 So, when we left off last week, we were looking at methods for determining an infinite series or infinite sums to figure out when they converge and when they don't converge. And so, when we came up with what I mentioned at the end is the last class or actually most of the last class is this idea of the integral test which says that if I have a sum, let's say, if I have a sum let's say n equals, it doesn't matter if 1 to the infinity of a n doesn't matter if and positive increasing, well if I have this then of n, the integral test says that if the integral from 1 to infinity of f of x converges, so does sum if the integral from 1 to infinity of f of x diverges, so this gives us a way for any series or any sum where we can integrate a formula for the terms provided that the terms are all positive and decreasing. So maybe this positive and decreasing seems like a strong restriction but in fact, you know, suppose we have something that increases for a little while and then it begins to decrease, then we just start here. So we don't need to start because if you apply the integral test for n equals 1,000 on all that matters is really the tail because the beginning is just some number, right? So for example, well so here's a stupid example but we can make it better. If I want the sum from n equals 4 to infinity of 1 over n minus 3 squared, we can easily see whether this converges or not. Or it did square, so let's make it q. So does this converge or not? And how would I tell? Nobody has a clue on Monday morning, so what would I do? Yeah, I don't need any, oh yeah, oh, not over, okay, yeah. So I just looked at the integral from 4 to infinity of 1 over x minus 3 q dx, which is, well, in my mind I think let u equal x minus 3, so du is dx, so this is 1 over u cubed, right? It's the integral of 1 over u cubed, so the integral of 1 over u cubed, so that's u to the minus 3, so that becomes minus 1 over 2 u to the minus 2. So this is 1 over 2 x minus 3 squared. This is the integral of u to the minus 3 du, so this is u to the minus 2 and then I divide by minus 3. So it's not 3, it's a 2. And we evaluate this from 4 to infinity, which is, well, the infinity part is actually a limit as x goes to infinity of minus 1 over 2 x minus 3 squared, this piece of chalk, that's this part, and then minus a minus 1 over 2, 4 minus 3 squared, so what's that? This piece goes to 0. Did I make a mistake? Okay, so since this converges and this does too, so did I make a mistake? No? I don't care about the balance for you. So are you saying you're having trouble with this step? Where are you? There you are. So okay, in my mind I'm thinking this, but since du is dx, I don't bother. If you prefer, I could of course let this be the integral from 1 to infinity of minus 1 over 2 u squared. That's fine too, but I was just, I mean this part I don't actually write down, I'm just thinking. But you can't do it by just thinking it right there. But I have to, I have to write it down because when I just think it, you can't receive it. I mean I have to try to open my head up. Usually it's just like blood and stuff. Okay, so the point is that we don't, we can't leave it out here. We don't need to start at n equals 1 because we're just looking at the end. And if you think about this, what the integral test is telling us is still the same picture. We have our series and I'm putting a box of height a whatever who ends or starts at the point 1, 2, 3. And it does some crazy stuff for a while. But after a while, we wanted to start looking like this. And so, and then this part we look at the integral of. And these are just numbers. And when we say does it converge, even if these first numbers are 100 billion, there's still a number. Converge means is it finite? So if I just ignore the first 100 billion terms, so what? We're caring about what happens here. That's really the question of this, the infinite part of infinite songs or infinite series is what really matters, what makes it mean that we have to do some work. Because our question is when do we add them up? Is it finite? Okay? So we can squeeze a little more information out of the integral test. Actually, let's just do this same one. So say we want to know, okay, so we know that this converges. We know that starting at n and going through infinity, 1 over n minus 3q adds up to something. Maybe we want to get a sense of what it adds up to. So, let's use this same picture. Notice that if I start, so this is the wrong picture. I'm just going to draw the same picture again but without the stuff. So here I'm starting at 4 and I'm going to put a little box of height 1 over n minus 3q by looking at the sum. And so starting with n is 4, I put a little box of height 1, which I guess I'm starting with a 4, it doesn't matter, whose height is 1. And