 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misseldine. In this lecture, we are beginning Part 38 of our series, which we want to talk more about the integral test, which we introduced that in the last few videos in Lecture 37. But I want to continue about that. And in fact, I want to talk about a very special case of the integral test that's special enough that actually gets its own name. This is what we'll commonly refer to as the p-test. So, first of all, what do we call a p-series? A p-series is a series of the form n equals the sum of n equals 1 to infinity of 1 over np. So p is some unspecified parameter, and that's why we call it a p-series. And so the p-test tells us that a p-series will be convergent exactly when p is greater than 1 and it'll be divergent when p is less than or equal to 1. And so let's see an argument about what's going on here. It turns out this is going to be using the integral test. So what we can see here is that by the integral test, the series n equals 1 to infinity of 1 over n to the p, this is going to be convergent. It's going to be convergent or divergent exactly when the integral 1 to infinity of 1 over x to the p dx is convergent or divergent. So what I'm trying to say here is that when the integral is convergent, that means the series is convergent, or when the integral is divergent, that'll mean that the series is divergent. That's what the integral test gives us. So we want to investigate what happens with the integral from 1 to infinity of 1 over x to the p there. Now this kind of depends on the choice of p. So some considerations. If p is equal to 1, that's sort of a special case. The integral from 1 to infinity here, you're going to get 1 over x dx. The antiderivative there will be the natural log of x as you go from 1 to infinity. And that gives you the natural log of infinity minus the natural log of 1. Although the natural log of 1 is equal to 0, there's really no reconciliation with the natural log of infinity. That's just going to be infinity as well. So infinity minus infinity is infinity. And we see that this integral will be divergent. And so when p equals 1, the p series will be divergent. That's an important case to remember. Now if p is not 1, then it turns out the integral from 1 to infinity here of 1 over x to the pdp, we could, or dx there, we could write this as the integral from 1 to infinity. We're going to get x to the negative pdx. And so by the power rule, we can use the power rule here. We're going to raise the power x to the 1, well, negative p plus 1, so we get 1 minus p power. You have to divide by this coefficient 1 minus p as we go from 1 to infinity. When we plug in 1, that's pretty simple. And we should also just plug in infinity, right? I'm just going to factor out the 1 minus p there. You're going to take this infinity to the 1 minus p minus 1 right there. Now we have to do a little bit better about this infinity to the 1 minus p, because it turns out that not all powers in infinity are infinite, right? This has a lot to do with whether we have a positive or negative exponent. So if we break this up a little bit further here, if 1 minus p is a positive number, greater than 0, this would imply that infinity to the 1 minus p, this is going to be infinite. You're taking the infinite power of infinity. Now if 1 minus p is greater than 0, that implies that 1 is less than p, like we see here. And so we're going to get divergence in this situation. Divergence. Combining this with the previous case, we can then say that, oh, when p is less than or equal to 1, we get divergence. Now the other direction, what if 1 minus p was less than 0? Then you're looking at infinity raised to a negative power. We can think of the reciprocal of 1 over infinity to the p minus 1. And that situation, you actually end up with a 0. That thing would be convergent. That thing converges. Now if 1 minus p is less than 0, that actually says that 1 is less than p. And so that actually brings us up to the dichotomy we had before. Our series, the p series, the sum of 1 over n to the p, it'll be convergent when p is greater than 1 like we just saw. And it'll be divergent when p is less than or equal to 1. And so it's a very simple test, but it's very, very powerful. We're going to see more in the future why these p series is so important. But we can get a very quick check going on right here. If we take the sum as n equals 1 to infinity of 1 over n cubed, this is a p series. We can see that p is equal to 3. 3 is greater than 1. And therefore this series will be convergent by the p test, which the p test is a special case of the integral test. But we'll often just refer to it as the p test. So this is a convergent series, which is pretty neat. That's all that one has to do. That was super slick. No anti-driven is necessary here. If we look at the next one, take the sum where n equals 1 to infinity of 1 over the square root of n. Now that one takes a little bit of effort, but not much, right? If you look at the sequence 1 over the square root of n, we can realize this as a p sequence. This becomes 1 over n to the 1 half. So your p value is equal to 1 half. That's less than or equal to 1. So this would actually be divergent. This is divergent by the p test. And so how nice is that? How simple is that? All we have to do is identify the parameter p for a p series and we're good to go. So for one third example of this, if we take the series where n equals 1 to infinity of 1 over n, this is likewise a p series. In this case, our p value is 1. This is often referred to as the harmonic series. The harmonic series is a very important example of a series. It is a p series, though, where p equals 1. And so this is the one that kind of sits on the boundary, right? It flips from convergence to divergence as we shift from bigger than 1 to less than 1. But what happens at 1? At 1, the harmonic series, we are divergent. We saw that before. And so this likewise would be divergent by the p test. And so determining the convergence of a p series is no harder than this. All one has to do is identify that has to just determine the p value. And then we see whether we're greater than 1 or less than or equal to 1 to determine convergence or divergence.