Understanding Power and p-Series

In this video, we will be looking at what's called the power series, which we will also refer to as the p-series. These are series in the form of 1 over n to a power. So if we were to expand it out, it looks something like this. One plus one-half to a power plus one-third to a power. Keep going as long as that power is greater than zero. When p is equal to one. We get the harmonic series that we saw in the first example in the previous video. That harmonic series expands out to be 1 plus 1 half plus 1 third plus 1 fourth and so forth. And we found out that that series diverged using the integral test. Remember, we set up that improper integral, and we got that the integral diverged, and so the harmonic series also diverged. But that's for when the power is equal to 1. What happens if our power does not equal to 1? Well, again, I'm going to use the integral test. So I'm going to let a function be 1 over x to a power. And I know this function is continuous and positive and decreasing for all x greater than or equal to 1. So I'm going to set up my improper integral, which I have here. Remember... x to a power on the bottom can be written as x to a negative power in the numerator. Now the reason why I can do this is because I'm going to use the power rule for integration because I know p does not equal to 1. Remember when p is equal to 1, then I get that natural log function. Everywhere else, I can use my power rule of integration. So exactly what is that power rule? Well, I'm going to have the limit as a approaches infinity. And then using that power rule, remember I add 1 to the power, and then I multiply by the reciprocal of that power. Well, that looks a little scary. Let's just rewrite it. Since this part is just going to be a constant, and the limit's not going to be involved, I'm going to pull that outside. I'm going to write this as 1 over negative p plus 1. multiplied by the limit as a approaches infinity of x to the negative p plus 1 power. So that's going to be 1 over negative p plus 1, the limit as a approaches infinity of a to the negative p plus 1, minus 1 to the negative p plus 1. So our question is what happens as a approaches infinity? Well we really need to look at that power. This limit is only going to exist if negative p plus 1 is negative. Now think about that. That means a to some negative power. What's that equivalent to? It's going to be 1 over a to some positive power. And as a goes to infinity, that would be 0. And so as long as negative p plus 1 is always negative, that's going to go to 0. That's just subtracting 1, and so our limit exists. Now what happens when negative p plus 1 is positive? Well, again we'd have a to a positive power and a as a gets bigger and bigger, this goes to infinity and the limit would not exist. So the only way that this integral converges is if negative p plus 1 is negative. How do we make sure that happens? Well, if negative p plus 1 is negative, that means that has to be less than 0. So I could rewrite this as if I add p to both sides, I'd have 1 is less than p. In other words, p has to be greater than 1. If p is greater than 1, then my associated integral converges. And if negative p plus 1 is positive, meaning greater than 0, then it's going to diverge. We can make these conclusions for our power series, otherwise known as the p-series. It converges if p is greater than 1, and it diverges if p is less than 1, because that would make this negative p plus 1 positive. But we also saw that it diverged in the harmonic series when p equal to 1. So we really need to be able to identify a p-series, not be confused with... with geometric and know when it converges and diverges. So let's see if we can apply some of our rules that we've learned, meaning we could use the geometric series test. the telescoping test, the nth term test, we have the integral test, and now we have the p-series test. You have to think about all of those and when does each apply to each problem. So let's look at this example. Does it look geometric? No. Does it look telescoping? No. What about the nth term test? Well, let's see. The limit as n approaches infinity. of 1 over 2n minus 1. Well that equals to 0 so that fails. Remember the only time the nth term test applies is when that nth term does not go to 0. So this fails. P series. Well I certainly don't have anything in this form of 1 over n to a power. So the only thing I have right now is my integral test. So let's use that. So the first thing using the integral test test, I'm going to let f of x be my function in terms of x for x greater than or equal to 1. And I know f of x is continuous and positive and decreasing on this interval. And I have to be really careful about that because it's not continuous at x equals a half right there, that denominator. That would give me a vertical asymptote at 1 half, but 1 half is not bigger than 1. So it's continuous. As long as x is bigger than 1, those terms are all positive. And as you notice, it's decreasing. As x gets bigger, the denominator gets bigger, the fraction gets smaller. Okay, so now I can say, let's take the limit as a approaches infinity. Let's do that integration. So that's going to be 1 half times the natural log of 2x minus 1. Just your natural log rule, let that be u, du is 2, so that's where this 1 half comes from. And so if I have the limit as a approaches infinity of 1 half times the natural log of 2a minus 1, minus one half times the natural log of, let's see if I put in one, that's going to be two minus one of one. So what happens as A approaches infinity, the natural log of something really huge, approaches infinity so this limit does not exist so we can therefore say our series diverges because of the integral test. So just another thing to practice. Right, let's look at number 5. Right, is it geometric? Now remember, geometric is some number raised to a power. Okay, not hashtag, but number. Where a p-series is 1 over n. raised to a power. Well number 5 looks very similar to a p-series. What about that 3? Well just like an integral we can take that 3 outside the sigma so we could write this as 3 times the series of 1 over n to the 5 thirds power. And then I know this is a p-series where p equals 5 thirds. I know that's greater than 1, so I can then conclude that the series converges because of the p-series test. Look at how you have to write down your reasonings and your conclusions. That's very important when working these problems. Okay, what about number 6? Is that a p-series? Well, I've got 1 over n to a number. That's pi. Okay, that's the same as writing this as 1 over n all raised to the pi power. Well, that's definitely a p-series. Pi is greater than 1. So therefore, the series converges because it is a power series. Now number seven, ooh, factorials over numbers. Hmm, if I go through my list, this isn't anything. I certainly don't want to integrate that. But what about that nth term test? The limit as n approaches infinity of my terms. What is this? Well, if you recall, factorials grow much faster than powers. So this limit does not exist. It goes to infinity. And yes, that's good, because that does not equal to zero. So therefore, the series diverges because of the nth term test. So again, as you are going through these problems, make you a list mentally or actually write it out. And think about what all types of series do we have so far? And yes, we're going to have a few more tests, but so far we have the geometric test, the telescoping test, the nth term test, the integral test, and the p-series test. You need to know when each of these converge, diverge, what's it a test for? Keep studying.

