Video: Limit Comparison Test

In this video, we will learn how to determine whether a series is convergent or divergent by comparing it to a series of known convergence using the limit comparison test.

17:22

Video Transcript

In this lesson, we’ll learn how to determine whether a series is convergent or divergent by comparing it to a series of known convergence using the limit comparison test. We’ll look at a number of examples of how we can apply this test. But first, let’s look at how the test works.

Remember that the series is the sum of the terms of an infinite sequence of numbers. If we have two series where the terms are 𝑎 𝑛 and 𝑏 𝑛, where 𝑎 𝑛 and 𝑏 𝑛 are greater than zero for all 𝑛. The limit comparison test tells us that if the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛 is the constant 𝐶, where 𝐶 is positive and finite, then either both series converge or both series diverge. Note that it doesn’t matter which of the two series is the numerator and which is the denominator in the limit. What this test says that as 𝑛 tends to ∞, the terms of the two series 𝑎 𝑛 and 𝑏 𝑛 behave in a similar way. So we can expect the series themselves to exhibit similar behaviour to each other. Note that the limit is often written as, the limit as 𝑛 tends to ∞ of 𝑎 𝑛 multiplied by one over 𝑏 𝑛 which is equal to 𝐶.

The limit comparison test is a variant on the direct comparison test, which we can use when we can’t apply the direct comparison test. In practice, how we use the limit comparison test is if we’re asked to determine the convergence of a particular series, say 𝑎 𝑛, summed from 𝑛 equals one to ∞. We bring in the series 𝑏 𝑛, summed from one to ∞, which we know the convergence of. We use the terms of the two series to find the limit 𝐶, which allows us to state the convergence or otherwise of our series, 𝑎 𝑛 from 𝑛 is one to ∞. Let’s apply this in an example.

Use the Limit comparison test to determine whether the series the sum from 𝑛 is one to ∞ of two over 𝑛 plus six is convergent or divergent.

We’ve been given the series whose terms we’ll call 𝑏 𝑛, which is the sum from 𝑛 is one to ∞ of two over 𝑛 plus six. And we’re asked to determine the convergence of this series using the limit comparison test. The limit comparison test tells us that for series with terms 𝑎 𝑛 and 𝑏 𝑛, where 𝑎 𝑛 and 𝑏 𝑛 are greater than zero. If the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛 is a constant 𝐶, where 𝐶 is greater than zero and finite, then either both series converge or both diverge. We’ve been given the series where the terms are two divided by 𝑛 plus six. And since the two is a constant, we can take this outside the series so that we have two times the sum from 𝑛 is one to ∞ of one over 𝑛 plus six.

To determine whether this series converges or diverges, we need to find another series with easily determined or known convergence, such that the limit comparison test applies. Notice that the limit comparison test is concerned with the terms of the series, not the series themselves. And it looks at the behaviour of the ratio of these terms as 𝑛 tends to ∞. So let’s look at the terms in our original series as 𝑛 tends to ∞.

Since we’ve taken the two outside, our term is now one over 𝑛 plus six. And as 𝑛 tends to ∞, this term is dominated by the 𝑛 in the denominator. The six becomes superfluous as 𝑛 gets very large. And our term has the same behaviour as one over 𝑛. We can therefore use the sum from 𝑛 is one to ∞ of one over 𝑛 as our second series. You might recognize this as the harmonic series. This is related to the wavelengths of the overtones of a vibrating string. So now we have our two series. We can think about applying the limit comparison test. But first, we need to check that the terms of each series are positive.

If our term is 𝑎 𝑛 is equal to one over 𝑛 and 𝑛 is positive from one to ∞, then one over 𝑛 is always positive. Similarly, if our term is one over 𝑛 plus six, since 𝑛 plus six is always positive, one over 𝑛 plus six is positive too. So both terms are positive, and we can use the limit comparison test.

