Graphics Reference
In-Depth Information
n
1
2
3
4
5
6
. . .
a
1
...
u
0
0
0
...
v
1
0
0
0
. . .
In each case the terms of the new sequences (
u
) and (
v
) areequal
either to 0 or to
a
, so clearly 0
u
a
, and 0
v
a
.
Now
a
1/
n
and
1/
n
is convergent from qn 27, so
(i)
u
is convergent (why?),
(ii)
v
is convergent (why?)
(iii)
(
u
v
) is convergent (why?),
(iv)
a
is convergent.
a
a
67 A series
is a convergent is said to be
absolutely
convergent
. The argument used in qn 66 can be generalised to
establish that
a series which is absolutely convergent is necessarily
convergent
. The clue, as we have seen, is to construct the series
formed by the positive terms and the series formed by the negative
terms.
If wedefine
u
for which
(
a
a
), what values may
u
take?
If wedefine
v
(
a
a
), what values may
v
take?
Now use the convergence of
a
to provethat
u
and
v
are
v
v
convergent and hence that
(
u
) is convergent. What is
u
equal to?
68 Use d'Alembert's ratio test (qn 43) to devise a condition for a series
to be absolutely convergent. When combined with qn 67, this gives
a stronger form of d'Alembert's ratio test.
a
a
69 Does the condition
k
1as
n
imply that the
series
a
is divergent?
70 UseCauchy's
n
th root test (qn 35) to devise a condition for a series
to be absolutely convergent. When combined with qn 67, this gives
a stronger form of Cauchy's
n
th root test.
Conditional convergence
A series
a
which is convergent, but not absolutely convergent, is
said to be
conditionally convergent
. The series
(
1)
/
n
is conditionally
convergent.
71 With thenotation of qn 67, usetheequations
a
u
v
and
a
u
v
to provethat, if
a
is conditionally convergent, then
