1. INFINITE SEQUENCES AND SERIES12
INFINITE SEQUENCES AND SERIES

2. and the Ratio and Root tests
INFINITE SEQUENCES AND SERIES
12.6
Absolute Convergence
and the Ratio and Root tests
In this section, we will learn about:
Absolute convergence of a series
and tests to determine it.

3. Given any series Σ an, we can consider the corresponding series
ABSOLUTE CONVERGENCE
Given any series Σ an, we can consider the corresponding series
whose terms are the absolute values of the terms of the original series.

4. ABSOLUTE CONVERGENCE
Definition 1
A series Σ an is called absolutely convergent if the series of absolute values Σ |an| is convergent.

5. Notice that, if Σ an is a series with positive terms, then |an| = an.
ABSOLUTE CONVERGENCE
Notice that, if Σ an is a series with positive terms, then |an| = an.
So, in this case, absolute convergence is the same as convergence.

6. is absolutely convergent because
ABSOLUTE CONVERGENCE
Example 1
The series
is absolutely convergent because
is a convergent p-series (p = 2).

7. We know that the alternating harmonic series
ABSOLUTE CONVERGENCE
Example 2
We know that the alternating harmonic series
is convergent.
See Example 1 in Section 11.5.

8. ABSOLUTE CONVERGENCE
Example 2
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.

9. CONDITIONAL CONVERGENCE
Definition 2
A series Σ an is called conditionally convergent if it is convergent but not absolutely convergent.

10. ABSOLUTE CONVERGENCE
Example 2 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.

11. If a series Σ an is absolutely convergent, then it is convergent.
ABSOLUTE CONVERGENCE
Theorem 3
If a series Σ an is absolutely convergent, then it is convergent.

12. Observe that the inequality
ABSOLUTE CONVERGENCE
Theorem 3—Proof
Observe that the inequality
is true because |an| is either an or –an.

13. If Σ an is absolutely convergent, then Σ |an| is convergent.
ABSOLUTE CONVERGENCE
Theorem 3—Proof
If Σ an is absolutely convergent, then Σ |an| is convergent.
So, Σ 2|an| is convergent.
Thus, by the Comparison Test, Σ (an + |an|) is convergent.

14. ABSOLUTE CONVERGENCE
Theorem 3—Proof
Then,
is the difference of two convergent series and is, therefore, convergent.

15. Determine whether the series
ABSOLUTE CONVERGENCE
Example 3
Determine whether the series
is convergent or divergent.
Fig , p. 747

16. ABSOLUTE CONVERGENCE
Example 3
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.

17. We can apply the Comparison Test to the series of absolute values:
ABSOLUTE CONVERGENCE
Example 3
We can apply the Comparison Test to the series of absolute values:

18. Since |cos n| ≤ 1 for all n, we have:
ABSOLUTE CONVERGENCE
Example 3
Since |cos n| ≤ 1 for all n, we have:
We know that Σ 1/n2 is convergent (p-series with p = 2).
Hence, Σ (cos n)/n2 is convergent by the Comparison Test.

19. ABSOLUTE CONVERGENCE
Example 3
Thus, the given series Σ (cos n)/n2 is absolutely convergent and, therefore, convergent by Theorem 3.

20. ABSOLUTE CONVERGENCE
The following test is very useful in determining whether a given series is absolutely convergent.

21. then the series is absolutely convergent (and therefore convergent).
THE RATIO TEST
Case i If
then the series is absolutely convergent (and therefore convergent).

22. then the series is divergent.
THE RATIO TEST
Case ii If
then the series is divergent.

23. the Ratio Test is inconclusive.
Case iii If
the Ratio Test is inconclusive.
That is, no conclusion can be drawn about the convergence or divergence of Σ an.

24. THE RATIO TEST
Case i—Proof
The idea is to compare the given series with a convergent geometric series.
Since L < 1, we can choose a number r such that L < r < 1.

25. the ratio |an+1/an| will eventually be less than r.
THE RATIO TEST
Case i—Proof
Since
the ratio |an+1/an| will eventually be less than r.
That is, there exists an integer N such that:

26. |an+1| < |an|r whenever n ≥ N
THE RATIO TEST
i-Proof (Inequality 4)
Equivalently,
|an+1| < |an|r whenever n ≥ N

27. |aN+2| < |aN+1|r < |aN|r2 |aN+3| < |aN+2| < |aN|r3
THE RATIO TEST
Case i—Proof
Putting n successively equal to N, N + 1, N + 2, in Equation 4, we obtain:
|aN+1| < |aN|r
|aN+2| < |aN+1|r < |aN|r2
|aN+3| < |aN+2| < |aN|r3

28. |aN+k| < |aN|rk for all k ≥ 1
THE RATIO TEST
i-Proof (Inequality 5)
In general,
|aN+k| < |aN|rk for all k ≥ 1

29. is convergent because it is a geometric series with 0 < r < 1.
THE RATIO TEST
Case i—Proof
Now, the series
is convergent because it is a geometric series with 0 < r < 1.

30. THE RATIO TEST
Case i—Proof
Thus, the inequality 5, together with the Comparison Test, shows that the series
is also convergent.

