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In-Depth Information
f
(
x
)
dx
s
so that, if (
s
) is convergent and so bounded, the integral is bounded
above, and so convergent by 4.35.
58
1/
r
dx
/
x
1
s
2
1/
n
.
59 0
1/
r
dx
/
x
1
0
1/
r
[2
x
]
1.
dx
x
ยท log
60
x
[log(log
x
)]
log(log
n
)
log(log2).
Now claim thecontrapositiveof theintegral tst, qn 57.
61 The sequence (
D
) is monotonic decreasing but positive, and so,
bounded below. Thus it is convergent from qn 4.34.
62
s
s
1/(2
n
1)
1/(2
n
2)
1/(2
n
1)(2
n
2)
0, so (
s
)
is strictly increasing.
s
s
1/(2
n
2)
1/(2
n
3)
1/(2
n
3)(2
n
2)
0,
so (
s
) is strictly decreasing. When
m
n
,
s
s
s
. When
m
n
,
s
s
s
. The length of the interval [
s
,
s
]is
1/(2
n
1) which tends to 0, so at most one point is contained in all the
intervals. But there is at least one point from the Chinese Box
Theorem, qn 4.42.
63 Argument reflects qn 62.
64 (1/(2
n
1)) is a monotonically decreasing null sequence of positive
terms. Apply the alternating series test. The sum is in fact
/4.
65
s
log 2
D
D
,so
s
log 2
.
a
66 (i) and (ii): usefirst comparison tst with
. (iii) From thesum rule
u
v
(qn 21) and thescalar rule(qn 19). (iv)
a
.
67
u
0 when
a
0,
u
a
when 0
a
.
v
a
when
a
0,
v
0 when 0
a
.
0
u
,
v
a
,so
u
and
v
are convergent by the first
comparison test. Now
(
u
v
) is convergent by the scalar rule and
thesum rul.
a
u
v
.
a
k
a
68
/
a
1
is absolutely convergent.
69 Yes, by the null sequence test.
