Question Video: Finding the Absolute Maximum and Minimum Values of a Polynomial Function on a Closed Interval | Nagwa Question Video: Finding the Absolute Maximum and Minimum Values of a Polynomial Function on a Closed Interval | Nagwa

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

Consider the power series β_(π = 0)^(β) ((3π₯)^(π)/(π + 5)). Find the interval of convergence of the power series. Find the radius of convergence of the power series.

05:27

### Video Transcript

Consider the power series the sum of three π₯ to the πth power over π plus five for π equals zero to β. 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, weβre going to recall the ratio test. The part of the ratio test weβre interested in says that, given a series the sum of π π, if the limit as π approaches β of the absolute value of π π plus one over π π is less than one, then the series is absolutely convergent. If we look at our series, we see that we can define π π as three π₯ to the πth power over π plus five. And then π π plus one is three π₯ to the power of π plus one over π plus one plus five.

And then of course that denominator simplifies nicely to π plus six. Our job is to establish where the limit as π approaches β 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 π plus one over π plus six times π plus five over three π₯ to the πth power. And we now see that we can divide through by three π₯ to the πth power.

Okay, so now we have the limit as π approaches β of the absolute value of three π₯ times π plus five over π plus six. We notice that three π₯ is independent of π. 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 π. Thatβs the highest power of π in our denominator. And so the denominator becomes π divided by π, which is one, plus five over π. And then the numerator becomes one plus six over π. And as π approaches β, five over π and six over π approach zero. And so the limit as π approaches β of the absolute value of one plus five over π over one plus six over π 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, weβve found the radius of convergence of our power series. Itβs π 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 weβve determined an interval of convergence. We certainly know that the power series converges for these values of π₯. But we donβt know what happens at the end points of our interval. Weβre going to clear some space and substitute π₯ equals negative one-third and π₯ equals one-third into our original power series. And weβll see if these series converge or diverge using an alternative test.

Weβll begin by letting π₯ be equal to negative one-third. Then our series is the sum from π equals zero to β of three times negative one-third to the πth power over π plus five. Which simplifies to the sum from π equals zero to β of negative one to the πth power over π plus five. And you might recognise this. Itβs an alternating series. This negative one to the πth power indicates so.

So weβ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 π, where π π is negative one to the πth power times π π or negative one to the power of π plus one times π π. And here π π must be greater than or equal to zero for all π. Then if the limit as π approaches β of π π equals zero and π π is a decreasing sequence, then we can say our series the sum of π π is convergent.

Letβs rewrite our series as the sum from π equals zero to β of negative one to the πth power times π plus five to the power of negative one. We now see itβs of the form given, where π is equal to π plus five to the power of negative one. Letβs check the limit as π approaches β of this π π is indeed zero.

Well, we know itβs equal to the limit as π approaches β of one over π plus five. And as π π gets larger, π plus five, the denominator of the fraction, gets larger and larger. So one over π plus five gets smaller and smaller and eventually approaches zero. So the limit of π π as π approaches β is indeed zero.

Next, we need to check whether π π is a decreasing sequence. To do so, letβs differentiate the entire function with respect to π. Using the general power rule, we see thatβs negative π plus five to the power of negative two, which is negative one over π plus five all squared. For all values of π between zero and β, 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.

Weβ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. Weβre now going to clear some space and test π₯ equals a third. This time, our series is the sum from π equals zero to β of three times a third to the πth power over π plus five, which is one to the πth power over π plus five, which is of course simply one over π 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.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions