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when
m
n
. Establish this inequality when
m
n
and when
m
n
, to show that every even partial sum is less
than every odd partial sum. Deduce that the intervals [
s
s
shows that
s
,
s
]
are nested, and that the intersection of all these intervals (the
Chinese Box Theorem, qn 4.42) contains just one number, the sum
of the series. We will find the limit in qn 65.
63 (
Leibniz
, 1682) The argument of qn 62 is easily generalised to
establish the convergence of any series of the form
(
1)
a
when (
a
) is a monotonic decreasing null sequence of positive terms.
This is called the
alternating series text
.
In this case, let
s
(
1)
a
.
Show that
s
s
0,
and that
s
0.
From theproposition
s
s
s
a
s
s
, show that the
intervals [
s
,
s
] are nested, and their intersection contains only
onenumbr.
Deduce that
(
1)
a
is convergent
64 Why is the series 1
...
(
1)
...
convergent?
65 We examine the series of qn 62, using the symbols
D
and
from
qn 56.
1
2
1
3
1
4
...
1
2
n
1
1
2
1
3
1
4
1
2
n
1
2
1
4
1
2
n
1
...
2
...
log 2
n
D
(log
n
D
)
log 2
D
D
.
Now usethefact that (
D
)
as
n
, to provethat thesum of the
series in qn 62 is log
2.
Absolute convergence
66 For the series
(
1)
/
n
both the subseries of positive terms and the
subseries of negative terms, taken separately, are divergent. In the
case of the series
(
1)
/
n
, the subseries of positive terms and the
subseries of negative terms, taken separately, are convergent.
For
(
1)
/
n
, the alternating series test will establish convergence,
but a more direct proof is available in such a case. We construct
two new series thus.
