Video: Comparison Test for Series

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 comparison test.

16:04

Video Transcript

Comparison Test for Series

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 comparison test. We’ll be looking at a variety of examples of how we can do this. First, let’s cover how the comparison test for series works. When using the comparison test, there are two series which we are concerned with. We can call these series the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛.

Now, in order to use the comparison test, we require both π‘Ž 𝑛 and 𝑏 𝑛 to be greater than or equal to zero for all 𝑛. We are also required to know whether the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges or diverges. Now, the comparison test tells us that if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Žπ‘› also converges. Conversely, if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 diverges, and π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 also diverges.

Now, it’s quite intuitive to see how the comparison test works. Since in the first part, we’re saying that if all our terms are smaller than or equal to the terms of a series which we know to be convergent, then our series also converges. And in the second part, we’re saying that if our terms are larger than or equal to the terms of a series which we know that diverges, then our series also diverges. Now, using the comparison test requires us to know a few convergent and divergent series.

Since we need to know some series which are convergent or divergent, so that’s the series 𝑏 𝑛, which we can use to compare to the series which we’re trying to find the convergence of. So, that’s the series of π‘Ž 𝑛. Now, one type of series which is often quite useful when using the comparison test is a 𝑝 series. A 𝑝 series has the form of the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝. And the 𝑝 series test will tell us that this series will diverge if 𝑝 is less than or equal to one. And it will converge if 𝑝 is greater than one. Let’s now move on and look at an example of how we can use the comparison test.

Use the comparison test to determine whether the series which is the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 over 𝑛 is convergent or divergent.

We can start answering this question by recalling what the comparison test tells us. The comparison test tells us that for the series which is the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, where π‘Ž 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. That if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is convergent, and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 is also a convergent. And if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is divergent, and π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 is also divergent.

Now, the series we’ve been given in the question is the sum from 𝑛 equals one to ∞ of the natural algorithm of 𝑛 over 𝑛. Therefore, we can let π‘Ž 𝑛 be equal to the natural logarithm of 𝑛 over 𝑛. Now, we need to find another series to compare this to. Looking at π‘Ž 𝑛, we can see that it looks similar to a 𝑝 series. Now, a 𝑝 series is a series of the form of the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝. And this 𝑝 series will diverge if 𝑝 is less than or equal to one and converge if 𝑝 is greater than one.

Now, the 𝑝 series, which our series looks most similar to, is the sum from 𝑛 equals one to ∞ of one over 𝑛. For this series, the value of 𝑝 is one. And therefore, the series will diverge. Before we can attempt to use the comparison test, we first need to check whether π‘Ž 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all 𝑛. Now, the values of 𝑛 which we’ll be using go from one to ∞. Therefore, 𝑛 is greater than or equal to zero.

Now, the natural logarithm of 𝑛 is an increasing function. And when 𝑛 is equal to one, the natural logarithm of 𝑛 is equal to zero. Therefore, we can say that for all our values of 𝑛, the natural logarithm of 𝑛 will be greater than or equal to zero. Since these two things are both greater than or equal to zero, this means that both π‘Ž 𝑛 and 𝑏 𝑛 will be greater than or equal to zero too.

Now, we’ve already said that the sum from 𝑛 equals one to ∞ of one over 𝑛 is divergent. Therefore, we need to try and satisfy the second part of the comparison test. Which means that we need to show that π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛. In other words, that’s showing that the natural logarithm of 𝑛 over 𝑛 is greater than or equal to one over 𝑛. Since 𝑛 is always greater than or equal to one, we can multiply both sides by 𝑛. Therefore, we have that the natural logarithm of 𝑛 is greater than or equal to one. And we need to try and prove this for all 𝑛.

Now, we can immediately see that this cannot be the case since the natural logarithm of one is equal to zero, which is less than one. However, this doesn’t mean that we cannot use the comparison test. Let’s consider the first few values of the natural logarithm of 𝑛. We have that the natural logarithm of two is equal to 0.693 and so on. And the natural logarithm of three is equal to 1.098 and so on.

Therefore, the natural logarithm of three is in fact greater than or equal to one. And as we said earlier, we know that the natural logarithm of 𝑛 is an increasing function. And so, our statement that the natural logarithm of 𝑛 is greater than or equal to one holds true for values of 𝑛 which are greater than or equal to three.

Now, we can rewrite our series which we were given in the question. We can take the first two terms out of the series. And we have that the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 over 𝑛 is equal to the natural logarithm of one over one plus the natural logarithm of two over two plus the sum from 𝑛 equals three to ∞ of the natural logarithm of 𝑛 over 𝑛. Now, we can compare this series with the 𝑝 series of the sum from 𝑛 equals three to ∞ of one over 𝑛.

