A power series is a series of the form , where for each and .
Such series are often used to define functions, such as
Obviously, if we are going to define new functions in this way, it is important to know when such a series converges. As indicated in the last equation, we generally think of the numbers as fixed numbers (or fixed coefficients) and the as a variable.
Power series are also used to define functions on the complex numbers too:
for . Although these notes are primarily about real numbers and real analysis, and our proofs are presented for the reals only, the area of power series is one important example where what we shall say also works equally well, and to great effect, for the complex numbers too.
The main result about power series is that each power series has a radius of convergence, , which is a real number such that:
This result, i.e., the existence of such a radius of convergence,
in fact works equally well for the complexes as for the reals.
radius arises from thinking about the set
of complex numbers with in the Argand
diagram: this set is a circular disc with certre and
The radius of convergence, could turn out to be any number between and , including both extremes and . Of course if that means that the power series converges for all possible values of . If it means that the power series diverges for all possible values of , except itself. (It is easy to see that all power series of the form converge for as there is at most one nonzero term in the series, the one for .)
One further point to bear in mind is that the theorem on radius of convergence says nothing about what actually happens on the perimeter of this circular disc, i.e., when . In fact, for some power series, the series converges everywhere on this boundary, and for others it converges nowhere on the boundary. For yet others, it converges at some points but not at others in a rather complicated way. This point is related to the ratio test, where the limit of the ratio is : in such cases, the situation may be highly complicated and a simple test cannot give the answer.
We shall consider a power series and prove it has a radius of convergence, , as defined above. The main work involved takes place in the following theorem.
Suppose and converges, and let with . Then converges absolutely.
We have as , by the null sequence test, so
So, as , we have for all
so converges by the comparison test, using comparison with the series
which converges as .
Surprisingly, perhaps, this does all the work we need, and to get the radius of convergence theorem we just need to combine the last theorem with an old result about completeness of the reals.
Theorem on Radius of Convergence.
For all power series there is a nonnegative radius of convergence such that the power series converges absolutely for all with and the power series diverges for all with .
Let be the set
There are two cases depending on whether or not is bounded.
Case 1: is unbounded. We show we can take in this case.
Let be arbitrary.
Then because is unbounded there is with and hence there is with and converging. Thus by the previous result converges absolutely.
Therefore converges absolutely for every , as required.
On the other hand,
Case 2: is bounded.
In this case, is a bounded nonempty set (nonempty because it contains ) and so by the supremum form of the completeness of reals there is a least upper bound of . We have to show converges for and diverges for .
Let with be arbitrary.
Then since is the least upper bound of and is smaller, is not an upper bound of so there is with and hence there is with and converging. Thus by the previous result converges absolutely.
So . Also,
Let with be arbitrary.
Then if converges we would have . But this is impossible as is an upper bound of and . Therefore does not converge.
The series has radius of convergence . It converges absolutely if , and diverges for all other including and .
The ratio of consecutive terms of is . This is less than if , in which case the series converges absolutely by the ratio test. It is greater than one if , in which case the series diverges by the ratio test. Therefore the radius of convergence is .
On the radius of convergence we have or and the series is either or . Both these diverge by the null sequence test.
The series has radius of convergence . It converges absolutely if , converges conditionally when and diverges for all other including .
The ratio of consecutive terms of is which converges to . Therefore, using the ratio test as in the last example, the radius of convergence is and the series converges absolutely when and diverges when . For our series is the harmonic series, which diverges, and for our series converges by the alternating series test, but does not converge absolutely as the series of absolute values of terms is again the harmonic series.
The series has radius of convergence . It converges absolutely if , and diverges for all other .
Again, the ratio of consecutive terms of is which converges to . Therefore, by the ratio test again, the radius of convergence is and the series converges absolutely when and diverges when . For our series is which converges and is made up of positive terms only, hence converges absolutely. For it also converges absolutely as the absolute value of the terms in this series are the same terms as in the case .
The series has radius of convergence . It converges absolutely for all .
The ratio of consecutive terms of is as , so by the ratio test the series converges absolutely for all .
The series has radius of convergence . It converges for and diverges for all .
For the sequence is not null. To see this, let and let be any odd positive integer with . Then , , , and so on, hence . This is the base case of an induction argument that shows that for . (The induction step is easy as for such .) Since it follows that for and is not a null sequence.
Hence from the null sequence test that the series does not converge absolutely for any . From this it follows by the theorem on power series given above that does not converge at all for any . For if this series did converge with then would also converge.