MATH1120 Calculus II
Solutions to Selected Exercises in Notes 10
October 31, 2013
Exercise 1.1 Since all the terms of the given series are nonnegative, we may add brackets
to, remove brackets from, and rearrange the terms to find the sum.
∞
X
1
1
1
1
1
n
1
=
+
+
+
+
+ ···
+
8n
8
82 82
83 83 83
n=1
1
8
1
82
1
83
+ ···
1 − 18
1 − 18
1 − 18
1
1
1
1
1
1
1
+ ···
=
+ 2 + 3 + ··· +
+
+
·
·
·
+
+
8 8
8
82 83
83 84
=
+
+
1
8
=
1− 81
1−
64
=
392
1
8
1
is decreasing, nonnegative and conx(ln x)2
tinuous for x ≥ 2. We may use the Integral Test to determine the convergence of the
series.
Z b
Z ∞
1
1
dx = lim
dx
2
b→∞ 2 x(ln x)2
x(ln x)
2
1 b
= lim −
b→∞
ln x 2
1
=
ln 2
Since the associated improper integral converges, so does the given series.
1
Exercise 2.2 Note that the function y =
is decreasing, continuous and
x(1 + ln2 x)
nonnegative for x ≥ 2. Thus the integral test applies.
Z ∞
Z b
1
1
dx = lim
dx
2
b→∞ 2 x(1 + ln2 x)
x(1 + ln x)
2
b
= lim tan−1 ln x 2
Exercise 2.1 Note that the function y =
b→∞
π
− tan−1 ln 2
2
Since the associated improper integral converges, so does the given series.
Exercise 2.3
=
1
2
(1) Note that for any positive number c, there exists a positive integer N such that
such that for all n > N ,
ln n < nc
1
Take c = . Then
8
(ln n)2
n
∞
X
1
5
3
2
converges as p =
1
<
n4
n
3
2
1
=
5
for all n > N
n4
∞
X
5
(ln n)2
> 1. By the Direct Comparison Test,
3
4
n2
n4
n=1
converges as well.
∞
X
1
1
1
< 2 for all n > 0 and
converges because p = 2 > 1,
(2) Since 2
n + 30
n
n2
n=1
∞
X
1
converges by the Direct Comparison Test.
n2 + 30
n=1
n=1
Exercise 2.4
(1) By picking the dominant term of each factor of each term of the given series, we
∞
∞
X
X
1
n·n
=
to be the series with which we compare the given series.
choose
n2 · n
n
n=1
n=1
n(n+1)
(n2 +1)(n−1)
lim
1
n→∞
n
n2 (n + 1)
n→∞ (n2 + 1)(n − 1)
= lim
= lim
n→∞
1+
1+
1
n2
1
n
1−
1
n
1
2
>0
=
Since
∞
X
1
diverges, so does the given series by the Limit Comparison Test.
n
n=1
3
(2) Note that
lim
ln 1 +
1
n2
n→∞
1
n2
1
= lim n ln 1 + 2
n→∞
n
2
1 n
= lim ln 1 + 2
n→∞
n
2!
1 n
= ln lim 1 + 2
n→∞
n
2
= ln e
=1
>0
(3) Note that
∞
X
n=1
∞
X
2
1
=
.
1 + 2 + ··· + n
n(n + 1)
n=1
2
n(n+1)
lim
1
n→∞
n2
2n2
n→∞ n(n + 1)
2
= lim
n→∞ 1 + 1
n
=2
>0
= lim
∞
X
1
converges, so does the original series by the Limit Comparison Test.
Since
n2
n=1
P
2
2
2
2
Observe that
= −
. The n-th partial sum sn of ∞
n=1 n(n+1) is
n(n + 1)
n
n+1
2
. Thus
2−
n+1
∞
X
2
= lim sn
n(n + 1) n→∞
n=1
= lim 2 −
n→∞
2
n+1
=2
∞
X
sin 1
1
1
> 0 for all n, and lim 1 n = 1 > 0. Since
diverges, so
n→∞
n
n
n
n=1
does the given series by the Limit Comparison Test.
(4) Note that sin
4
(5) By picking the dominant term of each factor of each term of the given series, we
∞
∞
X
X
n
1
choose
=
to be the series with which we compare the given
3
2
n ln n
n ln n
n=2
n=1
series.
n+1
(n3 +1) ln n
lim
1
n→∞
n2 ln n
(n + 1)n2
n→∞ n3 + 1
= lim
1 + n1
n→∞ 1 + 13
n
=1
>0
= lim
∞
X
∞
X
n+1
1
So both
and
have the same convergence property by
3
2
(n + 1) ln n
n ln n
n=2
n=1
the Limit Comparison Test. Note that
1
1
< 2 for all n > 1
2
n ln n
n
∞
X
1
1
and
converges, so does 2
by the Direct Comparison Test. Consequently
2
n
n ln n
n=1
the given series converges.