Our first step is to find the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛, with 𝑎 𝑛 is one over 𝑛 and 𝑏 𝑛 is one over 𝑛 plus six. That’s the limit as 𝑛 tends to ∞ of one over 𝑛 divided by one over 𝑛 plus six. That’s equal to the limit as 𝑛 tends to ∞ of one over 𝑛 multiplied by 𝑛 plus six over one. Which is the limit as 𝑛 tends to ∞ of 𝑛 plus six divided by 𝑛. And that’s equal to the limit as 𝑛 tends to ∞ of one plus six divided by 𝑛. We know that as 𝑛 tends to ∞, six divided by 𝑛 tends to zero. So that our limit is actually equal to one, which is a positive constant. So we’ve found our constant 𝐶, which is greater than zero. Then that’s the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛.

So all we need to know now is the convergence of our second series, the harmonic series. You probably know that this series diverges, but we can use the integral test to show this. The integral test tells us that if 𝑓 is continuous positive and decreasing on the interval one to ∞ and we let 𝑎 𝑛 equal to 𝑓 of 𝑛. Then if the improper integral from one to ∞ of 𝑓 of 𝑥 d𝑥 converges, so does the sum from 𝑛 is one to ∞ of 𝑎 𝑛. Alternatively, if the improper integral from one to ∞ of 𝑓 of 𝑥 with respect to 𝑥 diverges. Then so does the sum from 𝑛 is one to ∞ over 𝑎 𝑛. Before we can use this test, we need to check that our terms are positive and decreasing.

One over 𝑛 is positive for all positive 𝑛, so that 𝑎 𝑛 is greater than zero. We know also that for positive 𝑛, 𝑛 is less than 𝑛 plus one. And taking the reciprocal, we reversed the inequality. So that one over 𝑛 is greater than one over 𝑛 plus one. So that 𝑎 𝑛 is greater than 𝑎 𝑛 plus one. So that as 𝑛 increases, 𝑎 𝑛 decreases. So then, we have that the sum from 𝑛 is one to ∞ of one over 𝑛 and the integral between one and ∞ of one over 𝑥 with respect to 𝑥, either both converge or both diverge.

If we work out this integral, the integral of one over 𝑥 is the natural log of 𝑥. And we evaluate this between one and ∞. And that’s ∞ minus zero, which is ∞. Our integral therefore diverges. So our sum from 𝑛 is one to ∞ of one over 𝑛 also diverges. And hence, by the limit comparison test, our first series, which is the sum from 𝑛 is one to ∞ of two over 𝑛 plus six also diverges.

So we found a second series to use in the limit comparison test. And we found the limit as 𝑛 tends to ∞ of the ratio of the terms of the two series is a positive constant. We found that the second series diverges. And therefore, so does our original series.

Let’s try another example.

Using the limit comparison test, determine whether the series the sum from 𝑛 is one to ∞ of one over four to the power 𝑛 plus six is convergent or divergent.

We’ve been given a series the sum from 𝑛 is one to ∞ of one over four to the 𝑛 plus six. The terms of our series are therefore one over four to the 𝑛 plus six. And we’re asked to determine the convergence of this series using the limit comparison test. This test says that if we have two series with terms 𝑎 𝑛 and 𝑏 𝑛, which are positive, and the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛 is a constant 𝐶, which is greater than zero and less than ∞. Then either both series converge or both series diverge.

We have a series with terms one over four to the 𝑛 plus six. In order to use the limit comparison test, we need another series to compare in our ratio of terms. We’d like to find another series where the terms will make it easier to show that one term divided by the other as 𝑛 tends to ∞ is constant. We also need to know whether this second series converges or diverges. To find the terms of this second series, what we can do is look at the behaviour of our terms as 𝑛 tends to ∞. As 𝑛 tends to ∞ now in term one over four to the 𝑛 plus six, the six becomes superfluous, and our term turns to one over four to the power 𝑛.

This could help us since this is a term of a geometric series. So if we let the terms of our first series equal to 𝑏 𝑛, that’s one over four to the 𝑛 plus six is equal to 𝑏 𝑛, and the terms of our second series, one over four to the 𝑛 equal 𝑎 𝑛. We can then find the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛 within our limit comparison test. Note that it doesn’t matter which one is 𝑎 𝑛 and which one is 𝑏 𝑛, whichever makes the calculation easiest. So we’re going to find the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛, where 𝑎 𝑛 is one over four to the 𝑛 and 𝑏 𝑛 is one over four to the 𝑛 plus six.

