 Welcome back lecture 37. I think I think I looked at this yesterday. It was day 44 so today is day 45 of a 71 day semester. So we are certainly past the halfway point and headed toward the homestretch. So I gave the same message to my son. Now is not the time to coast or relax. Now is the time to start to put the foot on the accelerator because this is not the coasting or relaxing time of the semester. And that sound like something my dad should say. And he said oh, yeah dad. Okay. Yeah, that's really good. Where we eat? One thing I forgot to show you yesterday. I did mention it, but I forgot that I actually had this printed up. These models that we're working with whether it's a circuit with the L's and the R's and the 1 over C's or if it's a spring with the mass and the constant times the velocity which has to do with the viscosity of the fluid that the spring is in or the spring constant. These are kind of parallel equations. Second derivative term as a certain coefficient. First derivative term, certain coefficient. The original function, certain coefficient. And if there's a nice little chart in your supplement that talks about the parallel between these for the spring, the displacement or the amount of units it has traveled kind of its location. Q, the charge are really the same thing. So there's our original function in both x and q. Velocity, dx over dt, current derivative of the charge. So there's that term in the middle and then we've got the mass times acceleration and the L value times the derivative of the current which is second derivative of the charge. So you see a lot of parallels. It's the same mathematical model. So we've seen this before in mathematics whether it is y equals y0 e to the kt or a equals p times e to the rt. Different kind of visual appearances, but the same exact mathematical model. This is something growing or decaying exponentially. This is what is that? a equals p times e to the rt. Continuously compounded interest, right? But it is exactly the same mathematical model, just as these two are the same mathematical model. Alright, we're back to the book. Time permitting, we can go back and examine that, the end of that problem and see kind of where things went wrong on that number six problem from 7.9 web assign, but let's kind of do what we're supposed to do and see what time we have remaining to look at that problem. So in chapter 8, we're going to do sequences in series and we are about to experience one of the reasons why this particular course, Math 241, is perceived to be way more difficult than 141 or even 242 for that matter, is it is kind of choppy, is that we just finished doing a fairly significant introductory unit on second-order differential equations and now all of a sudden we switch horses completely, it seems like, and we go on to sequences in series. And that's not at all like, seemingly, what we did earlier in the semester, which is all the techniques of anti-differentiation, all that stuff that we did early. So it is kind of choppier in a sense, but they are loosely connected, not as strongly connected as the topics in Calc 1, and when you do Calc 3, you'll see that this course is not nearly as strongly connected as topics in Calc 3. So I think the choppiness factor enters into the perceived difficulty of this course. 8.1, which we'll look at today, and I think we can pretty much do justice to this today, is looking at sequences, and a sequence is just a list. So if it's got commas in there, we're talking about a sequence and not a series. If it's a series, we are talking about the indicated sum of these terms that are in the list. So it's just, first section is just a list. Now, why do we need to look at sequences if most of the rest, excuse me, rest of the chapter is talking about series, and when series converge and all kinds of convergence tests. In order for a series to converge, its sequence of partial sums has to converge. So if we don't first look at sequences, and what causes a sequence to converge, we're missing something basic in the background of what causes a series to converge. But when we look at the list, we're looking at what happens as we work our way out to the right. Those terms, way out there to the right, are they closing in on any one specific number? If they are, then it converges. If they are not, then it diverges. So we've already dealt with those things in terms of an integral, right? When an integral converges, we got a finite answer when it, we didn't get a finite answer, we said it diverged. So we will make some connections before we leave this chapter on convergent integrals and convergent series. In fact, one of the tests that we'll use is called an integral test. But that's later in chapter 8. So we have a list, a sub 1, or simply a, if you see a, that's the first term, a sub 2, you'll work your way up to a sub n, which would be the nth term of the sequence. The next term, just for notational purposes, would be a sub n plus 1. What would the preceding term be for a sub n? A sub n minus 1. So that's just a notational thing. So you can kind of pay attention to where the terms are in the series, in the sequence. If we have a sequence that is described, so the nth term of the sequence is described as n over 3n minus 1, we could generate that entire sequence and we could also decide if that sequence converges. So a sub 1 would be where n is 1, which is 1 half, a sub 2, where n is 2, and so on. So we've got this list going, what's the next term? We're not adding anything together yet. When n is 3, what do we get? 3 eighths. I don't know if it's obvious that as we progress from one term to the next, this particular sequence is convergent. So can you tell from the description the value to which we are converging? One third. One third, right? We've done limit problems like that. So we work our way out to the right as n approaches infinity. We examine kind of the shortcut. Normally we see x's here and x's here and x approaches infinity. It's the same limit problem, just n's in those positions. So the shortcut is, somebody tell me, what is the shortcut for this particular limit problem? Leading coefficients. We're going to use the leading coefficients when the degree is the same. So this is an n to the first. Down here we have an n to the first. So we've got 1n over 3n. Doesn't eventually the minus 1 become pretty insignificant as n approaches infinity. So as we look at these terms, as we work our way out to the right, they get closer and closer and closer to one third. Will we ever actually get to the value one third? No. I mean, you can put in n equals a thousand. And this would be 3 times a thousand minus 1, which is what? 2,000. 