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Use the comparison test to determine whether the series βˆ‘_(𝑛 = 1)^(∞) (𝑛³ + 1)/(𝑛⁡ + 3) is convergent or divergent.

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Video Transcript

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.

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