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Section 11.5 Alternating Series
1006 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
10. =√
3 + 2 0 for ≥ 1. {} is decreasing for ≥ 2 since
√
3 + 2
0=
(3 + 2)12(1)− · 12(3 + 2)−12(32)√
3 + 22 =
12(3 + 2)−12[2(3 + 2)− 33]
(3 + 2)1=
4− 3
2(3 + 2)32 0 for
3√
4 ≈ 16. Also, lim→∞
= lim→∞
√3 + 2
√2
= lim→∞
1+ 22
= 0. Thus, the series∞=1
(−1)√
3 + 2
converges by the Alternating Series Test.
11. =2
3 + 4 0 for ≥ 1. {} is decreasing for ≥ 2 since
2
3 + 4
0=
(3 + 4)(2)− 2(32)
(3 + 4)2=
(23 + 8− 33)
(3 + 4)2=
(8− 3)
(3 + 4)2 0 for 2. Also,
lim→∞
= lim→∞
1
1 + 43= 0. Thus, the series
∞=1
(−1)+1 2
3 + 4converges by the Alternating Series Test.
12. = − =
0 for ≥ 1. {} is decreasing for ≥ 1 since (−)0 = (−−) + − = −(1− ) 0 for
1. Also, lim→∞
= 0 since lim→∞
H= lim
→∞1
= 0. Thus, the series
∞=1
(−1)+1− converges by the Alternating
Series Test.
13. =
10 0 for ≥ 1. {} is decreasing for ≥ 1 since
10
0=
10(1)− · 10 ln 10
(10)2=
10(1− ln 10)
(10)2=
1− ln 10
10 0 for 1− ln 10 0 ⇒ ln 10 1 ⇒
1
ln 10≈ 04. Also, lim
→∞ = lim
→∞
10= lim
→∞
10H= lim
→∞
10 ln 10= 0. Thus, the series
∞=1
(−1)
10
converges by the Alternating Series Test.
14. =1
0 for ≥ 1. {} is decreasing since
1
0=
· 1(−12)− 1 · 12
=−1(1 + )
3 0 for
0. Also, lim→∞
= 0 since lim→∞
1 = 1. Thus, the series∞=1
(−1)−1 1
converges by the Alternating Series Test.
15. =sin+ 1
2
1 +√
=(−1)
1 +√. Now =
1
1 +√ 0 for ≥ 0, {} is decreasing, and lim
→∞ = 0, so the series
∞=0
sin+ 1
2
1 +√
converges by the Alternating Series Test.
16. = cos
2= (−1)
2= (−1). {} is decreasing for ≥ 2 since
(2−)0 = (−2− ln 2) + 2− = 2−(1− ln 2) 0 for 1
ln 2[≈14]. Also, lim
→∞ = 0 since
lim→∞
2H= lim
→∞1
2 ln 2= 0. Thus, the series
∞=1
cos
2converges by the Alternating Series Test.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 11.5 ALTERNATING SERIES ¤ 1007
17.∞=1
(−1) sin
. = sin
0 for ≥ 2 and sin
≥ sin
+ 1
, and lim
→∞sin
= sin 0 = 0, so the
series converges by the Alternating Series Test.
18.∞=1
(−1) cos
. lim→∞
cos
= cos(0) = 1, so lim
→∞(−1) cos
does not exist and the series diverges by the Test
for Divergence.
19.
!=
· · · · · · 1 · 2 · · · · · ≥ ⇒ lim
→∞
!=∞ ⇒ lim
→∞(−1)
!does not exist. So the series
∞=1
(−1)
!diverges
by the Test for Divergence.
20. =
√+ 1−√
1·√+ 1 +
√√
+ 1 +√
=(+ 1)− √+ 1 +
√
=1√
+ 1 +√ 0 for ≥ 1. {} is decreasing and
lim→∞
= 0, so the series∞=1
(−1)√
+ 1−√ converges by the Alternating Series Test.21. The graph gives us an estimate for the sum of the series
∞=1
(−08)
!of −055.
8 =(08)
8!≈ 0000 004, so
∞=1
(−08)
!≈ 7 =
7=1
(−08)
!
≈ −08 + 032− 00853 + 001706− 0002 731 + 0000 364− 0000 042 ≈ −05507
Adding 8 to 7 does not change the fourth decimal place of 7, so the sum of the series, correct to four decimal places,
is −05507.
22. The graph gives us an estimate for the sum of the series
∞=1
(−1)−1
8of 01.
6 =6
86≈ 0000 023, so
∞=1
(−1)−1
8≈ 5 =
5=1
(−1)−1
8
≈ 0125− 003125 + 0005 859− 0000 977 + 0000 153 ≈ 00988
Adding 6 to 5 does not change the fourth decimal place of 5, so the sum of the series, correct to four decimal places,
is 00988.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 11.5 ALTERNATING SERIES ¤ 1007
17.∞=1
(−1) sin
. = sin
0 for ≥ 2 and sin
≥ sin
+ 1
, and lim
→∞sin
= sin 0 = 0, so the
series converges by the Alternating Series Test.
