Video: Power Series and Radius of Convergence

In this video, we will learn how to define a power series and find the radius of convergence of the power series.

15:47

Video Transcript

In this video, we’re going to learn how to define a power series and how these differ from previous series we might have come across. We’ll also learn how to find the radius of convergence and the interval of convergence of these sides of series using the ratio or root test.

Up until this stage, you’ll likely have worked with infinite geometric series, harmonic and 𝑝 series, and alternating series. Now we look to define a new type of series, which we call a power series. Power series are useful in analysis since they arise as Taylor series of infinitely differentiable functions. They can be of the form shown. It’s the sum of 𝑐 𝑛 times π‘₯ to the 𝑛th power for values of 𝑛 between zero and ∞. Here π‘₯ is a variable, whilst the 𝑐 𝑛s are the coefficient of our series.

Consider now what happens if we define a fixed value for π‘₯. For example, let π‘₯ be equal to one. In this series, we obtain a series of constants, which we can then test for convergence or divergence using our usual tests.

Now it’s important to realise that a power series may converge for some values of π‘₯ and diverge for others. We also define the sum of the series as the function 𝑓 of π‘₯, which is equal to 𝑐 naught plus 𝑐 one π‘₯, and so on. Whose domain is the set of all π‘₯ for which the series converges. We might notice that 𝑓 looks a lot like a polynomial, though we should also observe that it is slightly different from one since it has an infinite number of terms.

We might observe what happens when we take 𝑐 𝑛 to be equal to one. The power series becomes the geometric series one plus π‘₯ plus π‘₯ squared plus π‘₯ cubed all the way up to π‘₯ to the 𝑛th power. Which we recall converges when the absolute value of π‘₯ is less than one and diverges when the absolute value of π‘₯ is greater than or equal to one. Let’s look to generalise.

A series of the form the sum of 𝑐 𝑛 times π‘₯ minus π‘Ž to the 𝑛th power for values of 𝑛 between zero and ∞ is called a power series in π‘₯ minus π‘Ž or a power series centred at π‘Ž or about π‘Ž. Notice that we’ve written 𝑐 naught for the term when 𝑛 equals zero. So we are assuming that π‘₯ minus π‘Ž to the power of zero is equal to one even when π‘₯ equals π‘Ž. Notice also that if π‘₯ equals π‘Ž, all terms are zero for 𝑛 is greater than or equal to one. So the power series always converges when π‘₯ is equal to π‘Ž. And we can use ideally the ratio or root tests to test for convergence of these series. Let’s see what that might look like.

For what values of π‘₯ is the series the sum of 𝑛 factorial times π‘₯ plus five to the 𝑛th power for values of 𝑛 between zero and ∞ convergent?

Let’s recall that we can use the ratio test to test for convergence. This says that suppose we have the series to find as the sum of π‘Ž 𝑛. Then if the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is less than one, the series is absolutely convergent. If the converse is true, if the limit is greater than one, then the series is divergent. And if the limit is equal to the one, the series may be divergent, conditionally convergent, or absolutely convergent.

Now since we’re looking to find the values of π‘₯ for which the series is convergent, we’re interested in this first case where the limit is less than one. And so we define π‘Ž 𝑛 to be equal to 𝑛 factorial times π‘₯ plus five to the 𝑛th power. And this means that π‘Ž 𝑛 plus one is 𝑛 plus one factorial times π‘₯ plus five to the power of 𝑛 plus one.

We’re looking to find when the limit as 𝑛 approaches ∞ of the absolute value of the quotient of these is less than one. So that’s 𝑛 plus one factorial times π‘₯ plus five to the power of 𝑛 plus one over 𝑛 factorial times π‘₯ plus five to the power of 𝑛. We begin by recalling that when we’re dividing two numbers whose base is the same, we can simply subtract the exponents. And this means that π‘₯ plus five to the power of 𝑛 plus one divided by π‘₯ plus five to the power of 𝑛 is simply π‘₯ plus five to the power of one or just π‘₯ plus five.