So we have the limit as 𝑛 tends to ∞ of the one over four to the 𝑛 divided by one over four to the 𝑛 plus six. And that’s equal to the limit as 𝑛 tends to ∞ of the one over four to the 𝑛 times four to the 𝑛 plus six divided by one. That’s the limit as 𝑛 tends to ∞ of four to the 𝑛 plus six divided by four to the 𝑛. Which is the limit as 𝑛 tends to ∞ of one plus six, divided by four to the 𝑛. Since one is a constant, that’s one plus the limit as 𝑛 tends to ∞ of six divided by four to the power 𝑛. And six is a constant. We could take this outside, so that gives us one plus six times the limit as 𝑛 tends to ∞ of one over four to the 𝑛. And as 𝑛 tends to ∞, one over four to the power 𝑛 tends to zero.

Our limit is therefore one which is a constant greater than zero. So we found a second series where the ratio of the terms of the two series as the 𝑛 tends to ∞ is a constant. Now, we need to find the convergence of that second series. Because by the limit comparison test, the convergence of the second series determines the convergence or divergence of the first series. Our second series is the sum from 𝑛 is one to ∞ of one over four to the power 𝑛. And we know that this is a geometric series. A geometric series is a series with the sum from 𝑛 is zero to ∞ of 𝑎 times 𝑟 to the power 𝑛. And that’s 𝑎 times 𝑟 to the zero plus 𝑎 times 𝑟 to the one plus 𝑎 times 𝑟 squared, et cetera. If the absolute value of 𝑟 is less than one, then the series sums to 𝑎 over one minus 𝑟. If the absolute value of 𝑟 is greater than or equal to one, then the series diverges.

In our series, 𝑎 is equal to one, and 𝑟 is equal to one over four. There is a difference, however. For the defined geometric series, 𝑛 begins with the value of zero. Whereas in our series, 𝑛 starts at one. Given that 𝑟, in our case, is less than one, when we come to work out the sum of our series, we’ll need to subtract this first term. So let’s work out the sum of our series, which is 𝑎 over one minus 𝑟. The first term in the defined series 𝑎 times 𝑟 to the power zero, in our case, will be one times one over four to the power zero, And that’s equal to one. So we’ll need to subtract this from the sum of our series.

That means we subtracted from 𝑎 over one minus 𝑟 where, in our case, 𝑎 is one and 𝑟 is one over four. The sum of our series is therefore the sum from 𝑛 is one to ∞ of one over four to the power 𝑛, which is equal to one over one minus one over four minus one. Where this negative one corresponds to 𝑎 times 𝑟 to the power zero. The sum of our second series is therefore four over three minus one which is one over three. Our second series, the sum from 𝑛 is one to ∞ of one over four to the power 𝑛, converges to the sum one over three. By the limit comparison test, if one of our series converges, then so does the other. This means that our original series, the sum from 𝑛 is one to ∞ of one over four to the power 𝑛 plus six, also converges.

Let’s look at one more example of the limit comparison test for the convergence or divergence of a series.

Use the Limit comparison test to determine whether the series the sum from 𝑛 is one to ∞ of three 𝑛 squared plus two 𝑛 plus one divided by 𝑛 to the power three plus one is convergent or divergent.

We’ve been given a series where the terms are 𝑎 𝑛 equal to three 𝑛 squared plus two 𝑛 plus one divided by 𝑛 to the power three plus one. We’re asked to use the limit comparison test to determine whether the series converges or diverges. The limit comparison test tells us that if we have two series where the terms 𝑎 𝑛 and 𝑏 𝑛 are positive and the limit as 𝑛 tends to ∞ of 𝑎 𝑛 divided by 𝑏 𝑛 is a constant 𝐶 such that 𝐶 is greater than zero and less than ∞, so it’s finite. Then either both the series converge or both series diverge. In order to use this test to determine whether the given series is convergent or divergent. Our plan of action is to try and find a second series such that the ratio of the terms 𝑎 𝑛 over 𝑏 𝑛 converges to a positive number.

