akdiverges. Explain why lim
n→∞Sn= ∞.
(b) Suppose that
∞ k=1
ak diverges. Use part (a) to show that
∞ k=1
bkdiverges.
39. Suppose that a(x) is continuous, positive, and decreasing for all x≥ 1, that ak= a(k) for all integers k ≥ 1, and that
∞
1 a(x) d x converges.
(a) Explain why the sequence of partial sums{Sn} is an in-creasing sequence.
(b) Explain why
n
1 a(x) d x≤∞
1 a(x) d x.
(c) Use parts (a) and (b) to show that the sequence of partial sums{Sn} converges.
40. (a) Where in the proof of the integral test (Theorem 8) is the assumption that a(x) is a decreasing function used?
(b) Suppose that the requirement that a(x) be decreas-ing for all x≥ 1 is replaced by the “weaker” require-ment that a(x) be decreasing for all x≥ 10. How does this change in assumptions affect the conclusions of Theorem 8?
42. Does the series1 2+ 1
(ln k)ln k converges.
[HINT: ln k> e2if k> 1619.]
In Exercises 45–52, determine whether the series converges. If it converges, find upper and lower bounds on its limit. Justify your answers.
53. Suppose that{ak} is a sequence of positive terms and that r is a constant such that ak+1/ak≤ r < 1 for all k ≥ 1.
[HINT: Use the comparison test and the formula for the sum of geometric series.]
(d) Show that Rn= ∞
k=n+1
ak≤ an+1
1− r.
54. Use Exercise 53 to find an integer N such that the partial sum SNapproximates the sum of the series
∞ n=1
n2/2n within 0.0005.
55. Use Exercise 53 to find an integer N such that the partial sum SN approximates the sum of the series
∞ n=1
(n!)2/(2n)!
within 0.0005.
11.4 11.4
11.4 A B S O L U T E C O N V E R G E N C E ; A L T E R N A T I N G S E R I E S
Not-necessarily-positive series The integral, comparison, and ratio tests apply only to nonnegative series. Some interesting series, however, have both positive and negative terms.
E X A M P L E 1 Does the alternating harmonic series
∞ k=1
(−1)k+1
k = 1 −1 2+1
3−1 4+1
5− · · · converge or diverge? To what limit?
S o l u t i o n As always, the question is how partial sums behave. Tabulating some of them shows a pattern.
Partial sums of 1−1 2+1
3−1 4+1
5− · · ·
n 1 2 3 4 5 6 7 8 9 10 . . .
Sn 1 0.500 0.833 0.583 0.783 0.617 0.760 0.635 0.746 0.646 . . .
n 51 52 53 54 55 56 57 58 59 60 . . .
Sn 0.703 0.684 0.702 0.684 0.702 0.684 0.702 0.685 0.702 0.685 . . .
Successive partial sums seem to hop back and forth across some limiting value. Plots of partial sums and terms (Figure 1) exhibit the same pattern:
{Sn}
{ak}
10 20 30 40 50
−0.75
−0.5
−0.25 0.25
0.5
Blue dots show partial sums;
black diamonds show terms.
0.75 1 1.25
FIGURE 1
Terms and partial sums for 1−1 2+1
3−1 4+1
5− · · ·
Because the terms alternate in sign, the partial sums successively rise and fall, alternately overshooting and undershooting the limiting value, which is apparently around 0.69. (It can be shown—with considerable effort—that the exact limit is ln 2≈ 0.69315.)
11.4 Absolute Convergence; Alternating Series 577
A b s o l u t e v s . c o n d i t i o n a l c o n v e r g e n c e
The alternating harmonic series illustrates the phenomenon of conditional convergence.
Although
1−1 2+1
3−1 4+1
5− · · ·
converges (as Example 1 suggests), the ordinary harmonic series 1+1
2+1 3+1
4+1 5+ · · · diverges, as we saw from the integral test.
E X A M P L E 2 Does
∞ k=1
sin k
k2 converge? Does
∞ k=1
|sin k|
k2 ? Estimate limits.
S o l u t i o n The first series, like the alternating harmonic series, has both negative and positive terms—although in no regular order this time. Plotting terms and partial sums (Figure 2) suggests (but doesn’t prove) that this series converges.
