NOTES
72. Use the divergence of the harmonic series to state the unfortu- unfortu-nate fact about the ratio of cards obtained to cards in the set as
n increases.
EXPLORATORY EXERCISES
1. Numerically investigate the p-series∞
k=1
1 k0.9 and ∞
k=1
1 k1.1 and for other values of p close to 1. Can you distinguish convergent from divergent series numerically?
2. You know that ∞
k=2
1
k diverges. This is the “smallest”
p-series that diverges, in the sense that 1 k < 1
kp for p< 1.
Show that ∞
k=2
1
k ln k diverges and 1 k ln k < 1
k. Show that
∞ k=2
1
k ln k ln (ln k) diverges and 1
k ln k ln (ln k)< 1 k ln k. Find a series such that ∞
k=2
ak diverges and ak< 1
k ln k ln (ln k). Is there a smallest divergent series?
7.4 ALTERNATING SERIES
So far, we have focused our attention on positive term series, that is, series all of whose terms are positive. Before we consider the general case, we spend some time in this section examining alternating series, that is, series whose terms alternate back and forth from positive to negative. There are several reasons for doing this. First, alternating series appear frequently in applications. Second, alternating series are surprisingly simple to deal with and studying them will yield significant insight into how series work.
An alternating series is any series of the form
∞ k=1
(−1)k+1ak= a1− a2+ a3− a4+ a5− a6+ · · · ,
where ak> 0, for all k.
EXAMPLE 4.1 The Alternating Harmonic Series
Investigate the convergence or divergence of the alternating harmonic series
∞ k=1
(−1)k+1
k = 1 − 1
2+1 3 −1
4 +1 5−1
6 + · · · .
Solution The graph of the first 20 partial sums seen in Figure 7.30 suggests that the series might converge to about 0.7. We now calculate the first few partial sums by hand.
Note that
S1= 1, S5= 7
12+1 5 = 47
60, S2= 1 −1
2 =1
2, S6= 47
60−1 6 = 37
60, S3=1
2 +1 3 = 5
6, S7= 37
60+1 7 = 319
420, S4=5
6 −1 4 = 7
12, S8= 319
420−1 8 =533
840
5 10 15 20
0.2 0.4 0.6 0.8 1.0
n Sn
FIGURE 7.30 Sn=n
k=1
(−1)k+1
k .
and so on. We have plotted these first 8 partial sums on the number line shown in Figure 7.31.
0.5 0.6 0.7 0.8 0.9 1
S2 S4 S6S8 S7S5 S3 S1
FIGURE 7.31 Partial sums of∞
k=1
(−1)k+1
k .
You should notice that the partial sums are bouncing back and forth, but seem to be zeroing-in on some value. (Could it be the sum of the series?) This should not be sur-prising, since as each new term is added or subtracted, we are adding or subtracting less than we subtracted or added (Why did we reverse the order here?) to get the pre-vious partial sum. You should notice this same zeroing-in in the accompanying table displaying the first 20 partial sums of the series. Based on the behavior of the par-tial sums, it is reasonable to conjecture that the series converges to some value be-tween 0.66877 and 0.71877.We can resolve the question of convergence definitively with Theorem 4.1.
n Sn n
k1
(−1)k1 k
1 1
2 0.5
3 0.83333 4 0.58333 5 0.78333 6 0.61667 7 0.75952 8 0.63452 9 0.74563 10 0.64563 11 0.73654 12 0.65321 13 0.73013 14 0.65871 15 0.72537 16 0.66287 17 0.7217 18 0.66614 19 0.71877 20 0.66877
THEOREM 4.1 (Alternating Series Test) Suppose that lim
k→∞ak= 0 and 0 < ak+1 ≤ akfor all k≥ 1. Then, the alternating series
∞
k=1(−1)k+1akconverges.
Make sure that you have a clear idea of what it is saying. In the case of an alternating series satisfying the hypotheses of the theorem, we start with 0 and add a1> 0 to get the first partial sum S1. To get the next partial sum, S2, we subtract a2from S1, where a2 < a1. This says that S2will be between 0 and S1. We illustrate this situation in Figure 7.32.
Continuing in this fashion, we add a3to S2to get S3. Since a3 < a2, we must have that S2< S3< S1. Referring to Figure 7.32, notice that
S2< S4< S6< · · · < S5< S3< S1.
a1 a2 a3 a4 a5 a6
0 S2 S4 S6 S5 S3 S1
FIGURE 7.32
Convergence of the partial sums of an alternating series.
In particular, this says that all of the odd-indexed partial sums (i.e., S2n+1, for n= 0, 1, 2,. . .) are larger than all of the even-indexed partial sums (i.e., S2n, for n= 1, 2,. . .). As the partial sums oscillate back and forth, they should be drawing closer and closer to some limit S, somewhere between all of the even-indexed partial sums and the odd-indexed partial sums,
S2< S4< S6< · · · < S < · · · < S5< S3< S1. (4.1) We illustrate the use of this new test in example 4.2.