One plus one-half to a power plus one-third to a power. Keep going as long as that power is greater than zero. When p is equal to one. We get the harmonic series that we saw in the first example in the previous video.

That harmonic series expands out to be 1 plus 1 half plus 1 third plus 1 fourth and so forth. And we found out that that series diverged using the integral test. Remember, we set up that improper integral, and we got that the integral diverged, and so the harmonic series also diverged.

But that's for when the power is equal to 1. What happens if our power does not equal to 1? Well, again, I'm going to use the integral test. So I'm going to let a function be 1 over x to a power. And I know this function is continuous and positive and decreasing for all x greater than or equal to 1. So I'm going to set up my improper integral, which I have here.

Remember... x to a power on the bottom can be written as x to a negative power in the numerator. Now the reason why I can do this is because I'm going to use the power rule for integration because I know p does not equal to 1. Remember when p is equal to 1, then I get that natural log function.

Everywhere else, I can use my power rule of integration. So exactly what is that power rule? Well, I'm going to have the limit as a approaches infinity. And then using that power rule, remember I add 1 to the power, and then I multiply by the reciprocal of that power.

Well, that looks a little scary. Let's just rewrite it. Since this part is just going to be a constant, and the limit's not going to be involved, I'm going to pull that outside.

I'm going to write this as 1 over negative p plus 1. multiplied by the limit as a approaches infinity of x to the negative p plus 1 power. So that's going to be 1 over negative p plus 1, the limit as a approaches infinity of a to the negative p plus 1, minus 1 to the negative p plus 1. So our question is what happens as a approaches infinity? Well we really need to look at that power.

This limit is only going to exist if negative p plus 1 is negative. Now think about that. That means a to some negative power.

What's that equivalent to? It's going to be 1 over a to some positive power. And as a goes to infinity, that would be 0. And so as long as negative p plus 1 is always negative, that's going to go to 0. That's just subtracting 1, and so our limit exists.

