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convergent by the first comparison test (qn 26) and the start rule (qn
16).
52
a
1
1/
n
1
b
2/
n
.
Choose
.
b
is divergent from qn 28, so
b
is divergent by the
scalar rule(qn 19). Now
n
N
0
b
a
,so
a
is divergent by
thefirst comparison tst (qn 26) and thestart rule(qn
16).
53 Since0
mb
a
,
a
convergent
mb
convergent by the first
comparison test
b
is convergent by the scalar rule.
a
Mb
b
Mb
Since0
,
convergent
convergent by the scalar
a
rule
convergent by thefirst comparison tst.
54
a
/
b
1/
n
. The test requires a positive number
m
such that
0
m
a
/
b
.
l
.
n
N
l
a
55 Choose
l
. Now apply thescond
comparison test (qn 53) and the start rule (qn 16).
0
/
b
56 When
n
x
n
1, 1/(
n
1)
1/
x
1/
n
, and see chapter 10. Sum
the inequalities for
n
1, 2, 3, . . .,
n
1.
s
1
log
n
s
1/
n
.
D
1,
D
0.807 . . .,
D
0.735 . . .,
D
0.697 . . .
1
n
dx
x
E
E
1
0
from thefirst inequality in thequstion.
dx
x
D
E
1/
n
.
So (
D
E
) is a null sequence by the squeeze rule.
57 (
f
(
n
)) is monotonic decreasing and bounded below, by 0, so it is
convergent (qn 4.34). Since
f
is decreasing,
n
x
n
1
f
(
n
1)
f
(
x
)
f
(
n
). Lower sum
integral
upper
sum. Add the inequalities for
n
1, 2, 3, . . .,
n
1. (
s
) is monotonic
increasing since
f
is non-negative. Since
s
f
(1)
f
(
x
)
dx
,
if the integral is bounded above, (
s
) is bounded above and so is
convergent by qn 4.35. Conversely
