Question Video: Finding the Radius of Convergence of a Given Macluarin Series Mathematics • Higher Education

A Maclaurin series is given by βˆ‘_(𝑛 = 1)^(∞) (βˆ’1)^(𝑛 + 1)(𝑛)π‘₯^(𝑛 βˆ’ 1). Find the radius of convergence for the series.

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Video Transcript

A Maclaurin series is given by the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times 𝑛 multiplied by π‘₯ to the power of 𝑛 minus one. Find the radius of convergence for the series.

We’re given a power series and we’re told this is the Maclaurin series of some function. We need to find the radius of convergence of this series. First, the fact that this is a Maclaurin series won’t change how we approach this problem. We just need to find the radius of convergence of this series. There’s a few different ways we could do this. One method we could use is to check if the series given to us is similar to a power series which we know already. However, this can be hard to see in this case, so, instead, we’ll try using the ratio test. And it’s usually a good idea to try using the ratio test for Maclaurin series or Taylor series. The reason is we often get a lot of cancellation when we find the ratio.

So let’s recall what the ratio test tells us. If we have a series the sum from 𝑛 equals one to ∞ of π‘Ž 𝑛 and we set 𝐿 equal to the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms π‘Ž 𝑛 plus one over π‘Ž 𝑛, then if this limit is less than one, our series must be convergent. However, if this limit is greater than one, then our series must be divergent. And it’s also worth pointing out if the value of 𝐿 is equal to one, then our ratio test is inconclusive. We want to use this as the series given to us in the question, so we need to set the value of π‘Ž 𝑛 equal to the summand of this series. That’s negative one to the power of 𝑛 plus one times 𝑛 multiplied by π‘₯ to the power of 𝑛 minus one.

So to apply the ratio test to this series, we need to find the value of 𝐿. 𝐿 is the limit as 𝑛 approaches ∞ of the absolute value of the ratio of successive terms, π‘Ž 𝑛 plus one over π‘Ž 𝑛. Let’s start with finding π‘Ž 𝑛 plus one. That means we need to write 𝑛 plus one into every version of 𝑛 in our expression for π‘Ž 𝑛. Doing this, we get negative one to the power of 𝑛 plus one plus one times 𝑛 plus one multiplied by π‘₯ to the power of 𝑛 plus one minus one, and we can simplify this. First, our exponent of negative one, 𝑛 plus one plus one, is equal to 𝑛 plus two. Next, the exponent of π‘₯, 𝑛 plus one minus one, is just equal to 𝑛. So we were able to simplify our expression for π‘Ž 𝑛 plus one to be negative one to the power of 𝑛 plus two multiplied by 𝑛 plus one times π‘₯ to the 𝑛th power.

Now we can just write in our expression for π‘Ž 𝑛. And of course, we know π‘Ž 𝑛 is equal to negative one to the power of 𝑛 plus one times 𝑛 multiplied by π‘₯ to the power of 𝑛 minus one. And now we can start simplifying. First, we can cancel 𝑛 plus one of the shared factors of negative one in our numerator and our denominator. This just leaves us with a factor of negative one in our numerator. Next, we want to cancel the shared factors of π‘₯ in our numerator and our denominator. To do this, we first need to recall we know our power series is convergent when π‘₯ is equal to zero. If we substitute this directly into our power series, our series is just the sum of zeros. So if we already know our power series is converging when π‘₯ is equal to zero, we can use the ratio test to find all of the values of π‘₯ where π‘₯ is not equal to zero. This means we can cancel 𝑛 minus one shared factors of π‘₯ in our numerator and our denominator. This just leaves us with a factor of π‘₯ in our numerator.

The last thing we’ll do is divide each term in 𝑛 plus one by our denominator of 𝑛. And, of course, 𝑛 plus one all over 𝑛 is one plus one over 𝑛, so this leaves us with the limit as 𝑛 approaches ∞ of the absolute value of negative one times one plus one over 𝑛 multiplied by π‘₯. And at this point, there’s a few different ways of evaluating this limit. We’ll just do this directly from our limit. Our limit is as 𝑛 is approaching ∞. We can see that negative one and π‘₯ are constants with respect to 𝑛. They are not varying as our value of 𝑛 varies.

The same is also true for one. However, we know as 𝑛 approaches ∞, one over 𝑛 is approaching zero. Its numerator remains constant; however, its denominator is growing without bound. So because the rest of this expression is constant, this means our limit evaluates to give us the absolute value of negative one times one times π‘₯. Of course, negative one times one times π‘₯ is equal to negative π‘₯. But when we take the absolute value, we don’t care about the sign of π‘₯. So the absolute value of negative π‘₯ is just the absolute value of π‘₯.

Remember the question is asking us to find the radius of convergence of this power series. And we know by using the ratio test, our series will be convergent when 𝐿 is less than one. But we just showed 𝐿 is the absolute value of π‘₯. So our power series is convergent when the absolute value of π‘₯ is less than one. And of course, the ratio test also tells us that our power series will be divergent when 𝐿 is greater than one, in other words, when the absolute value of π‘₯ is greater than one. Therefore, because our power series is convergent when the absolute value of π‘₯ is less than one and divergent when the absolute value of π‘₯ is greater than one, our radius of convergence is one.

Therefore, by using the ratio test, we were able to show the radius of convergence of the power series the sum from 𝑛 equals one to ∞ of negative one to the power of 𝑛 plus one times 𝑛 multiplied by π‘₯ to the power of 𝑛 minus one is just equal to one.

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