Graphics Reference
In-Depth Information
70 (
a
k
)
1
is absolutely convergent.
71 Since
a
u
v
,
u
a
v
and
v
u
a
. Now
a
is
convergent, so
are both convergent or both divergent by
the sum rule and the scalar rule. If both were convergent, then so
would
u
and
v
(
u
v
) beby thesum rul, and
u
v
a
. But
a
is
divergent, so both
u
and
v
are divergent and both are series of
non-negative terms.
72
s
0,
s
1/(
n
1). So (
s
)
0.
a
1/
n
,
a
1/
n
.
73 Denote the partial sum of
n
terms in qn 62 by
t
.
s
t
,
s
t
1/
n
,
s
t
1/2
n
,so(
s
) and (
t
) havethesame
1/(2
n
limit.
b
1/
n
,
b
1/2
n
,
b
1).
74
a
b
,
a
b
,
a
b
.
a
0.
b
log 2.
1
3
1
2
1
5
1
7
1
4
1
9
1
11
1
6
.
75
(i) 1
(iii) Yes, by the alternating series test.
(iv) Apply hint. L.H.S.
2/
n
which gives a decreasing null
sequence.
1
1
R.H.S.
3/4
n
1/4
n
2,
1
1
which decreases but is bounded below by 2
2.
When
n
12, L.H.S.
2
2.
(v) Use first comparison test. Series is divergent by comparison with
1/
n
.
77
u
and
v
are both convergent by comparison with
a
.
x
is a
rearrangement of
u
, and
y
is a rearrangement of
v
. But
u
and
v
are series of positive or zero terms, so
x
u
and
y
v
.
79
1,
divergent when
x
1. Divergent because terms not a null sequence
when
x
1.
a
/
a
x
(1
1/
n
)
x
. Series convergent when
x
80
a
x
n
x
.
81
a
x
(
n
)
x
. Conclusion as in qn 79.
