Question Video: Finding the Absolute Maximum and Minimum Values of a Polynomial Function on a Closed Interval Mathematics • Higher Education

Consider the power series βˆ‘_(𝑛 = 0)^(∞) ((3π‘₯)^(𝑛)/(𝑛 + 5)). Find the interval of convergence of the power series. Find the radius of convergence of the power series.

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Video Transcript

Consider the power series the sum of three π‘₯ to the 𝑛th power over 𝑛 plus five for 𝑛 equals zero to ∞. Find the interval of convergence of the power series and find the radius of convergence of the power series.

To test for convergence of our power series, we’re going to recall the ratio test. The part of the ratio test we’re interested in says that, given a series the sum of π‘Ž 𝑛, if the limit as 𝑛 approaches ∞ of the absolute value of π‘Ž 𝑛 plus one over π‘Ž 𝑛 is less than one, then the series is absolutely convergent. If we look at our series, we see that we can define π‘Ž 𝑛 as three π‘₯ to the 𝑛th power over 𝑛 plus five. And then π‘Ž 𝑛 plus one is three π‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus five.

And then of course that denominator simplifies nicely to 𝑛 plus six. Our job is to establish where the limit as 𝑛 approaches ∞ of the absolute value of the quotient of these is less than one. And we know that when dividing by a fraction, we multiply by the reciprocal of that very same fraction. So the inside of our limit becomes three π‘₯ to the power of 𝑛 plus one over 𝑛 plus six times 𝑛 plus five over three π‘₯ to the 𝑛th power. And we now see that we can divide through by three π‘₯ to the 𝑛th power.

Okay, so now we have the limit as 𝑛 approaches ∞ of the absolute value of three π‘₯ times 𝑛 plus five over 𝑛 plus six. We notice that three π‘₯ is independent of 𝑛. So we can take the absolute value of three π‘₯ outside our limit. And then we divide through both the numerator and denominator of the fraction by 𝑛. That’s the highest power of 𝑛 in our denominator. And so the denominator becomes 𝑛 divided by 𝑛, which is one, plus five over 𝑛. And then the numerator becomes one plus six over 𝑛. And as 𝑛 approaches ∞, five over 𝑛 and six over 𝑛 approach zero. And so the limit as 𝑛 approaches ∞ of the absolute value of one plus five over 𝑛 over one plus six over 𝑛 is simply the absolute value of one or one.

So we need to establish where the absolute value of three π‘₯ times one is less than one, or simply where the absolute value of three π‘₯ is less than one. Since three is a purely positive number, we can divide through both sides of our inequality by three without affecting those absolute value signs. And we find that the absolute value of π‘₯ must be less than one-third. So the series converges if the absolute value of π‘₯ is less than one-third. And conversely, we can say it will diverge if the absolute value of π‘₯ is greater than one-third. And in fact, we’ve found the radius of convergence of our power series. It’s 𝑅 equals one-third.

Now another way of writing the absolute value of π‘₯ being less than one-third is saying that π‘₯ must be greater than negative one-third and less than a third. And so we’ve determined an interval of convergence. We certainly know that the power series converges for these values of π‘₯. But we don’t know what happens at the end points of our interval. We’re going to clear some space and substitute π‘₯ equals negative one-third and π‘₯ equals one-third into our original power series. And we’ll see if these series converge or diverge using an alternative test.

We’ll begin by letting π‘₯ be equal to negative one-third. Then our series is the sum from 𝑛 equals zero to ∞ of three times negative one-third to the 𝑛th power over 𝑛 plus five. Which simplifies to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power over 𝑛 plus five. And you might recognise this. It’s an alternating series. This negative one to the 𝑛th power indicates so.

So we’re going to use the alternating series test to establish whether this converges or diverges. This series is to be used when we have a series the sum of π‘Ž 𝑛, where π‘Ž 𝑛 is negative one to the 𝑛th power times 𝑏 𝑛 or negative one to the power of 𝑛 plus one times 𝑏 𝑛. And here 𝑏 𝑛 must be greater than or equal to zero for all 𝑛. Then if the limit as 𝑛 approaches ∞ of 𝑏 𝑛 equals zero and 𝑏 𝑛 is a decreasing sequence, then we can say our series the sum of π‘Ž 𝑛 is convergent.

Let’s rewrite our series as the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times 𝑛 plus five to the power of negative one. We now see it’s of the form given, where 𝑏 is equal to 𝑛 plus five to the power of negative one. Let’s check the limit as 𝑛 approaches ∞ of this 𝑏 𝑛 is indeed zero.

Well, we know it’s equal to the limit as 𝑛 approaches ∞ of one over 𝑛 plus five. And as π‘Ž 𝑛 gets larger, 𝑛 plus five, the denominator of the fraction, gets larger and larger. So one over 𝑛 plus five gets smaller and smaller and eventually approaches zero. So the limit of 𝑏 𝑛 as 𝑛 approaches ∞ is indeed zero.

Next, we need to check whether 𝑏 𝑛 is a decreasing sequence. To do so, let’s differentiate the entire function with respect to 𝑛. Using the general power rule, we see that’s negative 𝑛 plus five to the power of negative two, which is negative one over 𝑛 plus five all squared. For all values of 𝑛 between zero and ∞, the denominator of our fraction is always positive. So negative one over a positive number gives us a negative number, which therefore means that we do indeed have a decreasing sequence.

We’ve satisfied all the criteria for our series. And we can therefore say that the series when π‘₯ is equal to negative one-third converges. We’re now going to clear some space and test π‘₯ equals a third. This time, our series is the sum from 𝑛 equals zero to ∞ of three times a third to the 𝑛th power over 𝑛 plus five, which is one to the 𝑛th power over 𝑛 plus five, which is of course simply one over 𝑛 plus five.

Now this is actually an example of a general harmonic series. And we of course know that these diverge. Our series therefore diverges for values of π‘₯ greater than or equal to negative one-third and less than one-third. And our interval is shown.

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