Section 11.6 The Ratio and Root Tests
34. Use any test to determine whether the series is absolutely convergent, conditionally convergent, or divergent.
∞
X
n=2
(−1)n
√n ln n [Hint: ln x <√ x.]
Solution:
SECTION 11.6 THE RATIO AND ROOT TESTS ¤ 1101 so the series ∞
=2
(−1)ln
converges by the Alternating Series Test. To determine absolute convergence, note that
(−1)ln
= ln
1
for ≥ 3, so ∞
=2
(−1)ln
is divergent by direct comparison with
∞
=2
1
, which is divergent.
Hence, ∞
=2
(−1)ln
is conditionally convergent.
28. lim
→∞
|| = lim→∞
1 − 2 + 3
= lim→∞
− 1
3 + 2 = lim
→∞
1 − 1
3 + 2 = 1
3 1, so the series
∞
=1
1 − 2 + 3
is absolutely convergent by the Root Test.
29. lim
→∞
+1
= lim→∞
(−9)+1
( + 1)10+2 · 10+1 (−9)
= lim→∞
(−9)
10( + 1)
= 9 10 lim
→∞
1
1 + 1 = 9
10(1) = 9
10 1, so the series
∞
=1
(−9)
10+1 is absolutely convergent by the Ratio Test.
30. lim
→∞
+1
= lim→∞
( + 1)52+2
10+2 ·10+1
52
= lim→∞
52( + 1)
10 = 5
2 lim
→∞
1 + 1
= 5
2(1) = 5
2 1, so the series
∞
=1
52
10+1 diverges by the Ratio Test. Or: Since lim
→∞= ∞, the series diverges by the Test for Divergence.
31. lim
→∞
|| = lim→∞
ln
= lim→∞
ln = lim
→∞
ln
= limH
→∞
1
1 = lim
→∞ = ∞, so the series
∞
=2
ln
diverges by the Root Test.
32.
sin(6) 1 + √
≤ 1 1 + √
1
32, so the series
∞
=1
sin(6) 1 + √
converges by direct comparison with the convergent series
∞
=1
1
32 ( =32 1). It follows that the given series is absolutely convergent.
33.
(−1)arctan
2
2
2 , so since ∞
=1
2
2 = 2
∞
=1
1
2 converges ( = 2 1), the given series ∞
=1
(−1)arctan
2 converges absolutely by the Direct Comparison Test.
34. ∞
=2
= ∞
=2
(−1)
√ ln = ∞
=2(−1). Now = 1
√ ln 0for ≥ 2, {} is decreasing for ≥ 2, and lim
→∞= 0, so the series∞
=2
(−1)
√ ln converges by the Alternating Series Test. Also, observe that
√(−1)
ln
= 1
√ ln 1
√√
= 1
for ≥ 2, so the series∞
=2
(−1)
√ ln
is divergent by direct comparison with
∞
=2
1
, which is a divergent (partial) series [ = 1 ≤ 1]. Thus,∞
=2
(−1)
√ ln is conditionally convergent.
35. By the recursive definition, lim
→∞
+1
= lim→∞
5 + 1 4 + 3
= 5
4 1, so the series diverges by the Ratio Test.
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
36. A seriesP an is defined by the equations
a1= 1 an+1= 2 + cos n
√n an
Determine whetherP an converges or diverges.
Solution:
152 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
34. lim
→∞
+1
= lim→∞
( + 1)52+2
10+2 · 10+1
52
= lim→∞
52( + 1)
10 = 5
2 lim
→∞
1 + 1
= 5
2(1) = 5
2 1, so the series
∞
=1
52
10+1 diverges by the Ratio Test. Or: Since lim
→∞= ∞, the series diverges by the Test for Divergence.
35. lim
→∞
|| = lim→∞ ln
= lim→∞
ln = lim
→∞
ln
= limH
→∞
1
1 = lim
→∞ = ∞, so the series
∞
=2
ln
diverges by the Root Test.
36.
sin(6) 1 + √
≤ 1 1 + √
1
32, so the series
∞
=1
sin(6) 1 + √
converges by comparison with the convergent -series
∞
=1
1
32 ( = 32 1). It follows that the given series is absolutely convergent.
