Power series

A power series in the variable x is a series of the form

,

where the coefficients , , ,… are real or complex numbers.

We are (in principle) allowed to put the variable x equal to any number we wish. For instance, with the power series

we may put x = 2 and we obtain the numerical series

Or, with the power series

(note that in this case all the coefficients are equal to 1) we obtain

if we put .

However, we have a problem: how do we know that the numerical series we obtain by putting in the power series

is convergent?

We can always put and then investigate the convergence of the numerical series, but this is a rather inefficient way of deciding if a particular value of x gives us a convergent series. So we come to the following question:

Given the power series

Theorem 1. For a power series

there are three possibilities:

1. The power series diverges for all ;

2. The power series converges for all values of x;

3. There is a positive number R such that converges for all values of x with and diverges for all values of x with .

At first sight, this looks like a very useless result, because it doesn't answer the question of which values of x are allowed. However, it is a very useful result: it tells us what sort of behavior we can expect, and what to look for in a power series. In particular, it tells us what is the decisive factor in our subsequent investigation: we need to find the number R, which is called the radius of convergence of the power series. One word of warning: the theorem tells us that, when we have found the radius of convergence R, then the series converges for and diverges whenever , but it doesn't say anything about the case when , that is, when and . The theorem is silent on this matter. In fact, we must investigate the cases

separately (that is, we put in or and we investigate the convergence of the

series that arise). This is a small price to pay when we know what R is.

So, given our theorem, how do we go about calculating R? One result is the following:

Theorem 2. Given the power series , suppose that one of the following limits exist:

, .

Then the following is true:

1. If then the power series converges for all values of x;

2. If , then the radius of convergence R of the power series is .

If either of the limits

, .

fails to exist, then the power series diverges for all values of .

This theorem is proved using the following result:

Theorem 3. Suppose is a numerical series and suppose that one of the following

limits exists

, .

If then the series converges absolutely. If then the series diverges. If then convergence or divergence of the series must be investigated using some other method.

Definition: Let denote a power series about . The Interval of Convergence (sometimes denoted by IOC) for is the set of all values of x for which the power series converges. It is related to the ROC by the following where R is the ROC for :

• If , then the interval of convergence is or, equivalently, .
• If , then the interval of convergence is the singleton set .
• If , then the series converges for , or . So the interval of convergence contains the interval together with any endpoints for which the power series converges.

Examples: Below the examples from the previous class are examined to determine their respective intervals of convergence. The answers are based on the previous work in determining the radius of convergence. In order to find the interval of convergence for a power series, one first finds the radius of convergence.

1) Consider the power series :

Since this power series is the Geometric Series, we know that the series converges if and only if . So the interval of convergence is , or equivalently, .

2) Consider the power series :

The radius of convergence for the power series is . So the interval of convergence is , or equivalently, .

3) Consider the power series :

The radius of convergence for the power series is zero. So the interval of convergence is the singleton set .

4) Consider the power series :

The radius of convergence is and the series converges for .

Check the endpoints:

For : which is the Harmonic Series and therefore diverges.

For : which is the Alternating Harmonic Series and therefore converges.. The interval of convergence is , or equivalently, .

5) Consider the power series :

The radius of convergence is and the series converges for .

Check the endpoints:

For : .Since does not exist and so , the series diverges by the Divergence Test.

For : . Since , the series diverges by the Divergence Test.

The interval of convergence is , or equivalently, .

6) Consider the power series :

The radius of convergence is and the series converges for .

Check the endpoints:

For : which is an alternating series with . Applying the Alternating series Test, . Further . So by the Alternating Series Test the series converges.

For: which is the p -series with . So the series converges.

The interval of convergence is , or equivalently, .

On the interval of convergence the sum of the power series may be thought of as a function. That is, if I represents the interval of convergence for the power series , then the function f may be defined as for all . If the interval of convergence is a singleton set (), then the function is of little interest. So when we are considering the sum of the series as a function, we generally assume that the radius of convergence is either a positive finite number or . This fact may be used to obtain other results as illustrated in the example below.

7) Consider the power series .

Normally, we see this series centered at with the sequence of coefficients defined as , and . However we could consider the power series . If we find the ROC using the Ratio Test, . So and the series has as its interval of convergence. Thus for all , we can define . Then

for all , or equivalently, .

8) Consider the power series and find the interval of convergence.

Apply the Ratio Test to find the ROC:

. So the series converges for . So the ROC is and .

Checking the endpoints:

For : . Since , the series diverges by the Divergence Test.

For : . Since , the series diverges by the Divergence Test. So the interval of convergence is .

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