then, right, so this is 1 over 4 minus 3q, so that's 1. And then I put a little box, I'm not going to draw this to scale because it will be way too short. Here, size a5, which is 1 over 8. And the width is 1. And I put the box 4, 5, so I put the corner here and so on. Right? Is it clear to people what I'm doing here? 1 over 6 minus 3 is 9, so 3q of 1 over 27 and so on. And I make these little boxes where I hook them up on one side. And now if I look at the integral here, it goes through those corners, which is, so I can look at the integral here, f of x, where I'm going to integrate from 4 to infinity. And that will be bigger than the sum. The sum is this area and the integral is this area under that. Since the series is decreasing, the integral, the area under the curve is bigger than the sum. Yes? No? Maybe? OK. So what does that tell us? That tells us that this sum, 1 over n minus 3q, which is, so just to emphasize, 1 plus, and then the next term is an 8 plus a 27 plus, so 6, 7 minus 3, 1 over 4q, et cetera. This sum is certainly less than a half because the integral was a half. This integral here is a half because we did that already. If I did it right. Yes. So this adds something wrong to it. What did I do wrong before? It can't be less than a half. Three halves. Yeah, I want to start it. I always do this. It's not less than a half, that's garbage. It's got to be less than, OK, what did I do wrong? Well, it is 1 plus the integral because the first term is 1. Yeah, so let's fix that because I drew the wrong thing, so this is the wrong, sorry. We'll fix it, don't worry. So let's draw the picture correctly. I drew the wrong picture, but I'll draw the right picture in a second. So let's do this one instead. So this goes through 1, here, here, here, and so now it's bigger, right? So I just went the wrong way. So now it's bigger than a half, so this is bigger than, bigger than a half, plus we can also look at the other integral, shift it by 1 to get, so that if I do this integral that I did wrong, this one, well, it's less than this one, which starts at 5. So it's less than, why is my brain off today? n plus 1 is smaller and n is bigger. So, oh, it's the tail. OK, sorry for confusing myself. All right, so suppose we want to look at plus 1 is smaller. OK, let me start over. Let's start with the integral. So here's my function starting at 4, and this integral here is something. I can now estimate this from below by putting my series here, or I can estimate it from above by throwing away this first term. So I can put the first term here, or I can put the first term here, and I'll get either bigger or small. So this means then, if you do it in general and then I'll do it in specific, that wherever I start, so if I want to do the sum of an from, I don't know, sum number k to infinity, and it will be bigger if I start at n and smaller if I start it in n plus 1 by k, that will be bigger and this will be small. So wherever I start, this integral will be bigger and this integral will be smaller, and it's just a matter of where I put my boxes, either in front or behind. All right, this integral here, the one that starts with k, which is bigger, means that I put my boxes this way, with them above the brackets. If I want the other one, I can think of shifting over by one and putting it below the brackets. It depends on whether I put the corner here or the corner here. Nobody has a clue what I found here? Okay. So that means that, for example, if I want to know how big is, let's just do one easier. 1 over n squared, well, I can say that certainly it's less than, so starting at 1, is less than the integral from 1 to infinity of 1 over x squared dx, which is 1. You've done that a million times. And it's bigger than the integral from 2 to infinity of 1 over x squared dx, which is a half. Is it clear to people that this is a half? Yes, I see nods. I don't see any head shakes. Shake your head if you're confused. Okay, so I know that this sum is between 1 and a half. Suppose I want a better answer. 1 and a half isn't very good. That's less than one decimal place. Suppose I want a better answer. What would I do? Suppose I want an answer within a tenth. What? Increase 2. Okay, yes. So what does that mean? That would mean that don't start at 1. Add up some numbers. This is 1 plus 4 plus a ninth plus 3. 