1. 12 INFINITE SEQUENCES AND SERIES

2. 12.4 The Comparison Tests In this section, we will learn: How to find the value of a series by comparing it with a known series. INFINITE SEQUENCES AND 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. 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. COMPARISON TESTS Series 1

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

6. 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). COMPARISON TESTS

7. 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: COMPARISON TESTS

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

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

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

11. Suppose that Σ a n and Σ b n are series with positive terms. i.If Σ b n is convergent and a n ≤ b n for all n, then Σ a n is also convergent. ii.If Σ b n is divergent and a n ≥ b n for all n, then Σ a n is also divergent. THE COMPARISON TEST

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

13. Since a i ≤ b i, we have s n ≤ t n. Hence, s n ≤ t for all n.  This means that {s n } is increasing and bounded above.  So, it converges by the Monotonic Sequence Theorem.  Thus, Σ a n converges. THE COMPARISON TEST—PROOF Part i

14. If Σ b n is divergent, then t n → ∞ (since {t n } is increasing).  However, a i ≥ b i ; so s n ≥ t n.  Thus, s n → ∞; so Σ a n diverges. THE COMPARISON TEST—PROOF Part ii

15. 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. SEQUENCE VS. SERIES

16. With every series Σ a n, there are associated two sequences: 1.The sequence {a n } of terms 2.The sequence {s n } of partial sums SEQUENCE VS. SERIES

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

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

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

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

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

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

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

24. Although the condition a n ≤ b n or a n ≥ b n in 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. NOTE 1

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

26. 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. COMPARISON TEST Example 2

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

28. 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. NOTE 2

29. For instance, consider  The inequality is useless as far as the Comparison Test is concerned.  This is because Σ b n = Σ (½) n is convergent and a n > b n. NOTE 2

30. Nonetheless, we have the feeling that Σ1/(2 n -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. NOTE 2

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

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

33. If Σ b n converges, so does Σ Mb n.  Thus, Σ a n converges by part i of the Comparison Test. LIMIT COMPARISON TEST—PROOF

34. If Σ b n diverges, so does Σ mb n.  Thus, Σ a n diverges by part ii of the Comparison Test. LIMIT COMPARISON TEST—PROOF

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

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

37. We obtain: COMPARISON TESTS Example 3

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

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

40. The dominant part of the numerator is 2n 2. The dominant part of the denominator is n 5/2.  This suggests taking: COMPARISON TESTS Example 4

41. COMPARISON TESTS Example 4 We obtain:

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

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

44. ESTIMATING SUMS We have used the Comparison Test to show that a series Σ a n converges by comparison with a series Σ b n.  It follows that we may be able to estimate the sum Σ a n by comparing remainders.

45. As in Section 11.3, we consider the remainder R n = s – s n = a n+1 + a n+2 + …  For the comparison series Σ b n, we consider the corresponding remainder T n = t - t n = b n+1 + b n+2 + … ESTIMATING SUMS

46. As a n ≤ b n for all n, we have R n ≤ T n.  If Σ b n is a p-series, we can estimate its remainder T n as in Section 11.3  If Σ b n is a geometric series, then T n is the sum of a geometric series and we can sum it exactly (Exercises 35 and 36).  In either case, we know that R n is smaller than T n ESTIMATING SUMS

47. Use the sum of the first 100 terms to approximate the sum of the series Σ 1/(n 3 +1). Estimate the error involved in this approximation. ESTIMATING SUMS Example 5

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

49. The remainder T n for the comparison series Σ 1/n 3 was estimated in Example 5 in Section 11.3 (using the Remainder Estimate for the Integral Test).  We found that: ESTIMATING SUMS Example 5

50. Therefore, the remainder for the given series satisfies: R n ≤ T n ≤ 1/2n 2 ESTIMATING SUMS Example 5

51. With n = 100, we have:  With a programmable calculator or a computer, we find that with error less than 0.00005 ESTIMATING SUMS Example 5