In this video, I use the Integral Test to show that the p-series, whose terms are 1/n^p, is convergent if p is greater than 1 and divergent for all other values of p. When p is less than 0, the ter...
Transcript
Let's go over example 2, and this is on the P series, and it states, for what values of p is the following series convergent, and we're given the summation n equals 1 to infinity of 1 over n to the power of p, hence, p series. So let's check some easy values for p, so if... If we have p is less than 0, then what we end up having is 1 to the 1 divided by n to the power of a negative power, which gets flipped, it becomes positive. So then what we have is limit, so each term limit as n approaches infinity of 1 over n to the p. This is going to be equal to infinity. This is going to approach infinity because, i .e., let's put an arrow. For example, if we had n to the power of, I mean, if we had something like 1 to the power of, 1 divided by n to the power of negative 10, this gets flipped upside down, so this becomes end to the power of 10. So the negative power we call, let's just flip it. up, you can see this in my earlier videos on power functions. Yes, so this is a positive, and then as you're going to infinity, you're going to have infinity the power of 10, or even if it's 0. It doesn't matter what is. It's going to be a positive power, it's going to go approach infinity because it's always positive and increasing. So, hence, if p is less than 0, that is divergent. And it's, yeah, so the terms are going to infinity themselves. And now, if p is equal to 0, let's see what we have here. So if p equals 0, then, then what we end up having is a limit as n approaches infinity of, well, 1 over n to the 0. And recall this is defined, yeah, my early video has defined it as equals to 1. So this just equals to, yeah, so this right here, put an arrow, equals to 1. So in other words, this whole thing is equals to 1. Limit of 1 over 1 as n, approach to infinity is just 1. The n is gone, like that. So then all the terms are once. So you're going to be summing here, one, one, one, one, so those are going to be divergent as well. So yes, in either case, the limit as n approaches infinity of 1 over and the power of p is not equal to 0 when p is equal 0 or p is less than 0. So the given series diverges by the test for divergence, also from my early video on infinite series, and it's from that, yeah, the same one here. Here's a screen cap from the... Calco is spoken the link to the Steam that notes again. So if the series right here, n equals 1 to infinity, the summation of A.N is convergent, then the limit as n-approach infinity of A-N has to equal to 0. So if it's convergent, this has to be going to zero as well. And there is the proof for reference. Again, you could see the earlier video to see it in more detail. And just to note, the opposite of this theorem is not true in general. So if the limit and approach infinity of AN is going to zero, we can't conclude that the whole series convergent. Because that harmonic series should before, it diverges even though that the terms are going to zero. And likewise, is the test for divergence. If the limit A.N does not exist, limit A.N as N approaches infinity, does not exist, or if limit N approaches infinity, A.N is not equal to zero, then the series, as N equals 1 to infinity, is divergent. So, yeah, so this is exactly what we have here. This one here is, it doesn't exist, goes to infinity, and this one equals to 1. So thus, by test to divergence, that series is diverged. So the test of divergence follows from theorem 6 because of the series is not divergent, then it is convergent, and so limit as an approach to infinity, AN equals 0. So yes, for our case, if p is greater than 0, then the function f of x equals 1 over x, p is clearly continuous, and it's also positive, yes, continuous, because there's no, you can't divide it by 0 there. One, yeah, you can't get a 0 at the denominator, so it's always continuous, it's always positive, because every number is just positive there, and it's decreasing, because as you increase the value of x right here, you're going to be dividing by bigger and bigger number, because this is positive, P is positive. And yeah, so it's always going to be, the bottom always gets bigger, so that's the one divided by, increasing means it's decreasing. So it's decreasing on the interval from 1 to infinity, and recall from my early video on improper integrals, which I mentioned above that, yeah, the one I went over earlier, so this one here, and let's just scroll up to that exact spot where I referenced that. over here. Actually, I scroll up more so all the way over here, and it's a summary of the example for there. So, this integral from 1 to infinity of 1 over x, the power of p, dx is convergent if p is greater than 1, and divergent if p is less than or equal to 1. So for values p is greater than 1, well, we know that's from my earlier video, that it is convergent. So it follows from the integral test, the series, summation of 1 divided by and the power of p converges if p is greater than 1 and diverges if it's between 0 and um yeah 0 and 1 there so if it's greater than 1 it's going to be divergent and if it's less than that and in our case uh yeah if it equals zero as well or if it's less than zero it's also diverges so uh going further and uh note that i briefly mentioned as well note that for p is equal to one this series is the harmonic series and in which i discussed in my previous infinite uh series video as well so yes make sure to watch my early videos because uh everything is follows up in all other all other videos So recall this one here, so if p is equal to 1, we have 1 over n to power of 1, so that's just harmonic series. We have 1 plus 1 plus 1 over 2 plus 1 over 3 plus 1 over 4 and so on, and then the proof that I showed it was actually divergent. So yes, so this is actually divergent there, even though that the actual 1 over n, the limit of that term goes to 0 as n goes to infinity. So note that in order to use integral tests, we need to be able to evaluate the integral from 1 to infinity of f x dx, and therefore we have to be able to find an anti-derivative of f. So yes, we need to enter the derivative. Frequently, this is difficult or impossible. Again, like that other one where I showed, I get the inverse 10, and that was a somewhat difficult one, too, because you have to find out a... Yeah, an antiderivative that isn't used that often, well, at least for me, but sometimes it's difficult or impossible to even find any antiderotiv. So we need other tests for convergence, too. The series and example two is called the P series, and it is important in the rest of this and later videos, so we summarize the results of example two. If for future reference as follows. So theorem 1, P series, the P series, right here, 1 over N to power of p, 4n is equal to 1, summation up to infinity, is convergent if p is greater than 1, and divergent if p is less than or equal to 1. So just keep that always in mind. And now let's look at example 3. So the series over here, summation from n equals 1 to infinity of 1 over n cubed, is going to be 1 over 1.1 cubed plus 1 over 2, plus 1 over 3, plus 1 over 4 cubed, plus, et cetera, keeps going on on. And this is going to be, well, the P test, p is greater than 1, p is 3. So it's convergent because it is a P series with p is equal to 3, which is greater than 1. So, yes, let's just use that test, and we can quickly solve that. But over here, the series, summation from n equals 1 to infinity of 1 over n to power of 1 over 3. So this 1 over 3 is going to be less than 1. So that's going to be obviously divergent. And you can also write it like this, cubed root. So a summation of n equals 1 to infinity of 1 over n cubed root is equal to 1 plus 1 over 2 cubed root. and then plus 1 over 3 cubed root, and then plus 1 over 3, 4 cubed root, plus et cetera. And this is divergent because it is a P series with p equals 1 over 3, and that's less than 1. So yes, you can just use that theorem and quickly see if it diverges or converges. And I want to make a note that we should not infer from the integral test that the sum of the series is equal to the value of the integral. In fact, as recall that I stated earlier, and I will prove in a much later video, the summation from n equals to 1 up to infinity of 1 over n squared, this is equal to pi squared over 6. Whereas the integral is going to be, well, way off, it's going to be the integral from 1 to infinity of 1 over x squared, d x, this is actually equal to 1. So they are not the same in this case. And therefore, in general, well, yeah, we can't say they're equal, so in general, we'll say that they're not equal. So the summation from n equals 1 to infinity of A, n. this series is not equal to in general from the integral from 1 to infinity of f x d x like that