Question Video: Finding the Integration of a Function Involving Using Integration by Substitution and Integration by Parts

Using the limit comparison test, determine whether the series βˆ‘_(𝑛 = 1)^(∞) 1/(4^(𝑛) + 6) is convergent or divergent.

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Video Transcript

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 π‘Ž 𝑛 and 𝑏 𝑛, which are positive, and the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 over 𝑏 𝑛 is a constant 𝐢, 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 π‘Ž 𝑛. We can then find the limit as 𝑛 tends to ∞ of π‘Ž 𝑛 over 𝑏 𝑛 within our limit comparison test. Note that it doesn’t matter which one is π‘Ž 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 π‘Ž 𝑛 over 𝑏 𝑛, where π‘Ž 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 π‘Ž times π‘Ÿ to the power 𝑛. And that’s π‘Ž times π‘Ÿ to the zero plus π‘Ž times π‘Ÿ to the one plus π‘Ž times π‘Ÿ squared, et cetera. If the absolute value of π‘Ÿ is less than one, then the series sums to π‘Ž over one minus π‘Ÿ. If the absolute value of π‘Ÿ is greater than or equal to one, then the series diverges.

In our series, π‘Ž is equal to one, and π‘Ÿ 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 π‘Ÿ, 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 π‘Ž over one minus π‘Ÿ. The first term in the defined series π‘Ž times π‘Ÿ 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 π‘Ž over one minus π‘Ÿ where, in our case, π‘Ž is one and π‘Ÿ 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 π‘Ž times π‘Ÿ 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.

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