EXAMPLE 4.2 Using the Alternating Series Test Reconsider the convergence of the alternating harmonic series ∞
k=1
(−1)k+1
k .
Solution Notice that
klim→∞ak= lim
k→∞
1 k = 0.
Further,
0< ak+1= 1 k+ 1 ≤ 1
k = ak, for all k ≥ 1.
By the Alternating Series Test, the series converges. (You can use the calculations from example 4.1 to arrive at an approximate sum.)
The Alternating Series Test is certainly the easiest test we’ve discussed so far for determining the convergence of a series. It’s straightforward, but you will sometimes need to work a bit to verify the hypotheses.
EXAMPLE 4.3 Using the Alternating Series Test Investigate the convergence or divergence of the alternating series ∞
k=1
(−1)k(k+ 3) k(k+ 1) .
5 10 15 20
0.5
1.0
1.5
2.0 Sn
n
FIGURE 7.33 Sn=n
k=1
(−1)k(k+ 3) k(k+ 1) .
Solution The graph of the first 20 partial sums seen in Figure 7.33 suggests that the series converges to some value around−1.5. The following table showing some select partial sums suggests the same conclusion.
n Sn n
k1
(−1)k(k 3) k(k 1)
50 −1.45545
100 −1.46066
200 −1.46322
300 −1.46406
400 −1.46448
n Sn n
k1
(−1)k(k 3) k(k 1)
51 −1.47581
101 −1.47076
201 −1.46824
301 −1.46741
401 −1.46699
We can verify that the series converges by first checking that
klim→∞ak= lim
k→∞
(k+ 3) k(k+ 1)
1 k2 1 k2
= lim
k→∞
1 k+k32
1+1k = 0.
Next, consider the ratio of the absolute value of two consecutive terms:
ak+1
ak
= (k+ 4) (k+ 1)(k + 2)
k(k+ 1)
(k+ 3) = k2+ 4k k2+ 5k + 6 < 1,
for all k≥ 1. From this, it follows that ak+1 < ak, for all k≥ 1 and so, by the Alternating Series Test, the series converges. Finally, from the preceding table, we can see that the series converges to a sum between−1.46448 and −1.46699. (How can you be sure that the sum is in this interval?)
EXAMPLE 4.4 A Divergent Alternating Series Determine whether the alternating series∞
k=1
(−1)kk
k+ 2 converges or diverges.
Solution First, notice that
klim→∞ak = lim
k→∞
k
k+ 2 = 1 = 0.
So, this alternating series is divergent, since by the kth-term test for divergence, the terms must tend to zero in order for the series to be convergent.
Estimating the Sum of an Alternating Series
We have repeatedly remarked that once you know that a series converges, you can always approximate the sum of the series by computing some partial sums. However, in finding an approximate sum of a convergent series, how close is close enough? Realize that answers to such questions of accuracy are not “one size fits all,” but rather, are highly context-sensitive.
For instance, if the sum of the series is to be used to find the angle from the ground at which you throw a ball to a friend, you might accept one answer. On the other hand, if the sum of that same series is to be used to find the angle at which to aim your spacecraft to ensure a safe reentry into the earth’s atmosphere, you would likely insist on greater precision (at least, we would).
So far, we have calculated approximate sums of series by observing that a number of successive partial sums of the series are within a given distance of one another. The underlying assumption here is that when this happens, the partial sums are also within that same distance of the sum of the series. Unfortunately, this is simply not true, in general (although it is true for some series). What’s a mathematician to do? For the case of alternating
series, we are quite fortunate to have available a simple way to get a handle on the accuracy.
Note that the error in approximating the sum S by the nth partial sum Snis S− Sn. Take a look back at Figure 7.32. Recall that we had observed from the figure that all of the even-indexed partial sums Sn of the convergent alternating series,∞
k=1(−1)k+1ak lie below the sum S, while all of the odd-indexed partial sums lie above S. That is, [as in (4.1)],
S2< S4< S6< · · · < S < · · · < S5< S3< S1. This says that for n even,
Sn ≤ S ≤ Sn+1. Subtracting Snfrom all terms, we get
0≤ S − Sn≤ Sn+1− Sn= an+1. Since an+1> 0, we have
−an+1≤ 0 ≤ S − Sn≤ an+1, or
|S − Sn| ≤ an+1, for n even. (4.2) Similarly, for n odd, we have that
Sn+1≤ S ≤ Sn. Again subtracting Sn, we get
−an+1= Sn+1− Sn ≤ S − Sn ≤ 0 ≤ an+1
or
|S − Sn| ≤ an+1, for n odd. (4.3) Since (4.2) and (4.3) (these are called error bounds) are the same, we have the same error bound whether n is even or odd. This establishes the result stated in Theorem 4.2.
THEOREM 4.2