Then we recall that 𝑛 plus one factorial is equal to 𝑛 plus one times 𝑛 times 𝑛 minus one times 𝑛 minus two, and so on. Well, that’s the same as 𝑛 plus one times 𝑛 factorial. So we can divide through by 𝑛 factorial. And so we have the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one times π‘₯ plus five. π‘₯ plus five is independent of 𝑛. So we can take the absolute value of π‘₯ plus five outside of our limit.

And then we notice that the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 plus one must itself be ∞. The absolute value of π‘₯ plus five multiplied by ∞ will always be greater than one, except if the absolute value of π‘₯ plus five itself is equal to zero. Now since the absolute value of π‘₯ plus five we know is equal to zero, we don’t actually need the absolute value signs. And so we solved by subtracting five from both sides. And we see that π‘₯ is equal to negative five. The series given will therefore only converge when π‘₯ is equal to negative five. Let’s now look to generalise the convergence of a power series.

For a given power series the sum of 𝑐 𝑛 times π‘₯ minus π‘Ž to the 𝑛th power, there are only three possibilities. The first is when the series converges only when π‘₯ is equal to π‘Ž. The second is when the series converges for all values of π‘₯. Or thirdly, there exists a positive number capital 𝑅 such that the series converges if the absolute value of π‘₯ minus π‘Ž is less than 𝑅 and diverges if the absolute value of π‘₯ minus π‘Ž is greater than 𝑅. We call this capital 𝑅 the radius of convergence of the power series.

By convention, for part one, we say that 𝑅 is equal to zero. And for part two, we say that 𝑅 is equal to ∞. Then we’re able to define the interval of convergence of a power series as the interval that consists of all values of π‘₯ for which the series converges. In our first possibility, the interval just consists of the single point at π‘Ž, whereas in our second possibility, it’s from negative ∞ to ∞. In our third, we rewrite the absolute value of π‘₯ minus π‘Ž being less than 𝑅 as π‘₯ minus π‘Ž is greater than negative 𝑅 and less than 𝑅.

Now when π‘₯ itself is an end point of the interval, in other words, when π‘₯ is equal to π‘Ž plus or minus capital 𝑅, the series might converge or diverge at one or both end points. And so the ratio and root test will always fail when π‘₯ is an end point of the interval of convergence.

In these cases, we need to check the end points using some alternative test. Let’s have a look at how to find the radius of convergence and interval of convergence of a power series.

Consider the power series the sum of three π‘₯ to the 𝑛th power over 𝑛 plus five for 𝑛 equals zero to ∞. Find the interval of convergence of the power series and find the radius of convergence of the power series.

To test for convergence of our power series, we’re going to recall the ratio test. The part of the ratio test we’re interested in says that, given a series the sum of π‘Ž 𝑛, if the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is less than one, then the series is absolutely convergent. If we look at our series, we see that we can define π‘Ž 𝑛 as three π‘₯ to the 𝑛th power over 𝑛 plus five. And then π‘Ž 𝑛 plus one is three π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus five.

And then of course that denominator simplifies nicely to 𝑛 plus six. Our job is to establish where the limit as 𝑛 approaches ∞ of the absolute value of the quotient of these is less than one. And we know that when dividing by a fraction, we multiply by the reciprocal of that very same fraction. So the inside of our limit becomes three π‘₯ to the power of 𝑛 plus one over 𝑛 plus six times 𝑛 plus five over three π‘₯ to the 𝑛th power. And we now see that we can divide through by three π‘₯ to the 𝑛th power.

Okay, so now we have the limit as 𝑛 approaches ∞ of the absolute value of three π‘₯ times 𝑛 plus five over 𝑛 plus six. We notice that three π‘₯ is independent of 𝑛. So we can take the absolute value of three π‘₯ outside our limit. And then we divide through both the numerator and denominator of the fraction by 𝑛. That’s the highest power of 𝑛 in our denominator. And so the denominator becomes 𝑛 divided by 𝑛, which is one, plus five over 𝑛. And then the numerator becomes one plus six over 𝑛. And as 𝑛 approaches ∞, five over 𝑛 and six over 𝑛 approach zero. And so the limit as 𝑛 approaches ∞ of the absolute value of one plus five over 𝑛 over one plus six over 𝑛 is simply the absolute value of one or one.

