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.


= 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 nth term of a convergent (geometric) series,  converges.

Now =  So  But, considering n odd,  = 



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,