We also need to know or be able to find out whether the second series converges or diverges. To find the second series, we can look at the behaviour of these terms of our first series as 𝑛 tends to ∞. However, let’s make sure our terms are positive before we start. Since 𝑛 is positive, both the numerator and the denominator are greater than zero. So in fact, our terms are positive. We can therefore use the limit comparison test. For large 𝑛 in this expression, the terms with the highest powers of 𝑛 dominate. This means in the numerator, three 𝑛 squared dominates. While in the denominator, 𝑛 to the power three dominates. So as 𝑛 tends to ∞, the terms of our series tend to three 𝑛 squared divided by 𝑛 to the power of three. And that’s equal to three over 𝑛.

We can use this as the term in our second series for the limit comparison test. Our second series is therefore the sum from 𝑛 is one to ∞ of three divided by 𝑛. Now, with 𝑎 𝑛 equal to three 𝑛 squared plus two 𝑛 plus one divided by 𝑛 to the power of three plus one and 𝑏 𝑛 equal to three divided by 𝑛, we can find the limit within our limit comparison test. That’s the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛. That’s equal to the limit as 𝑛 tends to ∞ of three 𝑛 squared plus two 𝑛 plus one divided by 𝑛 to the power of three plus one all divided by three divided by 𝑛. That’s equal to the limit as 𝑛 tends to ∞ of three 𝑛 squared plus two 𝑛 plus one divided by 𝑛 to the power of three plus one multiplied by 𝑛 divided by three.

That’s equal to the limit as 𝑛 tends to ∞ of three 𝑛 to the power of three plus two 𝑛 squared plus 𝑛 all divided by three 𝑛 to the power of three plus three. If we divide both the numerator and the denominator by three 𝑛 to the power of three. That gives us the limit as 𝑛 tends to ∞ of one plus two 𝑛 squared over three 𝑛 to the power of three plus 𝑛 over three 𝑛 to the power of three all divided by one plus three over three 𝑛 to the power of three. And cancelling down that gives us the limit as 𝑛 tends to ∞ of one plus two over three 𝑛 plus one over three 𝑛 squared all divided by one plus one over 𝑛 to the power of three.

Two divided by three 𝑛 tends to zero as 𝑛 tends to ∞. One over three 𝑛 squared tends to zero as 𝑛 tends to ∞. And one over 𝑛 to the power of three tends to zero as 𝑛 tends to ∞. This means that the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛 is equal to one. This is a positive constant, so satisfies our limit comparison test. So now if we can determine the convergence of our second series, this will establish the convergence or divergence of our first series.

Our second series is the sum from 𝑛 is one to ∞ of three divided by 𝑛. And since three is a constant multiply, we can take this outside. This gives us three times the sum from 𝑛 is one to ∞ of one over 𝑛. The sum from 𝑛 is one to ∞ of one over 𝑛 exactly the harmonic series. And we know that this series diverges. And this means by the limit comparison test that our first series diverges also. The series which is the sum from 𝑛 is one to ∞ of three 𝑛 squared plus two 𝑛 plus one divided by 𝑛 to the power of three plus one diverges.

We’ve looked at some examples of the limit comparison test. Now, let’s summarize with some key points. The limit comparison test tells us that if we have two series with positive terms 𝑎 𝑛 and 𝑏 𝑛 and the limit as 𝑛 tends to ∞ of the ratio of the two terms is a constant 𝐶, where 𝐶 is positive and finite, then either both series converge or both series diverge. To use the limit comparison test to determine the convergence or divergence of a given series.

We try to find a second series of known convergence or divergence and whose terms all satisfy the condition that the limit as 𝑛 tends to ∞ of the ratio of the terms is equal to a constant. We can often find the second series by looking at the behaviour of the terms of the given series as 𝑛 tends to ∞. Finally, we note that when we calculate the limit as 𝑛 tends to ∞ of 𝑎 𝑛 over 𝑏 𝑛, it doesn’t matter which we choose as 𝑎 𝑛 or 𝑏 𝑛.

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