{Sn}
{ak}
10 20 30 40 50
0.25 0.5 0.75 1
FIGURE 2
Terms and partial sums for
∞ k=1
sin k k2
The partial sums wander up and down just slightly but still appear to approach a horizontal asymptote, perhaps near y= 1.
The second series also seems to converge, as Figure 3 suggests. This time the limit appears to be near 1.25.
{Sn}
{ak}
10 20 30 40 50
0.25 0.5 0.75 1 1.25
FIGURE 3
Terms and partial sums for
∞ k=1
|sin k|
k2
The second series comes from the first by taking the absolute value of each term. The new series is nonnegative, and so the comparison test applies. Because
0≤ |sin k|
k2 ≤ 1 k2 for all k≥ 1 and∞
k=11/k2converges, so must∞
k=1|sin k| /k2.
Examples 1 and 2 illustrate the phenomena of conditional convergence and absolute Ifakis the first series,
|ak| is the second.
convergence, respectively. The formal definitions are as follows:
D E F I N I T I O N ( A b s o l u t e a n d c o n d i t i o n a l c o n v e r g e n c e ) Let
ak be any series.
r If|ak| converges, then
akconverges absolutely.
r If|ak| diverges but
akconverges, then
akconverges conditionally.
The wacky world of conditional convergence Conditionally convergent series have some surprising properties. Here is one of the oddest:
Let
ak be conditionally convergent, and let L be any real number. Then the terms of
ak can be reordered in such a way that the resulting series converges to L .
(For more details, see your instructor.) Notice how drastically this peculiar prop-erty of conditionally convergent series upsets the naive hope that addition is commutative even for infinite sums.
Pluses and minuses of pluses and minuses Let ∞
k=1ak be any series. If it happens that ak≥ 0 for all k, the advantage is simplicity: The partial sums are nondecreasing.
The disadvantage, as the harmonic series shows, is that the partial sums may tend to infinity.
Mixing positive and negative terms may cost something in simplicity, but it is an ad-vantage for convergence. As the alternating harmonic series shows, positive and negative terms can offset each other, thus helping the cause of convergence.
Absolute convergence implies ordinary convergence We saw in Chapter 10 that if
∞
1 | f (x)| dx converges, then so must∞
1 f(x) d x, and∞
1 f(x) d x ≤ 1∞| f (x)| dx. (See Theorem 2, page 534.) The same principle applies to infinite series:
T H E O R E M 10 If∞
k=1|ak| converges, then so does∞
k=1ak, and
∞ k=1
ak
≤∞
k=1
|ak|.
The theorem’s inequality should be believable—it is the infinite version of the fact that the absolute value of a sum can not exceed the sum of the absolute values. The idea
11.4 Absolute Convergence; Alternating Series 579
of a rigorous proof is to write the original series as a sum of two new series, one entirely positive and the other entirely negative. Using the comparison test, one can show that each of these new series converges.
The theorem shows that, as Figure 3 suggested, the series∞
k=1sin k/k2of Example 2 does indeed converge because∞
k=1|sin k| /k2does. Our limit estimates from Example 2 are also consistent with the theorem:
1≈
converge? (A power series is something like an “infinite polynomial”; we discuss power series carefully in the next section.)
S o l u t i o n First we use the ratio test to check for absolute convergence:
klim→∞
If|x| < 1, the original series converges absolutely. By Theorem 10, it also converges without absolute value signs.
If|x| ≥ 1, the series fails the nth term test and so diverges.
E s t i m a t i n g l i m i t s
Estimating a limit for any series—nonnegative or not—depends on keeping the upper tail small. Theorem 10, combined with earlier estimates, can help.
E X A M P L E 4 For the series∞
k=1(−1)k+1/k3, we find (using technology) that S100≈ 0.901542. How closely does S100approximate S, the true sum of the series?
S o l u t i o n Because we need only estimate R100, as follows:
|R100| =
The last integral is easy to calculate:
∞
and so the estimate S≈ S100≈ 0.901542 is good to at least four decimal places.
A l t e r n a t i n g s e r i e s
For series with both positive and negative terms, testing for absolute convergence is usually the best option. In the special (but surprisingly useful) case that the terms alternate in sign, we can sometimes do better.
D E F I N I T I O N An alternating series is one whose terms alternate in sign, that is, a series of the form
c1− c2+ c3− c4+ c5− c6+ · · · , where each ci is positive.