We have the same values of π‘Ž 𝑛 and 𝑏 𝑛 from before. And again, we’ve of course satisfied the condition that π‘Ž 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. And we have that the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is divergent. And we also now have that π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛. And so, from this, we can say that the sum from 𝑛 equals three to ∞ of the natural logarithm of 𝑛 over 𝑛 is divergent.

We can now relate this back to the original sum we were given in the question since its equal to two constant terms plus the series we’ve just shown to be divergent. Since it’s equal to the sum of two constants and a divergent series, we have found that the sum from 𝑛 equals one to ∞ of the natural logarithm of 𝑛 over 𝑛 is divergent.

Let’s now look at a second example of how we can use the comparison test.

Use the comparison test to determine whether the series which is the sum from 𝑛 equals one to ∞ of 𝑛 cubed plus one over 𝑛 to the five plus three is convergent or divergent.

We can start by recalling the comparison test. The comparison test tells us that for the series which is the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, where π‘Ž 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. That if firstly the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 converges. And secondly, if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 diverges, and π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 also diverges.

Now, before we use the comparison test on our series, it may be easier to split it up. We can say that the sum from 𝑛 equals one to ∞ of 𝑛 cubed plus one over 𝑛 to the five plus three is equal to the sum from 𝑛 equals one to ∞ of 𝑛 cubed over 𝑛 to the five plus three plus one over 𝑛 to the five plus three. And then, we can split this further by splitting it into two separate series, which will look like this.

We now have two series which we can perform the comparison test to. Let’s start with the second of this series. So, that’s the sum from 𝑛 equals one to ∞ of one over 𝑛 to the power of five plus three. And we can, in fact, compare this to a 𝑝 series. A 𝑝 series is a series which is of the form of the sum from 𝑛 equals one to ∞ of one over 𝑛 to the 𝑝. And the 𝑝 series will diverge if 𝑝 is less than or equal to one and converge if 𝑝 is greater than one.

We can see that our series looks quite similar to the 𝑝 series where 𝑝 is equal to five. So, that’s the sum from 𝑛 equals one to ∞ of one over 𝑛 to the five. We can apply the comparison test to these two series. Since in both of these series, we’re summing from 𝑛 equals one to ∞, this tells us that 𝑛 is always greater than or equal to one. Looking at the π‘Ž 𝑛 and the 𝑏 𝑛 from our series, when 𝑛 is greater than or equal to one, π‘Ž 𝑛 and 𝑏 𝑛 are both greater than or equal to zero. Therefore, we’ve satisfied the first condition of the comparison test.

Next, we need to check whether our sum from 𝑛 equals one to ∞ of one over 𝑛 to the five converges or diverges. Since this is a 𝑝 series where the value of 𝑝 is equal to five, and five is greater than one, this tells us that this series converges. Therefore, in order to use the comparison test, we need to satisfy the condition in the first part, which is that π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛. So, that’s that one over 𝑛 to the power of five plus three is less than or equal to one over 𝑛 to the power of five.

Since 𝑛 to the power of five plus three is greater than 𝑛 to the power of five for all 𝑛 greater than or equal to one, this tells us that this condition is, in fact, satisfied. Therefore, we’ve satisfied the last condition of the comparison test. And we can say that the sum from 𝑛 equals one to ∞ of one over 𝑛 to the five plus three converges. Now, we can focus on the other series. That’s the sum from 𝑛 equals one to ∞ of 𝑛 cubed over 𝑛 to the five plus three.

Let’s consider π‘Ž 𝑛. So, that’s 𝑛 cubed over 𝑛 to the power of five plus three. And we can say that this must be less than or equal to 𝑛 cubed over 𝑛 to the power of five since the denominator of the fraction on the left is bigger than the dominator of the fraction on the right. Now, 𝑛 cubed over 𝑛 to the power of five is simply one over 𝑛 squared. Therefore, this implies that we can compare the series on the left with the series which is the sum from 𝑛 equals one to ∞ of one over 𝑛 squared. And this series is again a 𝑝 series where 𝑝 is equal to two. Two is greater than one. And therefore, this series must converge.

Here, we have again satisfied all the conditions for the first part of the comparison test. We can, therefore, come to the conclusion that the sum from 𝑛 equals one to ∞ of 𝑛 cubed over 𝑛 to the power of five plus three also converges. We have now found two convergent series. Therefore, the sum of the two convergent series will also be convergent. Hence, we have that the sum from 𝑛 equal one to ∞ of 𝑛 cubed plus one ever 𝑛 to the five plus three is convergent.

