SECTION 8.4

1. SECTION 8.4 OTHER CONVERGENCE TESTS

2. INFINITE SEQUENCES AND SERIES • The convergence tests that we have looked at so far apply only to series with positive terms. • In this section we learn how to deal with series whose terms are not necessarily positive. 8.4

3. ALTERNATING SERIES • An alternating series is a series whose terms are alternately positive and negative. • Here are two examples: 8.4

4. ALTERNATING SERIES • From these examples, we see that the nth term of an alternating series is of the form an = (–1)n– 1bn or an = (–1)nbnwhere bn is a positive number. • In fact, bn = |an| 8.4

5. ALTERNATING SERIES • The following test states that, if the terms of an alternating series decrease toward 0 in absolute value, the series converges. 8.4

6. ALTERNATING SERIES TEST If the alternating series satisfies (i)bn+1≤ bn for all n (ii) then the series is convergent. 8.4

7. ALTERNATING SERIES • Before giving the proof, let’s look Figure 1, which gives a picture of the idea behind the proof. 8.4

8. ALTERNATING SERIES • First, we plot s1 = b1 on a number line. • To find s2,we subtract b2. • So, s2 is to the left of s1. 8.4

9. ALTERNATING SERIES • Then, to find s3, we add b3. • So,s3 is to the right of s2. • However, since b3 < b2, s3 is to the left of s1. 8.4

10. ALTERNATING SERIES • Continuing in this manner, we see that the partial sums oscillate back and forth. • Since bn→ 0, the successive steps are becoming smaller and smaller. 8.4

11. ALTERNATING SERIES • The even partial sums s2, s4, s6, … are increasing. • The odd partial sums s1, s3, s5, … are decreasing. 8.4

12. ALTERNATING SERIES • Thus, it seems plausible that both are converging to some number s, which is the sum of the series. • Therefore, in the following proof we consider the even and odd partial sums separately. 8.4

13. PROOF OF THE ALTERNATING SERIES TEST • First, we consider the even partial sums:s2 = b1–b2≥ 0 since b2 ≤ b1s4 = s2 + (b3–b4) ≥ s2 since b4 ≤ b3 • In general, s2n = s2n– 2 + (b2n– 1–b2n) ≥ s2n– 2 since b2n ≤ b2n– 1 • Thus, 0 ≤ s2 ≤ s4 ≤ s6 ≤ … ≤ s2n ≤ … 8.4

14. PROOF OF THE ALTERNATING SERIES TEST • However, we can also write:s2n = b1– (b2–b3) – (b4–b5) –…– (b2n– 2–b2n– 1) –b2n • Every term in brackets is positive. • So, s2n≤ b1for all n. • Thus, the sequence {s2n} of even partial sums is increasing and bounded above. • Therefore, it is convergent by the Monotonic Sequence Theorem. 8.4

15. PROOF OF THE ALTERNATING SERIES TEST • Let’s call its limit s, that is, • Now, we compute the limit of the odd partial sums: 8.4

16. PROOF OF THE ALTERNATING SERIES TEST • As both the even and odd partial sums converge to s, we have • See Exercise 46 in Section 8.1 • Thus, the series is convergent. 8.4

17. Example 1 • The alternating harmonic seriessatisfies(i) bn+1 < bn because(ii) • It is convergent by the Alternating Series Test. 8.4

18. ALTERNATING SERIES • Figure 2 illustrates Example 1 by showing the graphs of the terms an = (–1)n – 1/n and the partial sums sn. 8.4

19. ALTERNATING SERIES • Notice how the values of sn zigzag across the limiting value, which appears to be about 0.7 • In fact, it can be proved that the exact sum of the series is ln 2 ≈ 0.693. 8.4

20. Example 2 • The series is alternating. • However, • So, condition (ii) is not satisfied. 8.4

21. Example 2 SOLUTION • Instead, we look at the limit of the nth term of the series: • This limit does not exist. • So, the series diverges by the Test for Divergence. 8.4

22. Example 3 • Test the series for convergence or divergence. • SOLUTION • The given series is alternating. • So, we try to verify conditions (i) and (ii) of the Alternating Series Test. 8.4

23. Example 3 SOLUTION • Unlike the situation in Example 1, it is not obvious that the sequence given by bn = n2/(n3 + 1) is decreasing. • However, if we consider the related function f(x) = x2/(x3 + 1), we find that: 8.4

24. Example 3 SOLUTION • Since we are considering only positive x, we see that f ’(x) < 0 if 2 –x3 < 0, that is, x > . • Thus, f is decreasing on the interval ( , ∞). • This means that f(n + 1) < f(n) and, therefore, bn+1 < bn when n ≥ 2. • The inequality b2 < b1 can be verified directly. • However, all that really matters is that the sequence {bn} is eventually decreasing. 8.4