4 is a 16, et cetera. So go for a while. And then let's stop at, say, the tenth term. And now what's left? A bunch of terms that I didn't add. So I just get out my calculator and I add up these numbers. And then I stop. And I say how much am I missing? How much am I off? Well, I'm off by this much. But this, if I started 11, is between the integral from 11 to infinity of 1 over x squared dx and the integral from 12 to infinity of 1 over x squared dx. Well, this integral is an 11th and this integral is a 12th. So if I add up these numbers, I'm good to within an 11th. Somewhere between an 11th and an 12th. I probably should have stopped at 9, so I'm good to within the 10th. So that means that whatever these things add up to, and I just sit down and work them out, I get some answer, and that answer is good to one decimal place. Because it's within 11th and 12th from the correct answer. And you can play this game with any series that you can integrate like this. So, what else? Should I do one? It's okay. So the integral test is very powerful. It's also kind of retaining the net. But like a lot of things that we do with math, we do some hard work, and then we use that hard work to make things easier later. Let's, yeah? Is the integral test the same similar to the comparison test? No, it's similar, but not the same. So we don't have a comparison test yet. I'm going to do that next. But first I need to have some things to compare to. So, so far, the only series that we know converge is done by Adhoff methods. And we know this one. So we might as well do this for all powers. Right? So we can now say, and usually use the letter P there, P for power. So for example, let's start at one. Let's do them all at once. So any power here. So I'm going to, so, again, remember this is one plus one over two to the P plus one over three to the P plus one over four to the P. Just to emphasize what changes and what doesn't. So I want to answer for all of these whatever power I choose, which ones are going to work and which ones aren't. Well, we already did this. So this will converge exactly when the integral of one over X to the P from one to infinity converges. And just in case you forgot that problem, let's just do it out. So this is, so when I want to integrate X to the minus P, I get X to the minus P minus one. Wait a minute, that sounds wrong. I increased plus one. There was parentheses that I didn't write, but okay. Divided by one minus P. Evaluated from one to infinity. So this is the limit as X goes to infinity of X to the one minus P over one minus P minus one over P minus one over one minus P. Now if we take this limit, so if P is less than one, then this is negative. This power is negative. Right? And so this goes to zero. So this goes to zero if P is less than one. Think of P as being like three. So if P is like three, wait, bigger than one? Yeah, that's what I meant. So I'm sort of right, left, dyslexic, so I really scroll in qualities all the time. If P is like three, then the X is on the bottom. If P is like negative three, then the X is on the top. So if X gets big, if P is bigger than one, this goes to zero. And this diverges. P is bigger than one. And when P equals one, well that's the one case I didn't do, but we already know that. When P equals one, we have the harmonic series. It's a log. Right? So this is the number. See? Same problem. Okay. So if P is less than one, it diverges. If P is bigger than one, it converges. So all of that garbage is telling us that this converges when P is bigger than one and it diverges when P is less than or equal to one. So we did this before. We did this before with this integral, which I just did it again. And we see that. So these go by the name of a P series, P for power. So just to summarize, this converges exactly when P is bigger than one. And that makes sense if you think about it, because when N is really big, it's only going to converge if what you're adding on is small enough. Well this is small when P is big. When over N with P is small, when P is big. So, and let me just emphasize it diverges. So this is sort of your yardstick or, so he wanted to do comparison tests. Well now we have a whole pile of things to compare to. Namely, these guys. So any power. That makes our life a lot easier because we don't have to integrate all the time anymore. The next useful thing that we can do to understand when things converge. Again, I'm just, so A and, I'm just looking at positive series for now. So suppose I have something like the sum from N equals one to infinity of, let's do something easy first. One over N to the fifth power plus, plus ten N squared. We can just look at this and see that it converges. So this has to converge. One over ten to the fifth plus ten N squared is less than one over ten. One over N to the fifth. Right? You take a number and you add something positive on the denominator, it makes it smaller. I have N to the fifth people in the room and we're going to divide up, I don't know what we're going to divide up. We're going to divide up something and then I bring in ten N squared more people, everybody gets less. So since that's less than that, then for sure if I add up a bunch of less things, less things adds up, this adds up, then the other thing does too. So this