞ n p 5 + 2 n + 7 n 2 = 1 and lim n →∞ n ln n = 1 . 5. Use the Integral Test to show that X n =2 1 n ln n diverges and that X n =2 1 n (ln n ) p converges if p > 1. 6. Use #5 and CT to show that X n =2 1 n ln n diverges. 7. Use #5 and Alt Test to show that X ( - 1) n n ln n converges conditionally. In 8–10, show that each series converges conditionally. Be sure to show that b n is decreasing from some index onward by showing that the derivative of the associated f ( x ) is negative for x > x 0 . (Also, be sure to show that the series are not absolutely convergent.) 8. X ( - 1) n n n 2 + 4 Use LCT to show not absol conv 9. X ( - 1) n (ln n ) 3 n Use CT to show not absol conv 10. X ( - 1) n 2 1 n n last part: CT, n 2 n > 1 n In 11–27, test each series for conv or div. 11. X ± n n + 1 ² n 2 Root Test; L = 1 e < 1 12. X ± n + 1 n ² n 2 Test for Div 13. X ln n n 2 Recall ln n < n . So a n < n n 2 = 1 n 3 2 = b n . Use CT. 14. X ³ n 3 - 1 ´ LCT; b n = 1 n 15. X 1 (ln n ) ln n Rewrite a n . Then use CT. b n = 1 n 2 16. X

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Unformatted text preview: (2 n )! e n (3 n )! Ratio; L = 0 &amp;lt; 1 17. X 2 5 8 (3 n-1) (2 n )! Ratio; L = 0 &amp;lt; 1 18. X ( n !) 2 3 7 11 (4 n-1) Ratio; L = &amp;gt; 1 19. X (ln n ) 2 2 n Root; L = 1 2 &amp;lt; 1 20. X n ! n n Ct; b n = 2 n 2 21. X n n n ! n n n ! 1 . Use test for div. Or we may use 20 above and Ex 58 in 11.2. 22. X cos n n 2 + 3 n | a n | ? for absol conv 23. X sin 1 n n LCT; b n = 1 n 2 . See 1 above. 24. X tan 1 n n LCT; b n = 1 n 2 . See 1 above. 25. X e 1 n n 2 LCT b n = 1 n 2 26. Use Root Test for each. X 2 2 n n n X n 2 + 1 5 n X (2 n ) n n 2 n 27. X 2 n (2 n + 1)! CT; b n = 2 n n ! 28. Give the Maclaurin series for each function. e 5 x 3 x cos2 x 3 x 2 sin3 x 29. Find the interval of convergence for each series: X 2 n +1 ( x-5) n +1 n 9 2 , 11 2 X ( x + 8) n 3 n n ln n [-11 ,-5)...
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