So we need to establish where the absolute value of three π‘₯ times one is less than one, or simply where the absolute value of three π‘₯ is less than one. Since three is a purely positive number, we can divide through both sides of our inequality by three without affecting those absolute value signs. And we find that the absolute value of π‘₯ must be less than one-third. So the series converges if the absolute value of π‘₯ is less than one-third. And conversely, we can say it will diverge if the absolute value of π‘₯ is greater than one-third. And in fact, we’ve found the radius of convergence of our power series. It’s 𝑅 equals one-third.

Now another way of writing the absolute value of π‘₯ being less than one-third is saying that π‘₯ must be greater than negative one-third and less than a third. And so we’ve determined an interval of convergence. We certainly know that the power series converges for these values of π‘₯. But we don’t know what happens at the end points of our interval. We’re going to clear some space and substitute π‘₯ equals negative one-third and π‘₯ equals one-third into our original power series. And we’ll see if these series converge or diverge using an alternative test.

We’ll begin by letting π‘₯ be equal to negative one-third. Then our series is the sum from 𝑛 equals zero to ∞ of three times negative one-third to the 𝑛th power over 𝑛 plus five. Which simplifies to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power over 𝑛 plus five. And you might recognise this. It’s an alternating series. This negative one to the 𝑛th power indicates so.

So we’re going to use the alternating series test to establish whether this converges or diverges. This series is to be used when we have a series the sum of π‘Ž 𝑛, where π‘Ž 𝑛 is negative one to the 𝑛th power times 𝑏 𝑛 or negative one to the power of 𝑛 plus one times 𝑏 𝑛. And here 𝑏 𝑛 must be greater than or equal to zero for all 𝑛. Then if the limit as 𝑛 approaches ∞ of 𝑏 𝑛 equals zero and 𝑏 𝑛 is a decreasing sequence, then we can say our series the sum of π‘Ž 𝑛 is convergent.

Let’s rewrite our series as the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times 𝑛 plus five to the power of negative one. We now see it’s of the form given, where 𝑏 is equal to 𝑛 plus five to the power of negative one. Let’s check the limit as 𝑛 approaches ∞ of this 𝑏 𝑛 is indeed zero.

Well, we know it’s equal to the limit as 𝑛 approaches ∞ of one over 𝑛 plus five. And as π‘Ž 𝑛 gets larger, 𝑛 plus five, the denominator of the fraction, gets larger and larger. So one over 𝑛 plus five gets smaller and smaller and eventually approaches zero. So the limit of 𝑏 𝑛 as 𝑛 approaches ∞ is indeed zero.

Next, we need to check whether 𝑏 𝑛 is a decreasing sequence. To do so, let’s differentiate the entire function with respect to 𝑛. Using the general power rule, we see that’s negative 𝑛 plus five to the power of negative two, which is negative one over 𝑛 plus five all squared. For all values of 𝑛 between zero and ∞, the denominator of our fraction is always positive. So negative one over a positive number gives us a negative number, which therefore means that we do indeed have a decreasing sequence.

We’ve satisfied all the criteria for our series. And we can therefore say that the series when π‘₯ is equal to negative one-third converges. We’re now going to clear some space and test π‘₯ equals a third. This time, our series is the sum from 𝑛 equals zero to ∞ of three times a third to the 𝑛th power over 𝑛 plus five, which is one to the 𝑛th power over 𝑛 plus five, which is of course simply one over 𝑛 plus five.

Now this is actually an example of a general harmonic series. And we of course know that these diverge. Our series therefore diverges for values of π‘₯ greater than or equal to negative one-third and less than one-third. And our interval is shown.

In our final example, we’ll consider how we can use the root test to find convergence.

Consider the power series the sum from 𝑛 equals zero to ∞ of 𝑛 times two π‘₯ to the 𝑛th power. Determine the radius of convergence of the power series and determine the interval of convergence of the power series.