31. It follows that the series is convergent.
THE RATIO TEST
Case i—Proof
It follows that the series is convergent.
Recall that a finite number of terms doesn’t affect convergence.
Therefore, Σ an is absolutely convergent.

32. If |an+1/an| → L > 1 or |an+1/an| → ∞
THE RATIO TEST
Case ii—Proof
If |an+1/an| → L > 1 or |an+1/an| → ∞
then the ratio |an+1/an| will eventually be greater than 1.
That is, there exists an integer N such that:

33. This means that |an+1| > |an| whenever n ≥ N, and so
THE RATIO TEST
Case ii—Proof
This means that |an+1| > |an| whenever n ≥ N, and so
Therefore, Σan diverges by the Test for Divergence.

34. Part iii of the Ratio Test says that, if
NOTE
Case iii—Proof
Part iii of the Ratio Test says that, if
the test gives no information.

35. For instance, for the convergent series Σ 1/n2, we have:
NOTE
Case iii—Proof
For instance, for the convergent series Σ 1/n2, we have:

36. For the divergent series Σ 1/n, we have:
NOTE
Case iii—Proof
For the divergent series Σ 1/n, we have:

37. Therefore, if , the series Σ an might converge or it might diverge.
NOTE
Case iii—Proof
Therefore, if , the series Σ an might converge or it might diverge.
In this case, the Ratio Test fails.
We must use some other test.

38. Test the series for absolute convergence.
RATIO TEST
Example 4
Test the series for absolute convergence.
We use the Ratio Test with an = (–1)n n3 / 3n, as follows.

39. RATIO TEST
Example 4

40. RATIO TEST
Example 4
Thus, by the Ratio Test, the given series is absolutely convergent and, therefore, convergent.

41. Test the convergence of the series
RATIO TEST
Example 5
Test the convergence of the series
Since the terms an = nn/n! are positive, we don’t need the absolute value signs.

42. RATIO TEST
Example 5
See Equation 6 in Section 3.6
Since e > 1, the series is divergent by the Ratio Test.

43. NOTE
Although the Ratio Test works in Example 5, an easier method is to use the Test for Divergence.
Since it follows that an does not approach 0 as n → ∞.
Thus, the series is divergent by the Test for Divergence.

44. The following test is convenient to apply when nth powers occur.
ABSOLUTE CONVERGENCE
The following test is convenient to apply when nth powers occur.
Its proof is similar to the proof of the Ratio Test and is left as Exercise 37.

45. then the series is absolutely convergent (and therefore convergent).
THE ROOT TEST
Case i If
then the series is absolutely convergent (and therefore convergent).

46. then the series is divergent.
THE ROOT TEST
Case ii If
then the series is divergent.

47. the Root Test is inconclusive.
Case iii If
the Root Test is inconclusive.

48. ROOT TEST
If , then part iii of the Root Test says that the test gives no information.
The series Σ an could converge or diverge.

49. ROOT TEST VS. RATIO TEST
If L = 1 in the Ratio Test, don’t try the Root Test—because L will again be 1.
If L = 1 in the Root Test, don’t try the Ratio Test—because it will fail too.

50. Test the convergence of the series
ROOT TEST
Example 6
Test the convergence of the series
Thus, the series converges by the Root Test.

51. REARRANGEMENTS
The question of whether a given convergent series is absolutely convergent or conditionally convergent has a bearing on the question of whether infinite sums behave like finite sums.

52. REARRANGEMENTS
If we rearrange the order of the terms in a finite sum, then of course the value of the sum remains unchanged.
However, this is not always the case for an infinite series.

53. REARRANGEMENT
By a rearrangement of an infinite series Σ an, we mean a series obtained by simply changing the order of the terms.
For instance, a rearrangement of Σ an could start as follows: a1 + a2 + a5 + a3 + a4 + a15 + a6 + a7 + a20 + …

54. REARRANGEMENTS
It turns out that, if Σ an is an absolutely convergent series with sum s, then any rearrangement of Σ an has the same sum s.

55. REARRANGEMENTS
However, any conditionally convergent series can be rearranged to give a different sum.

56. REARRANGEMENTS
Equation 6
To illustrate that fact, let’s consider the alternating harmonic series
See Exercise 36 in Section 11.5

57. If we multiply this series by ½, we get:
REARRANGEMENTS
If we multiply this series by ½, we get:

58. Inserting zeros between the terms of this series, we have:
REARRANGEMENTS
Equation 7
Inserting zeros between the terms of this series, we have:

59. REARRANGEMENTS
Equation 8
Now, we add the series in Equations 6 and 7 using Theorem 8 in Section 11.2:

60. REARRANGEMENTS
Notice that the series in Equation 8 contains the same terms as in Equation 6, but rearranged so that one negative term occurs after each pair of positive terms.

61. However, the sums of these series are different.
REARRANGEMENTS
However, the sums of these series are different.
In fact, Riemann proved that, if Σ an is a conditionally convergent series and r is any real number whatsoever, then there is a rearrangement of Σ an that has a sum equal to r.
A proof of this fact is outlined in Exercise 40.