**
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 integersuch 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 *n*th 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,
*.* |