INFINITE SEQUENCES AND SERIES

1. 11 INFINITE SEQUENCES AND SERIES

2. INFINITE SEQUENCES AND SERIES 11.4 The Comparison Tests In this section, we will learn: How to find the value of a series by comparing it with a known series.

3. COMPARISON TESTS In the comparison tests, the idea is to compare a given series with one that is known to be convergent or divergent.

4. COMPARISON TESTS Series 1 Consider the series • This reminds us of the series . • The latter is a geometric series with a = ½ and r = ½ and is therefore convergent.

5. COMPARISON TESTS As the series is similar to a convergent series, we have the feeling that it too must be convergent. • Indeed, it is.

6. COMPARISON TESTS The inequality shows that our given series has smaller terms than those of the geometric series. • Hence, all its partial sums are also smaller than 1 (the sum of the geometric series).

7. COMPARISON TESTS Thus, • Its partial sums form a bounded increasing sequence, which is convergent. • It also follows that the sum of the series is less than the sum of the geometric series:

8. COMPARISON TESTS Similar reasoning can be used to prove the following test—which applies only to series whose terms are positive.

9. COMPARISON TESTS The first part says that, if we have a series whose terms are smallerthan those of a known convergentseries, then our series is also convergent.

10. COMPARISON TESTS The second part says that, if we start with a series whose terms are largerthan those of a known divergent series, then it too is divergent.

11. THE COMPARISON TEST Suppose that Σanand Σbn are series with positive terms. • If Σbn is convergent and an≤ bnfor all n, then Σanis also convergent. • If Σbn is divergent and an≥ bnfor all n, then Σanis also divergent.

12. THE COMPARISON TEST—PROOF Part i Let • Since both series have positive terms, the sequences {sn} and {tn} are increasing (sn+1 = sn + an+1 ≥ sn). • Also, tn→ t; so tn≤ tfor all n.

13. THE COMPARISON TEST—PROOF Part i Since ai≤bi, we have sn≤tn. Hence, sn≤t for all n. • This means that {sn} is increasing and bounded above. • So, it converges by the Monotonic Sequence Theorem. • Thus, Σanconverges.

14. THE COMPARISON TEST—PROOF Part ii If Σbnis divergent, then tn →∞ (since {tn} is increasing). • However, ai≥ bi; so sn≥ tn. • Thus, sn→∞; so Σan diverges.

15. SEQUENCE VS. SERIES It is important to keep in mind the distinction between a sequence and a series. • A sequence is a list of numbers. • A series is a sum.

16. SEQUENCE VS. SERIES With every series Σan, there are associated two sequences: • The sequence {an} of terms • The sequence {sn}of partial sums

17. COMPARISON TEST In using the Comparison Test, we must, of course, have some known series Σbnfor the purpose of comparison.

18. COMPARISON TEST Most of the time, we use one of these: • A p-series [Σ 1/npconverges if p > 1 and diverges if p≤ 1] • A geometric series [Σarn–1 converges if |r| < 1 and diverges if |r| ≥ 1]

19. COMPARISON TEST Example 1 Determine whether the given series converges or diverges:

20. COMPARISON TEST Example 1 For large n, the dominant term in the denominator is 2n2. • So, we compare the given series with the series Σ 5/(2n2).

21. COMPARISON TEST Example 1 Observe that since the left side has a bigger denominator. • In the notation of the Comparison Test, an is the left side and bn is the right side.

22. COMPARISON TEST Example 1 We know that is convergent because it’s a constant times a p-series with p = 2 > 1.

23. COMPARISON TEST Example 1 Therefore, is convergent by part i of the Comparison Test.

24. NOTE 1 Although the condition an≤ bnor an≥bnin the Comparison Test is given for all n, we need verify only that it holds for n≥ N, where N is some fixed integer. • This is because the convergence of a series is not affected by a finite number of terms. • This is illustrated in the next example.