The alternating harmonic series 1−1
2+1 3−1
4+1 5− · · ·
illustrates the best possibility. Because successive terms alternate in sign and decrease See Example 1, especially
Figure 1. in size, successive partial sums straddle smaller and smaller intervals. If the terms also tend to zero, then the partial sums narrow down on a limit. The following theorem makes these observations formal; it also gives a convenient error bound.
T H E O R E M 11 ( A l t e r n a t i n g s e r i e s t e s t ) Consider the series
∞ k=1
(−1)k+1ck= c1− c2+ c3− c4+ · · · , where
(i) c1≥ c2≥ c3≥ · · · ≥ 0; and (ii) lim
k→∞ck= 0.
Then the series converges, and its sum S lies between any two successive partial sums Snand Sn+1. In particular,
|S − Sn| < cn+1
for all n≥ 1.
A formal proof is slightly tricky, but the underlying idea is simple: Because the limit S lies between successive partial sums, adding another term to any partial sum always
“overshoots” the limit, which explains the final inequality.
U s i n g t h e t h e o r e m s : m i s c e l l a n e o u s e x a m p l e s
Combining Theorems 10 and 11 with results from earlier sections enables us to handle many not-necessarily-positive series, detecting convergence or divergence and estimating limits. The following examples illustrate some useful tricks of this trade.
E X A M P L E 5 (An alternating p-series: another look) What does Theorem 11 say about∞
k=1(−1)k+1/k3and its 100th partial sum S100≈ 0.9015422?
11.4 Absolute Convergence; Alternating Series 581
S o l u t i o n In this context, ck= 1/k3. Now Theorem 11 says not only that the series converges—which we already knew—but also that
|S − S100| < c101= 1
1013 ≈ 0.000001.
Thus, S100≈ 0.9015422 lies within 0.000001 of the true limit S. Equivalently, S lies between S100≈ 0.9015422 and S101≈ 0.9015432.
E X A M P L E 6 Does
∞ j=1
(−1)j j
j+ 1converge or diverge? Why?
S o l u t i o n The alternating series test looks tempting at first glance. But it does not apply because
jlim→∞
j
j+ 1= lim
j→∞
1
1+ 1/j = 1,
not zero as Theorem 11 requires. The nth term test does apply, however. (Maybe we should call it the “ j th term test” here.) Since
j
j+ 1→ 1 as j → ∞,
it follows that the j th term has no limit as j→ ∞. Therefore, the given series diverges.
Successive terms are alternately near 1 and−1, and so the terms diverge.
E X A M P L E 7 Does the series
1+ 2 + 3 + 4 + 5 −1 6+1
7−1 8+1
9− · · · converge? If so, find or estimate the limit.
S o l u t i o n The alternating series test does not apply right out of the box because the first five terms break the desired pattern. The problem isn’t fatal, however. Basic series algebra lets us group our terms into two blocks as follows:
(1+ 2 + 3 + 4 + 5) − 1
6−1 7+1
8−1 9+ · · ·
.
The first block is finite, and so convergence isn’t an issue; its sum is 15. The second block clearly satisfies all hypotheses of the alternating series test and so converges to some limit L. Any partial sum of the second block, moreover, differs from L by less than the magnitude of the next term (by the last line of Theorem 11).
The entire series therefore converges to S= 15 − L, and any partial sum differs from S by no more than the next term. The partial sum S9= 1 + 2 + · · · + 1/9 ≈ 14.962, for instance, overshoots the true limit by less than 1/10. In other words, the exact limit S satisfies 14.862 ≤ S ≤ 14.962.
E X A M P L E 8 Does
∞ n=1
sin n
n3+ n2+ n + 1 + cos n converge or diverge? Why?
S o l u t i o n The problem is easier than it looks. Hoping for absolute convergence, we start by taking absolute values:
sin n
n3+ n2+ n + 1 + cos n
= |sin n|
n3+ n2+ n + 1 + cos n.
The general appearance of numerator and denominator suggests a comparison. A simple inequality make the job much easier:
|sin n|
n3+ n2+ n + 1 + cos n ≤ 1 n3. Because the p-series
1/n3converges, so must the absolute-value version of the given series. Finally, Theorem 10 guarantees that the original series also converges.
B
A S I CE
X E R C I S E SExercises 1–4 are about the series in Example 7.
1. Does the series converge conditionally or absolutely? Justify