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