To test for convergence of this power series, we could use the ratio or the root test. Let’s look at the root test. The part of the test we’re interested in says suppose we have a series the sum of π‘Ž 𝑛. If the limit as 𝑛 approaches ∞ of the 𝑛th root of the absolute value of π‘Ž 𝑛, which can of course be alternatively written as the absolute value of π‘Ž 𝑛 to the power of one over 𝑛, is less than one. Then the series is absolutely convergent and hence convergent.

So we define π‘Ž 𝑛 for our series to be 𝑛 times two π‘₯ to the 𝑛th power. We’re looking to establish where the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 times two π‘₯ to the 𝑛th power to the power of one over 𝑛 is less than one. We know that finding an 𝑛th root doesn’t change the sign. So we can rewrite this and say that the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 to the power of one over 𝑛 times two π‘₯ to the 𝑛th power to the power of one over 𝑛 must be less than one. Two π‘₯ to the 𝑛th power to the power of one over 𝑛 is simply two π‘₯. And so our limit is as shown.

We know that two π‘₯ is independent of 𝑛. So we can take the absolute value of two π‘₯ outside of our limit. The problem we have now is that if we try to evaluate the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 to the power of one over 𝑛. We get the absolute value of ∞ to the power of one over ∞, which is ∞ to the power of zero. And we know that’s indeterminate form. But we might recall that we can write 𝑛 to the power of one over 𝑛 as 𝑒 to the power of the natural log of 𝑛 over 𝑛. And we know that as 𝑛 approaches ∞, the natural logarithm of 𝑛 over 𝑛 approaches zero. And so the limit as 𝑛 approaches ∞ of the absolute value of 𝑛 to the power of one over 𝑛 is 𝑒 to the power of zero, which is simply one.

Of course we don’t need the absolute value signs for one. And so we see we’re interested in when the absolute value of two π‘₯ is less than one. Since two is a purely positive number, we can divide through. And we see that the absolute value of π‘₯ is less than one-half. These are the values of π‘₯ for which the series converges. And we can therefore say that the radius convergence of our power series is one-half.

Another way of representing the absolute value of π‘₯ being less than one-half is to say that π‘₯ must be greater than negative one-half and less than one-half. And so we’ve determined an interval for convergence. But we do need to check if the power series converges or diverges at the end points of our interval. In other words, when π‘₯ is equal to negative one-half or π‘₯ is equal to one-half.

We’re going to plug these into our original power series and see if those series converge or diverge using alternative tests. Let’s clear some space. We’ll begin by letting π‘₯ equal negative one-half. The series is the sum from 𝑛 equals zero to ∞ of 𝑛 times two minus negative one-half to the 𝑛th power. And this becomes 𝑛 times negative one to the power of 𝑛.

To establish whether this series converges or diverges, we could use the alternating series test with 𝑏 𝑛 equals 𝑛. Since the limit as 𝑛 approaches ∞ of 𝑏 𝑛 is not equal to zero, we can say that this series ultimately diverges. So let’s try when π‘₯ is equal to one-half. We have the sum from 𝑛 equals zero to ∞ of 𝑛 times two times a half to the 𝑛th power. That’s 𝑛 times one to the 𝑛th power. And of course one to the 𝑛th power will always be one. So we have the sum from 𝑛 equals zero to ∞ of 𝑛.

This time, we can actually apply the 𝑛th term divergence test. The limit as 𝑛 approaches ∞ of 𝑛 is ∞. That’s not equal to zero. And so, once again, when π‘₯ is equal to a half, our series diverges. And so the interval of convergence of our power series is the open interval negative one-half to one-half.

In this video, we saw that a series of the form the sum of 𝑐 𝑛 times π‘₯ minus π‘Ž to the 𝑛th power for values of 𝑛 between zero and ∞ is called a power series about π‘Ž. We also saw that we can use the ratio test and root test to find the radius of convergence and interval of convergence of our power series. But that it’s also important that we should test the end points of the interval for convergence or divergence.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.