Video: Using the Comparison Test

Use the Limit comparison test to determine whether the series βˆ‘_(𝑛 = 1)^(∞) (2/(𝑛 + 6)) is convergent or divergent.

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Video Transcript

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 π‘Ž 𝑛 and 𝑏 𝑛, where π‘Ž 𝑛 and 𝑏 𝑛 are greater than zero. If the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 over 𝑏 𝑛 is a constant 𝐢, where 𝐢 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 π‘Ž 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 π‘Ž 𝑛 over 𝑏 𝑛, with π‘Ž 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 𝐢, which is greater than zero. Then that’s the limit as 𝑛 tends to ∞ of π‘Ž 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 π‘Ž 𝑛 equal to 𝑓 of 𝑛. Then if the improper integral from one to ∞ of 𝑓 of π‘₯ dπ‘₯ converges, so does the sum from 𝑛 is one to ∞ of π‘Ž 𝑛. Alternatively, if the improper integral from one to ∞ of 𝑓 of π‘₯ with respect to π‘₯ diverges. Then so does the sum from 𝑛 is one to ∞ over π‘Ž 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 π‘Ž 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 π‘Ž 𝑛 is greater than π‘Ž 𝑛 plus one. So that as 𝑛 increases, π‘Ž 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 π‘₯ with respect to π‘₯, either both converge or both diverge.

If we work out this integral, the integral of one over π‘₯ is the natural log of π‘₯. 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.

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