37. Since 0 ≤ 1
3 ≤
3 =
1
3
and ∞
=1
1
3 is a convergent -series [ = 3 1], ∞
=1
1
3 converges, and so
∞
=1
(−1)1
3 is absolutely convergent.
38. The function () = 1
ln is continuous, positive, and decreasing on [2 ∞).
∞
2
1
ln = lim
→∞
2
1
ln = lim
→∞[ln(ln )]2= lim
→∞[(ln(ln ) − ln(ln 2)] = ∞, so the series
∞
=2
(−1)
ln diverges by the Integral Test. Now {} =
1
ln
with ≥ 2 is a decreasing sequence of positive terms and lim
→∞= 0. Thus,
∞
=2
(−1)
ln converges by the Alternating Series Test. It follows that
∞
=2
(−1)
ln is conditionally convergent.
39. By the recursive definition, lim
→∞
+1
= lim→∞
5 + 1 4 + 3
= 5
4 1, so the series diverges by the Ratio Test.
40. By the recursive definition, lim
→∞
+1
= lim→∞
2 + cos
√
= 0 1, so the series converges absolutely by the Ratio Test.
41. The series ∞
=1
cos
= ∞
=1(−1)
, where 0for ≥ 1 and lim
→∞ = 1 2.
lim→∞
+1
= lim→∞
(−1)+1+1
+ 1 ·
(−1)
= lim→∞
+ 1 = 1
2(1) = 1
2 1, so the series ∞
=1
cos
is
absolutely convergent by the Ratio Test.
42. lim
→∞
+1
= lim→∞
(−1)+1( + 1)!
( + 1)+112· · · +1 ·12· · · (−1)!
= lim→∞
(−1)( + 1)
+1( + 1)+1
= lim→∞
+1( + 1)
= lim
→∞
1
+1
+ 1
= lim
→∞
1
+1
1
1 + 1
= lim
→∞
1
+1(1 + 1) = 1
1 2 = 2
1 so the series ∞
=1
(−1)!
123· · · is absolutely convergent by the Ratio Test.
° 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.c
39. For which of the following series is the Ratio Test inconclusive (that is, it fails to give a definite answer)?
(a)
∞
P
n=1 1 n3 (b)
∞
P
n=1 n 2n (c)
∞
P
n=1 (−3)n−1
√n (d)
∞
P
n=1
√n 1+n2
Solution: SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ¤ 153 43. (a) lim
→∞
1( + 1)3 13
= lim→∞
3
( + 1)3 = lim
→∞
1
(1 + 1)3 = 1. Inconclusive (b) lim
→∞
( + 1) 2+1 · 2
= lim→∞
+ 1 2 = lim
→∞
1 2+ 1
2
= 1
2. Conclusive (convergent) (c) lim
→∞
√(−3)
+ 1 ·
√ (−3)−1
= 3 lim→∞
+ 1 = 3 lim
→∞
1
1 + 1 = 3. Conclusive (divergent)
(d) lim
→∞
√ + 1
1 + ( + 1)2 · 1 + 2
√
= lim→∞
1 + 1
· 12+ 1 12+ (1 + 1)2
= 1. Inconclusive
44. We use the Ratio Test:
lim→∞
+1
= lim→∞
[( + 1)!]2/[( + 1)]!
(!)2/()!
= lim→∞
( + 1)2
[( + 1)] [( + 1) − 1] · · · [ + 1]
Now if = 1, then this is equal to lim
→∞
( + 1)2 ( + 1)
= ∞, so the series diverges; if = 2, the limit is
lim→∞
( + 1)2 (2 + 2)(2 + 1)
= 1
4 1, so the series converges, and if 2, then the highest power of in the denominator is larger than 2, and so the limit is 0, indicating convergence. So the series converges for ≥ 2.
45. (a) lim
→∞
+1
= lim→∞
+1 ( + 1)! · !
= lim→∞
+ 1
= || lim→∞
1
+ 1 = || · 0 = 0 1, so by the Ratio Test the series ∞
=0
! converges for all .
(b) Since the series of part (a) always converges, we must have lim
→∞
! = 0by Theorem 11.2.6.
