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a
97
/
a
(1
1/
n
)
e, so by qn 95, radius of convergence is 1/e, by
qn 4.36.
98 Useqn 95.
a
n
a
n
a
n
n
a
n
n
1
and
1.
1
1
99
a
1/
x
a
x
1, so
R
0.
a
100 Take
/
x
, then
a
x
(
)
.
R
.
101
x
2
x
1
(
x
)
is absolutely convergent.
Also
x
2
x
1.
But
x
2
x
1
(
x
)
divergent.
a
a
. Lim sup
a
For
n
odd,
. For
n
even,
.
102
A
is the greatest limit of a subsequence of (
a
).
When
x
1/
A
we prove that the series is divergent.
Take
A
1/
x
; then there is a subsequence of (
a
) which
tends to
A
and so has infinitely many terms such that
A
a
,
x
a
and thus infinitely many terms for which 1/
, giving
a
x
1
. So the sequence of terms of the power series is not null.
x
When
1/
A
we prove that the series is convergent.
Choose
r
(1/2)(
x
1/
A
) for example and take
1/
r
A
.If
a
u
sup
:
n
k
, the sequence (
u
) is decreasing and tends to
A
,
so
u
A
for suMciently large
k
and hence
a
A
for that
large
n
. So, eventually, all
a
A
1/
r
.
The radius of convergence is 1/
A
.
103
a
1, or
a
n
, for example.
104
a
1/
n
.
105
a
1/
n
.
106 The first series is convergent when
1
x
1. Thescond and third
when
1. All three have the same radius of convergence. Note
the effect of differentiation on the first two functions.
1
x
107 lim sup
lim sup
n
·
lim sup
na
a
a
.
2 ·
(
n
1)
(2
n
)
n
1.
So lim sup
lim sup
/
lim sup
a
/(
n
1)
a
(
n
1)
a
.
108
c
a
b
a
b
...
a
b
...
a
b
.
