# Question Video: Using the Comparison Test to Determine the Divergence or Convergence of a Series Mathematics • Higher Education

Use the comparison test to determine whether the series ∑_(𝑛 = 1) ^(∞) 4/(3𝑛³ + 𝑛² + 5𝑛 + 2) is convergent or divergent.

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### Video Transcript

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

The question gives us an infinite series, and it tells us to use the comparison test to determine whether this series is convergent or divergent. To use the comparison test, we either want to find a smaller series which is divergent or a bigger series which is convergent. The first step to using the comparison test is to look at our series and determine whether we think it will be convergent or divergent.

So let’s take a look at our series. Our numerator remains constant at four. And the largest term in our denominator is three 𝑛 cubed. As 𝑛 gets larger and larger, this will be the dominant term in our denominator. This gives us our motivation. As 𝑛 is approaching ∞, the terms in our series are getting closer and closer to four divided by three 𝑛 cubed. In fact, this is just a constant multiple of one over 𝑛 cubed, so we would expect this series to behave similarly to the infinite sum of one over 𝑛 cubed.

And we know that this series is a 𝑝-series where 𝑝 is equal to three, so it’s convergent. So using this logic, we would expect our series to be convergent. So let’s try and prove our series is convergent by using the comparison test. The convergent version of the comparison tells us if we have two series the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 and the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 where our sequences 𝑎 𝑛 and 𝑏 𝑛 are greater than or equal to zero for all values of 𝑛 and we also have the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 is convergent and 𝑎 𝑛 is less than or equal to 𝑏 𝑛 for all values of 𝑛, then we can conclude the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 must also be convergent.

We want to use the comparison test on the series given to us in the question, so we’ll set 𝑎 𝑛 to be four divided by three 𝑛 cubed plus 𝑛 squared plus five 𝑛 plus two. And it’s worth pointing out this sequence is greater than zero for all values of 𝑛. The values of 𝑛 in our sum run from one to ∞, so 𝑛 is always positive. This means our denominator is always positive, so 𝑎 𝑛 is positive. So the first requirement for the comparison test is true; 𝑎 𝑛 is greater than or equal to zero for all of our values of 𝑛.

We now need to find our sequence 𝑏 𝑛. We need 𝑏 𝑛 to be greater than or equal to zero for all values of 𝑛. We need the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 to be convergent. And we need 𝑎 𝑛 to be less than or equal to 𝑏 𝑛 for all of our values of 𝑛. To do this, let’s start with our sequence 𝑎 𝑛. This is a very complicated-looking sequence. In particular, our denominator looks very complicated.

We want to find the sequence 𝑏 𝑛 which is greater than or equal to our sequence 𝑎 𝑛. If we were to remove two from our denominator, we can ask, what would happen? We can see we’re now dividing by a smaller positive number. So we’ve made our sequence bigger. In fact, we can keep doing this. We can remove five 𝑛. It’s positive, so now we’re dividing by a smaller positive number. And we can do this one more time. This gives us that our sequence 𝑎 𝑛 is less than or equal to four over three 𝑛 cubed.

And now we’re ready to try and use the comparison test. We’ll set our sequence 𝑏 𝑛 to be four divided by three 𝑛 cubed. First, we know that four 𝑛 cubed is positive for all of our values of 𝑛. Next, we already explained that our sequence 𝑎 𝑛 will be less than or equal to our sequence 𝑏 𝑛 for all of these values of 𝑛. This is because both sequences share the same positive numerator of four. However, our sequence 𝑏 𝑛 is dividing by a smaller positive number.

The last thing we need to show to use our comparison test is that the sum from 𝑛 equals one to ∞ of 𝑏 𝑛 converges. To do this, we’ll start by taking our constant factor of four over three outside of our series. This gives us four over three times the sum from 𝑛 equals one to ∞ of one divided by 𝑛 cubed. This is a constant multiple of a 𝑝-series where 𝑝 is equal to three, so we know that this is convergent.

This means we’ve shown all of our prerequisites are true for the comparison test. This means we can conclude the sum from 𝑛 equals one to ∞ of 𝑎 𝑛 is also convergent. Therefore, by using the comparison test, we were able to show the sum from 𝑛 equals one to ∞ of four divided by three 𝑛 cubed plus 𝑛 squared plus five 𝑛 plus two is convergent.