18.∞=1
(−1) cos
. lim→∞
cos
= cos(0) = 1, so lim
→∞(−1) cos
does not exist and the series diverges by the Test
for Divergence.
19.
!=
· · · · · · 1 · 2 · · · · · ≥ ⇒ lim
→∞
!=∞ ⇒ lim
→∞(−1)
!does not exist. So the series
∞=1
(−1)
!diverges
by the Test for Divergence.
20. =
√+ 1−√
1·√+ 1 +
√√
+ 1 +√
=(+ 1)− √+ 1 +
√
=1√
+ 1 +√ 0 for ≥ 1. {} is decreasing and
lim→∞
= 0, so the series∞=1
(−1)√
+ 1−√ converges by the Alternating Series Test.21. The graph gives us an estimate for the sum of the series
∞=1
(−08)
!of −055.
8 =(08)
8!≈ 0000 004, so
∞=1
(−08)
!≈ 7 =
7=1
(−08)
!
≈ −08 + 032− 00853 + 001706− 0002 731 + 0000 364− 0000 042 ≈ −05507
Adding 8 to 7 does not change the fourth decimal place of 7, so the sum of the series, correct to four decimal places,
is −05507.
22. The graph gives us an estimate for the sum of the series
∞=1
(−1)−1
8of 01.
6 =6
86≈ 0000 023, so
∞=1
(−1)−1
8≈ 5 =
5=1
(−1)−1
8
≈ 0125− 003125 + 0005 859− 0000 977 + 0000 153 ≈ 00988
Adding 6 to 5 does not change the fourth decimal place of 5, so the sum of the series, correct to four decimal places,
is 00988.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
1008 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
23. The series∞=1
(−1)+1
6satisfies (i) of the Alternating Series Test because
1
(+ 1)6
1
6and (ii) lim
→∞1
6= 0, so the
series is convergent. Now 5 =1
56= 0000064 000005 and 6 =
1
66≈ 000002 000005, so by the Alternating Series
Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to add the first 5 terms to get the
sum to the desired accuracy.)
24. The series∞=1
(− 13)
=
∞=1
(−1)1
3satisfies (i) of the Alternating Series Test because
1
(+ 1)3+1
1
3and
(ii) lim→∞
1
3= 0, so the series is convergent. Now 5 =
1
5 · 35≈ 00008 00005 and 6 =
1
6 · 36≈ 00002 00005,
so by the Alternating Series Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to
add the first 5 terms to get the sum to the desired accuracy.)
25. The series∞=1
(−1)−1
22satisfies (i) of the Alternating Series Test because
1
(+ 1)22+1
1
22and (ii) lim
→∞1
22= 0,
so the series is convergent. Now 5 =1
5225= 000125 00005 and 6 =
1
6226≈ 00004 00005, so by the Alternating
Series Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to add the first 5 terms to
get the sum to the desired accuracy.)
26. The series∞=1
− 1
=
∞=1
(−1)1
satisfies (i) of the Alternating Series Test because
1
(+ 1)+1
1
and
(ii) lim→∞
1
= 0, so the series is convergent. Now 5 =
1
55= 000032 000005 and 6 =
1
66≈ 000002 000005, so
by the Alternating Series Estimation Theorem, = 5. (That is, since the 6th term is less than the desired error, we need to add
the first 5 terms to get the sum to the desired accuracy.)
27. 4 =1
8!=
1
40,320≈ 0000 025, so
∞=1
(−1)
(2)!≈ 3 =
3=1
(−1)
(2)!= −1
2+
1
24− 1
720≈ −0459 722
Adding 4 to 3 does not change the fourth decimal place of 3, so by the Alternating Series Estimation Theorem, the sum of
the series, correct to four decimal places, is −04597.
28. 6 =1
36 · 6! =1
524,880≈ 0000 001 9, so
∞=1
(−1)
3!≈ 5 =
5=1
(−1)
3!= − 1
3+ 1
18− 1
162+ 1
1944− 1
29,160 ≈ −0283 471
Adding 6 to 5 does not change the fourth decimal place of 5, so by the Alternating Series Estimation Theorem, the sum of
the series, correct to four decimal places, is −02835.
29.∞=1
(−1)−2 ≈ 5 = − 1
2+
2
4− 3
6+
4
8− 5
10≈ −0105 025. Adding 6 = 612 ≈ 0000 037 to 5 does not
change the fourth decimal place of 5, so by the Alternating Series Estimation Theorem, the sum of the series, correct to four
decimal places, is −01050.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 11.5 ALTERNATING SERIES ¤ 1009
30.∞=1
(−1)−1
4≈ 6 =
1
4− 1
2 · 42+
1
3 · 43− 1
4 · 44+
1
5 · 45− 1
6 · 46≈ 0223136. Adding 7 =
1
7 · 47≈ 0000 0087 to 6
does not change the fourth decimal place of 6, so by the Alternating Series Estimation Theorem, the sum of the series, correct
to four decimal places, is 02231.
