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11.3 The Integral Test 775 EXERCISES 11.3 2n Determining Convergence or Divergence 10. a q ln n q 11. a n 2n q 12. a n 5n 2n Which of the series in Exercises 1–30 converge, and which diverge? n=2 n=1 3 n=1 4 + 3 n = 1 2n A 2n + 1 B q q q Give reasons for your answers. (When you check an answer, remem- -2 1 2n 13. a 14. a 15. a ber that there may be more than one way to determine the series’ con- n=0 n + 1 n = 1 2n - 1 n=1 n + 1 vergence or divergence.) n 18. a a1 + n b q q q 1 1 16. a 17. a 2n q q q 1 n ln n 1. a n 2. a e -n 3. a n=2 n=1 n = 1 10 n=1 n=1 n + 1 q q 1 1 q 5 q 3 q 19. a n 20. a n n = 1 sln 2d n = 1 sln 3d -2 4. a 5. a 6. a n=1 n + 1 n=1 n2n n=1 q s1>nd q 1 q 1 q -8 q ln n 21. a 22. a 2 7. a - n 8. a n 9. a n n = 3 sln nd2ln2 n - 1 n = 1 ns1 + ln nd n=1 8 n=1 n=2 Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 776 Chapter 11: Infinite Sequences and Series q q 1 1 39. Logarithmic p-series 23. a n sin n 24. a n tan n n=1 n=1 a. Show that q n q e 2 q 25. a 2n 26. a n dx s p a positive constantd n=1 1 + e n=1 1 + e q q L2 xsln xd p -1 8 tan n n 27. a 2 28. a 2 converges if and only if p 7 1 . n=1 1 + n n=1 n + 1 q q b. What implications does the fact in part (a) have for the 29. a sech n 30. a sech2 n convergence of the series n=1 n=1 q 1 a p ? Theory and Examples n = 2 nsln nd For what values of a, if any, do the series in Exercises 31 and 32 Give reasons for your answer. converge? 40. (Continuation of Exercise 39.) Use the result in Exercise 39 to de- 31. a a b 32. a a b q q a 1 1 2a termine which of the following series converge and which di- - - n=1 n + 2 n + 4 n=3 n - 1 n + 1 verge. Support your answer in each case. 33. a. Draw illustrations like those in Figures 11.7 and 11.8 to show that q 1 q 1 the partial sums of the harmonic series satisfy the inequalities a. a b. a 1.01 n = 2 nsln nd n = 2 nsln nd q q n+1 1 1 1 1 Á + 1 c. a d. a ln sn + 1d = x dx … 1 + 2 + n n ln sn 3 d nsln nd3 L1 n=2 n=2 n 41. Euler’s constant Graphs like those in Figure 11.8 suggest that as 1 … 1 + x dx = 1 + ln n. n increases there is little change in the difference between the sum L1 1 1 T b. There is absolutely no empirical evidence for the divergence 1 + + Á + n 2 of the harmonic series even though we know it diverges. The partial sums just grow too slowly. To see what we mean, and the integral suppose you had started with s1 = 1 the day the universe was n formed, 13 billion years ago, and added a new term every 1 ln n = x dx . second. About how large would the partial sum sn be today, L1 assuming a 365-day year? To explore this idea, carry out the following steps. 34. Are there any values of x for which g n = 1s1>snxdd converges? q Give reasons for your answer. a. By taking ƒsxd = 1>x in the proof of Theorem 9, show that g n = 1 an q 35. Is it true that if is a divergent series of positive numbers 1 1 ln sn + 1d … 1 + + Á + n … 1 + ln n then there is also a divergent series g n = 1 bn of positive numbers q 2 with bn 6 an for every n? Is there a “smallest” divergent series of positive numbers? Give reasons for your answers. or 36. (Continuation of Exercise 35.) Is there a “largest” convergent se- 1 1 ries of positive numbers? Explain. 0 6 ln sn + 1d - ln n … 1 + + Á + n - ln n … 1 . 2 37. The Cauchy condensation test The Cauchy condensation test says: Let 5an6 be a nonincreasing sequence (an Ú an + 1 for all n) Thus, the sequence of positive terms that converges to 0. Then gan converges if and 1 1 only if g2na2n converges. For example, gs1>nd diverges because an = 1 + 2 + Á + n - ln n g2n # s1>2n d = g1 diverges. Show why the test works. is bounded from below and from above. 38. Use the Cauchy condensation test from Exercise 37 to show that q b. Show that 1 a. a diverges; 1 n+1 1 n = 2 n ln n 6 x dx = ln sn + 1d - ln n , q 1 n + 1 Ln b. a p converges if p 7 1 and diverges if p … 1 . n=1 n and use this result to show that the sequence 5an6 in part (a) is decreasing. Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley 11.3 The Integral Test 777 Since a decreasing sequence that is bounded from below con- expression with a simple law of formulation has ever been found verges (Exercise 107 in Section 11.1), the numbers an defined in for g . part (a) converge: 42. Use the integral test to show that 1 1 1 + + Á + n - ln n : g . q -n2 2 ae n=0 The number g , whose value is 0.5772 Á , is called Euler’s con- stant. In contrast to other special numbers like p and e, no other converges. Copyright © 2005 Pearson Education, Inc., publishing as Pearson Addison-Wesley

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