25. Example 3 SOLUTION • Condition ii is readily verified: • Thus, the given series is convergent by the Alternating Series Test. 8.4

26. ESTIMATING SUMS • A partial sum sn of any convergent series can be used as an approximation to the total sum s. • However, this is not of much use unless we can estimate the accuracy of the approximation. • The error involved in using s≈ snis the remainder Rn = s–sn. 8.4

27. ESTIMATING SUMS • The next theorem says that, for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than bn+1. • This is the absolute value of the first neglected term. 8.4

28. ALTERNATING SERIES ESTIMATION THEOREM If s = Σ(–1)n–1bnis the sum of an alternating series that satisfies (i)0 ≤ bn+1 ≤ bn and (ii) then |Rn| = |s–sn| ≤ bn+1 8.4

29. ALTERNATING SERIES ESTIMATION THEOREM PROOF • From the proof of the Alternating Series Test, we know that s lies between any two consecutive partial sums sn and sn+1. • It follows that: |s–sn| ≤ |sn+1–sn| = bn+1 8.4

30. ALTERNATING SERIES ESTIMATION THEOREM • You can see geometrically why the theorem is true by looking at Figure 1. • Notice that s–s4 < b5, |s–s5| < b6, and so on. • Notice also that s lies between any two consecutive partial sums. 8.4

31. Example 4 • Find the sum of the seriescorrect to three decimal places. • By definition, 0! = 1. 8.4

32. Example 4 SOLUTION • First, we observe that the series is convergent by the Alternating Series Test because:(i)(ii) 8.4

33. Example 4 SOLUTION • To get a feel for how many terms we need to use in our approximation, let’s write out the first few terms of the series: 8.4

34. Example 4 SOLUTION • Notice that and 8.4

35. Example 4 SOLUTION • By the Alternating Series Estimation Theorem, we know that: | s–s6 | ≤ b7 < 0.0002 • This error of less than 0.0002 does not affect the third decimal place. • So, we have s ≈ 0.368 correct to three decimal places. 8.4

36. NOTE • The rule that the error (in using sn to approximate s) is smaller than the first neglected term is, in general, valid only for alternating series that satisfy the conditions of the Alternating Series Estimation Theorem. • The rule does not apply to other types of series. 8.4

37. ABSOLUTE CONVERGENCE • Given any series Σan, we can consider the corresponding serieswhose terms are the absolute values of the terms of the original series. 8.4

38. Definition • Notice that if Σan is a series with positive terms, then |an| = an and so absolute convergence is the same as convergence. A series Σan is called absolutely convergent if the series of absolute values Σ|an| is convergent. 8.4

39. Example 5 • The seriesis absolutely convergent becauseis a convergent p-series (p = 2). 8.4

40. Example 6 • We know that the alternating harmonic seriesis convergent. • See Example 1. 8.4

41. Example 6 • However, it is not absolutely convergent because the corresponding series of absolute values is: • This is the harmonic series (p-series with p = 1) and is, therefore, divergent. 8.4

42. Definition • Example 6 shows that the alternating harmonic series is conditionally convergent. • Thus, it is possible for a series to be convergent but not absolutely convergent. • However, the next theorem shows that absolute convergence implies convergence. A series Σanis called conditionally convergent if it is convergent but not absolutely convergent. 8.4

43. Theorem 1 If a series Σanis absolutely convergent, then it is convergent. • PROOF • Observe that the inequality • is true because |an| is either an or –an. 8.4

44. Theorem 1 PROOF • If Σan is absolutely convergent, then Σ|an|is convergent. • So, Σ2|an|is convergent. • Thus, by the Comparison Test, Σ(an + |an|) is convergent. • Then, is the difference of two convergent series and is, therefore, convergent. 8.4

45. Example 7 • Determine whether the seriesis convergent or divergent. 8.4

46. Example 7 SOLUTION • The series has both positive and negative terms, but it is not alternating. • The first term is positive. • The next three are negative. • The following three are positive—the signs change irregularly. 8.4

47. Example 7 SOLUTION • We can apply the Comparison Test to the series of absolute values: 8.4

48. Example 7 SOLUTION • Since |cos n| ≤ 1 for all n, we have: • We know that Σ1/n2 is convergent (p-series with p = 2). • Hence, Σ(cos n)/n2 is convergent by the Comparison Test. • Thus, the given series Σ(cos n)/n2 is absolutely convergent and, therefore, convergent by Theorem 1. 8.4

49. THE RATIO TEST • The following test is very useful in determining whether a given series is absolutely convergent. 8.4

50. THE RATIO TEST • If , then the series is absolutely convergent (and therefore convergent). • If , then the series is divergent. • If , the Ratio Test is inconclusive; that is, no conclusion can be drawn about the convergence or divergence of Σan. 8.4