31.∞=1
(−1)−1
= 1− 1
2+
1
3− 1
4+ · · ·+ 1
49− 1
50+
1
51− 1
52+ · · · . The 50th partial sum of this series is an
underestimate, since∞=1
(−1)−1
= 50 +
1
51− 1
52
+
1
53− 1
54
+ · · · , and the terms in parentheses are all positive.
The result can be seen geometrically in Figure 1.
32. If 0,1
(+ 1) ≤
1
({1} is decreasing) and lim
→∞1
= 0, so the series converges by the Alternating Series Test.
If ≤ 0, lim→∞
(−1)−1
does not exist, so the series diverges by the Test for Divergence. Thus,
∞=1
(−1)−1
converges ⇔ 0.
33. Clearly =1
+ is decreasing and eventually positive and lim
→∞ = 0 for any . So the series
∞=1
(−1)
+ converges (by
the Alternating Series Test) for any for which every is defined, that is, + 6= 0 for ≥ 1, or is not a negative integer.
34. Let () =(ln)
. Then 0() =
(ln)−1
(− ln)
2 0 if so is eventually decreasing for every . Clearly
lim→∞
(ln)
= 0 if ≤ 0, and if 0 we can apply l’Hospital’s Rule [[+ 1]] times to get a limit of 0 as well. So the series
∞=2
(−1)−1 (ln)
converges for all (by the Alternating Series Test).
35.
2 =
1(2)2 clearly converges (by comparison with the -series for = 2). So suppose that
(−1)−1
converges. Then by Theorem 11.2.8(ii), so does
(−1)−1 +
= 21 + 1
3+ 1
5+ · · · = 2
1
2− 1. But this
diverges by comparison with the harmonic series, a contradiction. Therefore,
(−1)−1
must diverge. The Alternating
Series Test does not apply since {} is not decreasing.
36. (a) We will prove this by induction. Let () be the proposition that 2 = 2 − . (1) is the statement 2 = 2 − 1,
which is true since 1− 12
=1 + 1
2
− 1. So suppose that () is true. We will show that (+ 1) must be true as a
consequence.
2+2 − +1 =
2 +
1
2+ 1+
1
2+ 2
− +
1
+ 1
= (2 − ) +
1
2+ 1− 1
2+ 2
= 2 +1
2+ 1− 1
2+ 2= 2+2
which is (+ 1), and proves that 2 = 2 − for all .
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
1010 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
(b) We know that 2 − ln(2) → and − ln → as → ∞. So
2 = 2 − = [2 − ln(2)] − ( − ln) + [ln(2) − ln], and
lim→∞
2 = − + lim→∞
[ln(2)− ln] = lim→∞
(ln 2 + ln− ln) = ln 2.
11.6 Absolute Convergence and the Ratio and Root Tests
1. (a) Since lim→∞
+1
= 8 1, part (b) of the Ratio Test tells us that the series
is divergent.
(b) Since lim→∞
+1
= 08 1, part (a) of the Ratio Test tells us that the series
is absolutely convergent (and
therefore convergent).
(c) Since lim→∞
+1
= 1, the Ratio Test fails and the series
might converge or it might diverge.
2. =1√ 0 for ≥ 1, {} is decreasing for ≥ 1, and lim
→∞ = 0, so
∞=1
(−1)−1
√
converges by the Alternating
Series Test. To determine absolute convergence, note that∞=1
1√diverges because it is a -series with = 1
2≤ 1. Thus, the
series∞=1
(−1)−1
√
is conditionally convergent.
3. =1
5+ 1 0 for ≥ 0, {} is decreasing for ≥ 0, and lim
→∞ = 0, so
∞=0
(−1)
5+ 1converges by the Alternating
Series Test. To determine absolute convergence, choose =1
to get
lim→∞
= lim
→∞1
1(5+ 1)= lim
→∞5+ 1
= 5 0, so
∞=1
1
5+ 1diverges by the Limit Comparison Test with the
harmonic series. Thus, the series∞=0
(−1)
5+ 1is conditionally convergent.
4. 0 1
3 + 1
1
3for ≥ 1 and
∞=1
1
3is a convergent -series ( = 3 1), so
∞=1
1
3 + 1converges by comparison and
the series∞=1
(−1)
3 + 1is absolutely convergent.
5. 0
sin2
1
2for ≥ 1 and
∞=1
1
2is a convergent geometric series ( = 1
2 1), so
∞=1
sin2
converges bycomparison and the series
∞=1
sin
2is absolutely convergent.
6. lim→∞
+1
= lim→∞
(−3)+1
[2(+ 1) + 1]!· (2+ 1)!
(−3)
= lim→∞
(−3)1
(2+ 3)(2+ 2)
= 3 lim→∞
1
(2+ 3)(2+ 2)
= 3(0) = 0 1
so the series∞=0
(−3)
(2+ 1)!is absolutely convergent by the Ratio Test.
c° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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