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Definition. A power series in z
is a
series of the form
where  are complex constants.
The
operations of addition and subtraction of power series are
defined in a natural
way.
For multiplication we use
Cauchy’s Product.
We
define the product of
and
 as the power series
 , where
Theorem 1.
Suppose there exists some
 such that
 converges. Then for each z such
that   , the series
 converges absolutely.
Proof.
Let
r be a real number such that
 . Since
converges,
 .
Thus, { }is bounded. So there exists a constant
M>0 such that
 for all n.
Thus
 =  where
 and so the series
converges absolutely by the comparison
test with a convergent geometric series.
Theorem 2. A
power series must satisfy one of the following
1)
The series converges for
 only.
2)
The series converges for all z.
3)
There exists an R>0 such that the series converges
for  and diverges for
We
call the radius of convergence of the power series
the
extended nonnegative number R (Note that in the
first case we can take R=0 and in the second
R=∞).
Proof.
Without loss of generality assume that
 =0. Assuming that 1) and 2) do
not hold, it is enough to show that 3) does. Then the
series converges at some value
 and diverges at some other value
 . Let S denote the set of all
numbers µ such that the series converges for
 µ; S is nonempty by using Theorem 1.
Moreover, S contains no numbers larger than
 and so S is bounded from above.
By Real Analysis, S must have a least upper bound
which we call R. That R has the remaining
properties of 3) follows from Theorem
1.
Definitions.
Let
 be a sequence in
R.
The limit superior of
, denoted by
 ,
is
the extended real number defined by
 The limit inferior
of
, denoted by
 , is the extended real number
defined by
Theorem (Cauchy-Hadamard Rule).
R=
(with the convention that
 and
 ).
To
prove the Cauchy-Hadamard Rule, we first state and
prove the following
Lemma (Cauchy’s Root Test,
1821).
Let  be a sequence in
C.
Then
a)
If  then
 converges.
b)
If  then
diverges.
Proof.
a)
Choose  such that
 . Then, in particular,
 So there exists an integer such that
  for otherwise the previous step is
contradicted. Thus
 or
 . Since r<1, converges and so
 converges. It follows that
converges absolutely and hence
converges.
b)
Suppose now that
 . Then
 So for every
 . In particular,
 . Thus there exists an integer
 such that
 for otherwise the previous step is
contradicted. Again since for every
we have
 . Therefore there exists an integer
 such that
 . Proceeding like this one can show
that there exists a subsequence
 of the sequence
 such that
 (i.e,
 and so
 and consequently
diverges.
Before proving the Cauchy-Hadamard Rule, let us
mention the corresponding:
d’Alembert Ratio Test
(1768). Let
be a sequence in
C
with  . Then
a)
If  then
converges.
b)
If  then
diverges.
whose proof follows by putting
 in the following inequality (proof left
as an exercise)
and
applying the root test.
Proof of the Cauchy-Hadamard Rule.
Let
 . Then
 .
By
the root test, if |z|L<1, then
converges while if |z|L>1, then
diverges. That is,
if
 , then
converges and
if
 , then
diverges.
Consequently R=
.
Remark 1.
Note
that R is not necessarily
 as we shall see next:
It
is enough to show that
 can happen. Indeed, let
 denote the series where
 if n is even and
 if n is odd.
Since  , the nth term of a convergent
(geometric) series,
converges.
Now
 = So
 But, considering n odd,
 =
Consequently,
=0<2=
Remark 2.
If
 and
 exist, then they must be equal (and as
a result R=):
Let
c= and d= and without loss of generality assume
that c<d and get a contradiction. Choose
 such that
 and let
 = and consider the series
On
the one hand,  and the series
 is convergent while, on the other hand,
 and the series
is divergent.
Remark 3.
In
general, we have
  :
Let
L= . Then
 L.
By
the ratio test, when |z|L<1,
 converges. That is,
When
 ,
converges. Consequently,
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