Next, let’s note another common type of series used in the comparison tests. These are geometric series. These are series which are of the form of the sum from 𝑛 equals one to ∞ of π‘Ž multiplied by π‘Ÿ to the power of 𝑛 minus one. Now, these geometric series converge if the absolute value of π‘Ÿ is less than one and diverge if the absolute value of π‘Ÿ is greater than or equal to one. Let’s now look at an example of how we can use geometric series in the comparison test.

Use the comparison test to decide whether the series which is the sum from 𝑛 equals one to ∞ of 237 over 𝑛 plus 1.1 to the 𝑛 is convergent or divergent.

Let’s start by recalling the comparison test. The comparison test tells us that for the series which is the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛, and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛, where π‘Žπ‘› and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. That firstly if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges, and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the sum 𝑛 equals one to ∞ of π‘Ž 𝑛 also converges. And secondly, if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 diverges, and π‘Ž 𝑛 is greater or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 also diverges.

Now, our series has been given to us in the question, which gives that π‘Ž 𝑛 is equal to 237 over 𝑛 plus 1.1 to the 𝑛. Now, it looks as though we may be able to compare our series to a 𝑝 series. And this 𝑝 series is 237 multiplied by the sum from 𝑛 equals one to ∞ of one over 𝑛. Now, for this 𝑝 series, 𝑝 is equal to one. Therefore, this series diverges. Therefore, we’re looking at the second part of the comparison test. And we need to try and prove that π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛. So, that’s trying to prove that 237 over 𝑛 plus 1.1 to the 𝑛 is greater than or equal to 237 over 𝑛.

Now, the only differences in these two fractions is the denominators, in particular the term of 1.1 to the 𝑛. Since 𝑛 is always greater than or equal to one, therefore, 1.1 to the 𝑛 is always greater than zero. This tells us that the denominator of the fraction on the left will be greater than the denominator of the fraction at the right. And because of this, our inequality here does not hold, and so we cannot use this 𝑝 series to apply the comparison test. We’ll need to try a different series.

We can, in fact, try a geometric series. Geometric series is of the form of the sum of π‘Ž π‘Ÿ to the 𝑛. And a geometric series is convergent if the absolute value of π‘Ÿ is less than one and divergent if the absolute value of π‘Ÿ is greater than or equal to one. We now need to find a geometric series to compare our series to. Looking at our series, we have 1.1 to the 𝑛 in the denominator. Therefore, let’s try comparing our series to the sum from 𝑛 equals one to ∞ of 237 over 1.1 to the 𝑛.

For both of these series, since 𝑛 is greater than or equal to one, we have that both π‘Ž 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all 𝑛. Now, we can rewrite our geometric series in a form which is easier to find its convergence. And that is the sum from 𝑛 equals one to ∞ of 237 multiplied by one over 1.1 to the power of 𝑛. Here, we can see that π‘Ÿ is equal to one over 1.1. Since one over 1.1 is less than one, we have that this geometric series is convergent. Hence, we are looking at the first part of the comparison test.

We next need to try and prove the π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛. So, that is that 237 over 𝑛 plus 1.1 to the 𝑛 is less than or equal to 237 over 1.1 to the 𝑛. Since 𝑛 is greater than or equal to one, we have that the denominator of 𝑛 plus 1.1 to the 𝑛 is greater than the denominator of 1.1 to the 𝑛. And therefore, our inequality holds. Hence, we have satisfied the condition of the second part of the comparison test. And so, we reach our conclusion, which is that the sum from 𝑛 equals one to ∞ of 237 over 𝑛 plus 1.1 to the 𝑛 is convergent.

We have now covered a variety of examples of how we can use the comparison test to find convergent or divergent series. Let’s recap some key points of the video.

Key Points

The comparison test for series tells us that for the series which is the sum for 𝑛 equals one to ∞ of π‘Ž 𝑛 and the sum for 𝑛 equals one to ∞ of 𝑏 𝑛, where π‘Ž 𝑛 and 𝑏 𝑛 are both greater than or equal to zero for all 𝑛. That firstly if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is convergent, and π‘Ž 𝑛 is less than or equal to 𝑏 𝑛 for all 𝑛, then the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 is convergent. And secondly, if the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is divergent, and π‘Ž 𝑛 is greater than or equal to 𝑏 𝑛 for all 𝑛, then the sum for 𝑛 equals one to ∞ of π‘Ž 𝑛 is divergent. When using the comparison test, the most common types of series we compare to are 𝑝 series and geometric series.

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