converges to some number. That's what it means to converge. It adds up to some number. We can figure out exactly what number by using the integral test. I don't want to bother. Not exactly, but approximately. So I don't want to bother. So this is less than some number so it converges. This is called the comparison test. So we write it more explicitly. It works just like the comparison test that we used for integrals and it works for exactly the same reason that the comparison test that we used for integrals. So again, it says if, well I don't want to give it up, so I have positive terms and then it says that if, so I have also some BN, and if the ANs are always bigger than some other BN, then the BNs converge then so do and also if, let's just turn around. Whoa. I'm really backwards today. Sorry. If the ANs are bigger than the BNs and the BNs diverge then so do the ANs. It's exactly the same as the comparison test for integrals. If the thing you're integrating is bigger than something else and the integral exists then so does this something else. If the thing you're integrating is smaller than something else and the small thing blows up then so is the big thing. So this one says if we have something holding us down that doesn't go to infinity, we don't go to infinity either. If we have something pushing us up that goes to infinity, we go to infinity too. I'm sorry for my trouble with inequalities. Okay. So let's do another example of that. Say we have something like this. Does this converge or diverge? You can use the integral test if you want. The integral isn't too hard. Or you can use comparison tests if you think of something you can compare to. Yeah. Okay. Why? Well what's AN and what's BN? So now this is not saying, this is saying if I have two different series, two different sums that I want to compare. So the example that I used here, this is my BN and this is my AN. And since my BNs are bigger than my ANs, my BNs converge then so do the ANs. You have to pick something. So we have two choices here. Okay. We want to compare this to one over AN. Alright. So how does this compare, how does log of N compare to one over AN? Remember A is bigger than one. So if I pick an N, ten, is the log of ten over ten bigger or smaller than one ten? Okay. Pick eleven. Log of eleven over eleven bigger or smaller than eleven. Okay. So this is bigger than one over N. This is bigger than one over N for N bigger than E. So the first term is not. The first two terms are not bigger. We don't care about the first two terms because it's just a few terms. The later terms are bigger. When N is one, I have zero which is certainly less than one. When N is two I have the log of two which is slightly less than one over two which is smaller than N. But when N is three, the log of three is bigger than one. So then it starts getting bigger. So this diverges since this is true and we know the harmonic series diverges. It's not true yet. Just bigger than three. Yes. So I mean if this is an exam question and there will be questions like this on the exam for sure. This answer is the kind of answer you want to write. So notice that it's not true in the beginning but that it becomes true. The question would be what about this? Does it converge or diverge and why? It's not significant. You might lose a point if you neglect N equals one and two. But if you just test N equals one and two, you might get it wrong because you'll try to test it the other way. Notice that you can also if you prefer do this integral. And this diverges too. So if you prefer you could do the integral test. I don't care. They're both right. You have to do this integral. It's not a hard integral. You make the substitution u equals the log. This is one over u du. The integral of one over u du is the log. So this is the log of the log. And as x goes to infinity, it blows up. So you can do that instead if you prefer. It's fine. Yeah. How many points are valid? I don't understand. You want the one. So you really want to know what happens for N big. And how big does N need to be? So all that matters is the comparison is what happens when N is big. And when N is big, this is bigger than that. When N is small, so what? Do you need me to do more comparison problems? Let me put up one that actually the comparison test doesn't work very well to stay here. I may have to fiddle with comparison tests a little bit. So I have this one here. N to the fifth plus 10 N squared. So that one we know converges. What if I change that plus to minus? And now since I have to do well N, let's just say N is bigger than 10. Still screws up to 10. Let's get rid of the 10. I would like to do a comparison test to 1 over N to the fifth. Look at this. This power matters for N big. This doesn't matter. But I can't. Why can't I? If it's a good