46. (a) = +1+ +2+ +3+ +4+ · · · = +1
1 + +2
+1
++3
+1
++4
+1 + · · ·
= +1
1 ++2
+1
++3
+2
+2
+1
+ +4
+3
+3
+2
+2
+1 + · · ·
= +1(1 + +1+ +2+1+ +3+2+1+ · · · ) ()
≤ +1
1 + +1+ +12 + +13 + · · ·
[since {} is decreasing] = +1
1 − +1
(b) Note that since {} is increasing and → as → ∞, we have for all . So, starting with equation (),
= +1(1 + +1+ +2+1+ +3+2+1+ · · · ) ≤ +1
1 + + 2+ 3+ · · ·
= +1
1 − . 47. (a) 5=
5
=1
1
2 = 1 2 +1
8+ 1 24 + 1
64 + 1
160 = 661
960 ≈ 068854. Now the ratios
= +1
= 2
( + 1)2+1 =
2( + 1)form an increasing sequence, since
+1− = + 1
2( + 2)−
2( + 1) = ( + 1)2− ( + 2)
2( + 1)( + 2) = 1
2( + 1)( + 2) 0. So by Exercise 46(b), the error in using 5is 5≤ 6
1 − lim
→∞
= 1 6 · 26 1 − 12 = 1
192 ≈ 000521.
° 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.c
41. (a) Show that
∞
P
n=0 xn
n! converges for all x.
(b) Deduce that lim
n→∞
xn
n! = 0 for all x.
Solution:
SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ¤ 153 43. (a) lim
→∞
1( + 1)3 13
= lim→∞
3
( + 1)3 = lim
→∞
1
(1 + 1)3 = 1. Inconclusive (b) lim
→∞
( + 1) 2+1 ·2
= lim→∞
+ 1 2 = lim
→∞
1 2+ 1
2
= 1
2. Conclusive (convergent) (c) lim
→∞
√(−3)
+ 1 ·
√ (−3)−1
= 3 lim→∞
+ 1 = 3 lim
→∞
1
1 + 1 = 3. Conclusive (divergent)
(d) lim
→∞
√ + 1
1 + ( + 1)2 ·1 + 2
√
= lim→∞
1 + 1
· 12+ 1 12+ (1 + 1)2
= 1. Inconclusive
44. We use the Ratio Test:
lim→∞
+1
= lim→∞
[( + 1)!]2/[( + 1)]!
(!)2/()!
= lim→∞
( + 1)2
[( + 1)] [( + 1) − 1] · · · [ + 1]
Now if = 1, then this is equal to lim
→∞
( + 1)2 ( + 1)
= ∞, so the series diverges; if = 2, the limit is
lim→∞
( + 1)2 (2 + 2)(2 + 1)
= 1
4 1, so the series converges, and if 2, then the highest power of in the denominator is larger than 2, and so the limit is 0, indicating convergence. So the series converges for ≥ 2.
45. (a) lim
→∞
+1
= lim→∞
+1 ( + 1)!· !
= lim→∞
+ 1
= || lim→∞
1
+ 1 = || · 0 = 0 1, so by the Ratio Test the series ∞
=0
! converges for all .
(b) Since the series of part (a) always converges, we must have lim
→∞
! = 0by Theorem 11.2.6.
46. (a) = +1+ +2+ +3+ +4+ · · · = +1
1 ++2
+1
+ +3
+1
++4
+1 + · · ·
= +1
1 ++2
+1
++3
+2
+2
+1
++4
+3
+3
+2
+2
+1 + · · ·
= +1(1 + +1+ +2+1+ +3+2+1+ · · · ) ()
≤ +1
1 + +1+ 2+1+ 3+1+ · · ·
[since {} is decreasing] = +1
1 − +1
(b) Note that since {} is increasing and → as → ∞, we have for all . So, starting with equation (),
= +1(1 + +1+ +2+1+ +3+2+1+ · · · ) ≤ +1
1 + + 2+ 3+ · · ·
= +1
1 − . 47. (a) 5 =
5
=1
1
2 = 1 2+ 1
8+ 1 24 + 1
64 + 1
160 = 661
960 ≈ 068854. Now the ratios
= +1
= 2
( + 1)2+1 =
2( + 1) form an increasing sequence, since
+1− = + 1
2( + 2)−
2( + 1) = ( + 1)2− ( + 2)
2( + 1)( + 2) = 1
2( + 1)( + 2) 0. So by Exercise 46(b), the error in using 5is 5≤ 6
1 − lim
→∞
= 1 6 · 26 1 − 12 = 1
192 ≈ 000521.