Now what happens when negative p plus 1 is positive? Well, again we'd have a to a positive power and a as a gets bigger and bigger, this goes to infinity and the limit would not exist. So the only way that this integral converges is if negative p plus 1 is negative. How do we make sure that happens?

Well, if negative p plus 1 is negative, that means that has to be less than 0. So I could rewrite this as if I add p to both sides, I'd have 1 is less than p. In other words, p has to be greater than 1. If p is greater than 1, then my associated integral converges. And if negative p plus 1 is positive, meaning greater than 0, then it's going to diverge. We can make these conclusions for our power series, otherwise known as the p-series.

It converges if p is greater than 1, and it diverges if p is less than 1, because that would make this negative p plus 1 positive. But we also saw that it diverged in the harmonic series when p equal to 1. So we really need to be able to identify a p-series, not be confused with... with geometric and know when it converges and diverges.

So let's see if we can apply some of our rules that we've learned, meaning we could use the geometric series test. the telescoping test, the nth term test, we have the integral test, and now we have the p-series test. You have to think about all of those and when does each apply to each problem. So let's look at this example.

Does it look geometric? No. Does it look telescoping? No. What about the nth term test?

Well, let's see. The limit as n approaches infinity. of 1 over 2n minus 1. Well that equals to 0 so that fails. Remember the only time the nth term test applies is when that nth term does not go to 0. So this fails.

P series. Well I certainly don't have anything in this form of 1 over n to a power. So the only thing I have right now is my integral test. So let's use that. So the first thing using the integral test test, I'm going to let f of x be my function in terms of x for x greater than or equal to 1. And I know f of x is continuous and positive and decreasing on this interval.

And I have to be really careful about that because it's not continuous at x equals a half right there, that denominator. That would give me a vertical asymptote at 1 half, but 1 half is not bigger than 1. So it's continuous. As long as x is bigger than 1, those terms are all positive.

And as you notice, it's decreasing. As x gets bigger, the denominator gets bigger, the fraction gets smaller. Okay, so now I can say, let's take the limit as a approaches infinity.

Let's do that integration. So that's going to be 1 half times the natural log of 2x minus 1. Just your natural log rule, let that be u, du is 2, so that's where this 1 half comes from. And so if I have the limit as a approaches infinity of 1 half times the natural log of 2a minus 1, minus one half times the natural log of, let's see if I put in one, that's going to be two minus one of one.

So what happens as A approaches infinity, the natural log of something really huge, approaches infinity so this limit does not exist so we can therefore say our series diverges because of the integral test. So just another thing to practice. Right, let's look at number 5. Right, is it geometric?

Now remember, geometric is some number raised to a power. Okay, not hashtag, but number. Where a p-series is 1 over n.

raised to a power. Well number 5 looks very similar to a p-series. What about that 3?

Well just like an integral we can take that 3 outside the sigma so we could write this as 3 times the series of 1 over n to the 5 thirds power. And then I know this is a p-series where p equals 5 thirds. I know that's greater than 1, so I can then conclude that the series converges because of the p-series test.

Look at how you have to write down your reasonings and your conclusions. That's very important when working these problems. Okay, what about number 6?

Is that a p-series? Well, I've got 1 over n to a number. That's pi. Okay, that's the same as writing this as 1 over n all raised to the pi power. Well, that's definitely a p-series.

Pi is greater than 1. So therefore, the series converges because it is a power series. Now number seven, ooh, factorials over numbers. Hmm, if I go through my list, this isn't anything.

I certainly don't want to integrate that. But what about that nth term test? The limit as n approaches infinity of my terms. What is this?

Well, if you recall, factorials grow much faster than powers. So this limit does not exist. It goes to infinity. And yes, that's good, because that does not equal to zero. So therefore, the series diverges because of the nth term test.

So again, as you are going through these problems, make you a list mentally or actually write it out. And think about what all types of series do we have so far? And yes, we're going to have a few more tests, but so far we have the geometric test, the telescoping test, the nth term test, the integral test, and the p-series test. You need to know when each of these converge, diverge, what's it a test for? Keep studying.

Transcript for:Understanding Power and p-Series

Transcript for:
Understanding Power and p-Series