question then you don't know what the hell is going on there. Sorry. So suppose, okay so let's do this one again with the plus first. With the plus I want to compare with 1 over N to the fifth. So if I compare with 1 over N to the fifth, what do I see? I see that 1 over N to the fifth plus N squared. I don't want to do partial fractions. You can. I don't want to integrate. This is less than 1 over N to the fifth with the plus. If I take an N to the fifth and I add more at the bottom it gets smaller. So that means since 1 over N to the fifth converges because of this P series test, so does N to the fifth plus N squared. This is exactly the same example as that one except I got rid of the 10. This is just straightforward comparison test. I found something bigger that converges. Happy times. I'm going to move over here now. I'm going to change the problem to make it a little harder and then I'm in trouble. And if you want to do partial fractions you are welcome to do that. So it's almost the same problem but this doesn't work anymore. So I want to do this. So this is wrong. I want to say well okay 1 over N to the fifth minus N squared that's kind of like 1 over N to the fifth and then I look at it and I think about it and I say okay well that's bigger than 1 over N to the fifth because it is because I took a little away from the bottom and this converges and this tells me nothing. I'm bigger than something that converges. Nice for you. It says nothing. So I would claim all this converges too because they're kind of the same. I still want to be able to say they're kind of the same. This converges. So this converges but this is wrong. This doesn't help. So how can I make it help? Well I can work harder. So this is bigger than 1 over N squared for N bigger than 1. So that's fine. And then it's good. So this is fine. So here's one good way. Oh wait smaller than. I'm really backwards today. So that's good and that's fine. What else could I do? Well I can make it, yeah. Yeah but the comparison only works for positive terms. Now you can turn it, and if we turn both over it doesn't work. So the comparison only works for things that are positive. You can make it work for things that are both positive and negative with a little effort that we'll do next time. We'll maybe fry it. Yeah. And to the fifth minus 1 over N squared if it comes to be a series that converges. You could do partial fractions and integrate. Somebody over here really wanted to do partial fractions. So you could do partial fractions and integrate. That's okay too. The thing that I haven't told you that you could do is something that's, I guess I'll leave it here and I'll go back over there. So the other thing you could do, and again, there's many choices here. So this is one of the things that annoys students about this, actually this whole class, is just like with integration we often have several ways to do the same problem. Or several things to try. And some will work and some won't. Okay, so the last thing that we can try is I can try 1 over N squared minus 1, minus N squared. That's a, sorry, this is a 5, right? With 1 over N to the fifth. I want to somehow compare these two things. Because in my heart I know this doesn't matter. When N is big, N squared doesn't matter. So I want to compare these two. So let's look at the limit as N goes to infinity of one of them divided by the other one. Geez. Can't get a number now. If I take this limit, so this is N to the fifth minus N squared over N to the fifth limit, this is 1 minus 1 over N cubed is 1. What does that tell me? If I look at the limit of the ratio of two things, what does this say, and it's 1, what does this say to me in words? What? They're almost equal. So for N big, 1 over N to the fifth minus N squared is almost the same as 1 over N to the fifth. This means that in fact I can compare the two series. If the limit of the ratio is 1 to a number, then they're comparable. And so I'm not going to prove this, but we can change the comparison test to a new kind of comparison test called the limit comparison test, which I can say in one minute. So again, A-N, B-N, positive series. And then we say if the limit as N goes to infinity of 1 over the other is not 0 and it's not infinity, then the two series do the same thing. The limit comparison test says if I have a series of positive terms something that looks kind of like it, if for big things their ratio is not 0 it doesn't blow up, then they do the same thing. So in this case, it doesn't really matter what's here because the limit will kill it. 1 less than 5 in power. I'll come back to that. So we have now three tests, yeah. Does it matter which? No, because I said it's not 0 or infinity. So pink one, put it on the top. The other one on the bottom. If you get a number, you're good.