° 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.c
1
42. Let P an be a series with positive terms and let rn= an+1/an. Suppose that lim
n→∞rn= L < 1, so P an converges by the Ratio Test. As usual, we let Rn be the remainder after n terms, that is,
Rn = an+1+ an+2+ an+3+ · · ·
(a) If {rn} is a decreasing sequence and rn+1< 1, show, by summing a geometric series, that Rn ≤ an+1
1 − rn+1
(b) If {rn} is an increasing sequence, show that
Rn ≤ an+1
1 − L Solution:
SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS ¤ 153 43. (a) lim
→∞
1( + 1)3 13
= lim→∞
3
( + 1)3 = lim
→∞
1
(1 + 1)3 = 1. Inconclusive (b) lim
→∞
( + 1) 2+1 ·2
= lim→∞
+ 1 2 = lim
→∞
1 2+ 1
2
= 1
2. Conclusive (convergent) (c) lim
→∞
√(−3)
+ 1 ·
√ (−3)−1
= 3 lim→∞
+ 1 = 3 lim
→∞
1
1 + 1 = 3. Conclusive (divergent)
(d) lim
→∞
√ + 1
1 + ( + 1)2 ·1 + 2
√
= lim→∞
1 + 1
· 12+ 1 12+ (1 + 1)2
= 1. Inconclusive
44. We use the Ratio Test:
lim→∞
+1
= lim→∞
[( + 1)!]2/[( + 1)]!
(!)2/()!
= lim→∞
( + 1)2
[( + 1)] [( + 1) − 1] · · · [ + 1]
Now if = 1, then this is equal to lim
→∞
( + 1)2 ( + 1)
= ∞, so the series diverges; if = 2, the limit is
lim→∞
( + 1)2 (2 + 2)(2 + 1)
= 1
4 1, so the series converges, and if 2, then the highest power of in the denominator is larger than 2, and so the limit is 0, indicating convergence. So the series converges for ≥ 2.
45. (a) lim
→∞
+1
= lim→∞
+1 ( + 1)!· !
= lim→∞
+ 1
= || lim→∞
1
+ 1 = || · 0 = 0 1, so by the Ratio Test the series ∞
=0
! converges for all .
(b) Since the series of part (a) always converges, we must have lim
→∞
! = 0by Theorem 11.2.6.
46. (a) = +1+ +2+ +3+ +4+ · · · = +1
1 ++2
+1
+ +3
+1
++4
+1 + · · ·
= +1
1 ++2
+1
++3
+2
+2
+1
++4
+3
+3
+2
+2
+1 + · · ·
= +1(1 + +1+ +2+1+ +3+2+1+ · · · ) ()
≤ +1
1 + +1+ 2+1+ 3+1+ · · ·
[since {} is decreasing] = +1
1 − +1
(b) Note that since {} is increasing and → as → ∞, we have for all . So, starting with equation (),
= +1(1 + +1+ +2+1+ +3+2+1+ · · · ) ≤ +1
1 + + 2+ 3+ · · ·
= +1
1 − . 47. (a) 5 =
5
=1
1
2 = 1 2+ 1
8+ 1 24 + 1
64 + 1
160 = 661
960 ≈ 068854. Now the ratios
= +1
= 2
( + 1)2+1 =
2( + 1) form an increasing sequence, since
+1− = + 1
2( + 2)−
2( + 1) = ( + 1)2− ( + 2)
2( + 1)( + 2) = 1
2( + 1)( + 2) 0. So by Exercise 46(b), the error in using 5is 5≤ 6
1 − lim
→∞
= 1 6 · 26 1 − 12 = 1
192 ≈ 000521.
° 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.c
2