25. COMPARISON TEST Example 2 Test the given series for convergence or divergence:

26. COMPARISON TEST Example 2 This series was tested (using the Integral Test) in Example 4 in Section 11.3 • However, it is also possible to test it by comparing it with the harmonic series.

27. COMPARISON TEST Example 2 Observe that ln n > 1 for n≥ 3. So, • We know that Σ 1/nis divergent (p-series with p = 1). • Thus, the series is divergent by the Comparison Test.

28. NOTE 2 The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. • If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, the Comparison Test doesn’t apply.

29. NOTE 2 For instance, consider • The inequality is useless as far as the Comparison Test is concerned. • This is because Σbn = Σ (½)nis convergent and an > bn.

30. NOTE 2 Nonetheless, we have the feeling that Σ1/(2n -1) ought to be convergent because it is very similar to the convergent geometric series Σ (½)n. • In such cases, the following test can be used.

31. LIMIT COMPARISON TEST Suppose that Σanand Σbn are series with positive terms.If where c is a finite number and c > 0, either both series converge or both diverge.

32. LIMIT COMPARISON TEST—PROOF Let m and M be positive numbers such that m < c < M. • Since an/bn is close to c for large n, there is an integer N such that

33. LIMIT COMPARISON TEST—PROOF If Σbnconverges, so does ΣMbn. • Thus, Σanconverges by part i of the Comparison Test.

34. LIMIT COMPARISON TEST—PROOF If Σbndiverges, so does Σmbn. • Thus, Σandiverges by part ii of the Comparison Test.

35. COMPARISON TESTS Example 3 Test the given series for convergence or divergence:

36. COMPARISON TESTS Example 3 We use the Limit Comparison Test with:

37. COMPARISON TESTS Example 3 We obtain:

38. COMPARISON TESTS Example 3 This limit exists and Σ 1/2nis a convergent geometric series. • Thus, the given series converges by the Limit Comparison Test.

39. COMPARISON TESTS Example 4 Determine whether the given series converges or diverges:

40. COMPARISON TESTS Example 4 The dominant part of the numerator is 2n2. The dominant part of the denominator is n5/2. • This suggests taking:

41. COMPARISON TESTS Example 4 We obtain:

42. COMPARISON TESTS Example 4 Σbn = 2 Σ 1/n1/2 is divergent (p-series with p = ½ < 1). • Thus, the given series diverges by the Limit Comparison Test.

43. COMPARISON TESTS Notice that, in testing many series, we find a suitable comparison series Σbnby keeping only the highest powers in the numerator and denominator.

44. ESTIMATING SUMS We have used the Comparison Test to show that a series Σanconverges by comparison with a series Σbn. • It follows thatwe may be able to estimate the sum Σan by comparing remainders.

45. ESTIMATING SUMS As in Section 11.3, we consider the remainder Rn = s – sn = an+1 + an+2 + … • For the comparison series Σbn, we consider the corresponding remainder Tn = t - tn = bn+1 + bn+2 + …

46. ESTIMATING SUMS As an≤ bnfor all n, we have Rn≤ Tn. • If Σbnis a p-series, we can estimate its remainder Tnas in Section 11.3 • If Σbnis a geometric series, then Tn is the sum of a geometric series and we can sum it exactly (Exercises 35 and 36). • In either case, we know that Rn is smaller than Tn

47. ESTIMATING SUMS Example 5 Use the sum of the first 100 terms to approximate the sum of the series Σ 1/(n3+1). Estimate the error involved in this approximation.

48. ESTIMATING SUMS Example 5 Since the given series is convergent by the Comparison Test.

49. ESTIMATING SUMS Example 5 The remainder Tn for the comparison series Σ 1/n3was estimated in Example 5 in Section 11.3 (using the Remainder Estimate for the Integral Test). • We found that:

50. ESTIMATING SUMS Example 5 Therefore, the remainder for the given series satisfies: Rn≤ Tn ≤ 1/2n2