1. INFINITE SEQUENCES AND SERIES12
INFINITE SEQUENCES AND SERIES

2. INFINITE SEQUENCES AND SERIES
The convergence tests that we have looked at so far apply only to series with positive terms.

3. 12.5 Alternating Series INFINITE SEQUENCES AND SERIES
In this section, we will learn:
How to deal with series
whose terms alternate in sign.

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

5. an = (–1)n – 1bn or an = (–1)nbn where bn is a positive number.
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)nbn
where bn is a positive number.
In fact, bn = |an|

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

7. ALTERNATING SERIES TEST
If the alternating series
satisfies
bn+1 ≤ bn for all n
then the series is convergent.

8. ALTERNATING SERIES
Before giving the proof, let’s look at this figure—which gives a picture of the idea behind the proof.
Fig , p. 746

9. First, we plot s1 = b1 on a number line.
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.
Fig , p. 746

10. However, since b3 < b2, s3 is to the left of s1.
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.
Fig , p. 746

11. 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.
Fig , p. 746

12. The even partial sums s2, s4, s6, . . . are increasing.
ALTERNATING SERIES
The even partial sums s2, s4, s6, are increasing.
The odd partial sums s1, s3, s5, are decreasing.
Fig , p. 746

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

14. ALTERNATING SERIES TEST—PROOF
First, we consider the even partial sums:
s2 = b1 – b2 ≥ 0 since b2 ≤ b1
s4 = s2 + (b3 – b4) ≥ s2 since b4 ≤ b3

15. ALTERNATING SERIES TEST—PROOF
In general,
s2n = s2n – 2 + (b2n – 1 – b2n) ≥ s2n – 2
since b2n ≤ b2n – 1
Thus, 0 ≤ s2 ≤ s4 ≤ s6 ≤ … ≤ s2n ≤ …

16. ALTERNATING SERIES TEST—PROOF
However, we can also write:
s2n = b1 – (b2 – b3) – (b4 – b5) – …
– (b2n – 2 – b2n – 1) – b2n
Every term in brackets is positive.
So, s2n ≤ b1 for all n.

17. ALTERNATING SERIES TEST—PROOF
Thus, the sequence {s2n} of even partial sums is increasing and bounded above.
Therefore, it is convergent by the Monotonic Sequence Theorem.

18. ALTERNATING SERIES TEST—PROOF
Let’s call its limit s, that is,
Now, we compute the limit of the odd partial sums:

19. ALTERNATING SERIES TEST—PROOF
As both the even and odd partial sums converge to s, we have
See Exercise 80(a) in Section 11.1
Thus, the series is convergent.

20. The alternating harmonic series
ALTERNATING SERIES
Example 1
The alternating harmonic series
satisfies
bn+1 < bn because
It is convergent by the Alternating Series Test.

21. ALTERNATING SERIES
The figure illustrates Example 1 by showing the graphs of the terms an = (–1)n – 1/n and the partial sums sn.
Fig , p. 747

22. 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 ≈ (Exercise 36).
Fig , p. 747

23. The series is alternating.
ALTERNATING SERIES
Example 2
The series is alternating.
However,
So, condition ii is not satisfied.

24. Instead, we look at the limit of the nth term of the series:
ALTERNATING SERIES
Example 2
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.

25. Test the series for convergence or divergence.
ALTERNATING SERIES
Example 3
Test the series for convergence or divergence.
The given series is alternating.
So, we try to verify conditions i and ii of the Alternating Series Test.

26. ALTERNATING SERIES
Example 3
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:

27. Thus, f is decreasing on the interval ( , ∞).
ALTERNATING SERIES
Example 3
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 ( , ∞).

28. ALTERNATING SERIES
Example 3
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.

29. Condition ii is readily verified:
ALTERNATING SERIES
Example 3
Condition ii is readily verified:
Thus, the given series is convergent by the Alternating Series Test.

30. 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 ≈ sn is the remainder Rn = s – sn.

31. 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.

32. ALTERNATING SERIES ESTIMATION THEOREM
If s = Σ (–1)n-1bn is the sum of an alternating series that satisfies
0 ≤ bn+1 ≤ bn
then |Rn| = |s – sn| ≤ bn+1

33. ALTERNATING SERIES ESTIMATION THM.—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

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

35. Find the sum of the series
ESTIMATING SUMS
Example 4
Find the sum of the series
correct to three decimal places.
By definition, 0! = 1.

36. ESTIMATING SUMS
Example 4
First, we observe that the series is convergent by the Alternating Series Test because: i.
ii.

37. ESTIMATING SUMS
Example 4
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:

38. ESTIMATING SUMS
Example 4
Notice that and

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

40. 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.