Graphics Reference
In-Depth Information
b
for all
r
A
B
a
for all
n
.
0
a
b
both (
A
) and (
B
) aremonotonic increasing.
(
B
) convergent
(
B
) bounded
(
A
) bounded
(
A
) convergent by
qn 4.35.
a
divergent
(
A
)
(
B
)
b
divergent.
27
1
(
n
1
n
.
1)
28
101
1.
(i) With thenotation of qn 26,
A
(
n
1)
1
as
n
.
(iii)
n
(
n
1)
2
n
n
(
n
1)
n
)
(
n
n
).
1/(2
1/(
1)
29
2
0
1/
n
1/
n
1/
n
is convergent, by the first comparison
test.
0
n
1/
n
1/
n
0
1/
is divergent, by the first
comparison test.
30 Partial sums areunbounded, so
1/
n
is divergent.
1/
n
1/
n
1/
n
31
1
0
is divergent.
32
1
1
0. 2
1
2
1
, though wehaveonly proved
this for rational
so far, from 2.20. The result for all real numbers
follows from qn 11.34. Thus 1/2
1. So all partial
sums
1/(1
1/2
).
33 0
(
n
/(2
n
1))
(
)
. Now useqn 5 and thefirst comparison tst.
34
n
3
8
n
72
(180
172)
n
72
36(5
n
2)
43ยท 4
n
.Asa
geometric progression
(
)
is convergent (qn 9), so
(4
n
/(5
n
2))
is convergent by the first comparison test (qn 26) for
n
3. From the
start rule(qn 16)
(4
n
/(5
n
2))
is convergent.
35 There is an
N
such that
n
N
a
k
(1
k
). So
n
N
0
a
(1
k
)
1. Now, as a geometric progression
(qn 9)
is convergent. So by the first comparison test (qn 26)
a
is
convergent for
n
N
. From thestart rule(qn 16)
a
is convergent.
36 (
and (
(
n
/2
))
(
n
(0.8)
))
0.8, using qn 3.59 in each case.
37
a
1
a
1
(
a
) is not null. Null sequence test (qn 11).