2,999. Is that pretty close to one third? Very close to one third. If you put in 100,000, you're going to get even closer. So it is a convergent sequence. Again, we're not addressing in this section. We're not adding any terms together. We're just looking at the value of that term as it progresses in this sequence way out to the right. Series, believe it or not, will also be able to converge. We'll be adding an infinite number of terms together. And we do get a number, so we get a convergent series. But this is just what happens to this nth term way out to the right. I'm going to try to use some examples that show some things that we're going to encounter throughout this chapter. This has a couple different things that we're going to encounter. What do you think the negative one to the n does to this sequence as it progresses to the right? Part of this sequence. What does it do? It alternates the signs. And if you put in some different values, the first term would be negative one to the first. The second term, negative one squared. The third term, negative one cubed. So the alternating of signs could in fact become an issue. It kind of depends on what else is in that sequence. So negative one to the first, we get a negative, negative one squared. We get a positive, negative one cubed. We're back to negative. And they're going to alternate based on whether we have an even or an odd power of negative one. So let's look at the rest of this progression. So the numerator is increasing by one, one digit that we add to n. So it's getting larger. Isn't the denominator getting larger at a much larger rate? So we're going two over two, three over four. It may not be evident there, but now we've got four over eight. The next one would be what? Five over 16. So doesn't the denominator win in a sense? So when the denominator eventually wins, what happens to the part that I've circled here? It goes to zero. It goes to zero. So we've got stuff that may not be that evident when we start. So we've got, I'm going to go ahead and keep it two over two, three over four, four over eight. And I think we said it was going to be what? Five over 16. It may not be clear from the ones that we have thus far, but eventually the values of this particular sequence close in on zero. So the alternating signs don't become an issue that they might in another sequence because we're oscillating on either side of zero. So you can approach zero from either direction, positive and negative. And in a sense, if we're down here at one and up here, and then for the next one we're here and we're here, we are closing in on the value zero, but we're just doing so from positive values on the what? Even terms and negative values on the odd terms. It is possible to converge in that fashion. But if you just had an alternating sequence, let's say the sequence is just a to the n equals negative one to the n and that's it, that doesn't have any hope of convergence, right? Because you're going from negative one to one to negative one to one and so you're stuck in those two. Are we closing in on any one specific number? That would certainly by itself be divergent, even though this one is convergent. It converges to zero. It does so though. Some terms are positive, some terms are negative. You may encounter sequences or later descriptions of terms that we're going to add together, where it tells you that you can only use certain values. And this is all of these are functions of integers. The domain is the set of integers. So if we don't want to start with all integers, we better somehow put a little disclaimer here that says we don't want to start with all integers. We want n to be integers greater than or equal to two. So give me some terms of this particular sequence. When n is two, what do we get? Two. When n is three? Three and a half. When n is four? Four thirds. Four thirds, right? And we continue with that same process. I don't know if it's clear from these numbers, but we can always go back to the original description. Does this have a limit? As we pick integer values and keep working our way out to the right. One? Right? So it is a convergent sequence. One of the sequences mentioned, and this is hard to get an explicit formula. This would be an explicit formula that tells you if you want to find the 23rd term, you just put in 23 for n and if you want to find the 1,000th term, you put in 1,000 for n. Sometimes we don't have the luxury of an explicit description, and that would be the case with a Fibonacci sequence. So they actually do have a variety of ways that they model something, a Fibonacci sequence. So written out, it looks something like this. Has anybody dealt with a Fibonacci enough to tell how to arrive at 8th term, an individual term? What would the next term be? 21? Is that right? Where do you get 21? Add the two predecessors. So there's not a formula, so if we wanted to come out here and get the 39th term, there's not a formula we can put 39 in and generate the 39th term. So if we want the 39th term, we better have the 38th and the 37th. So we have to, in this setting, describe the first term, and it's not enough to just describe the first term. We also have to give a number to the second term. Now we can get the pattern to get the n plus 2 term. How do we get that? Don't we add the term before it, which would be n plus 1 to the term before that one, which would be n, right? So if we take the nth value of the nth term, add it to the n plus first term, we come up with that one. So if you don't like that notation, and you want to write it as a sub n, so it's described in terms of the nth term, we need the two predecessors, which are what? n minus 1 and n minus 2. So they both say the same thing. It just depends on how you prefer to write it. Most of the time, the preference is to write it in terms of the nth term, a sub n. All right, this is in the form of a definition. Let's see if I have those things ready. A sequence, a sub n, you can put the sequence in curly braces, since it's just a list anyway, it's a set. We separate the elements in the set with commas. So the sequence a sub n has the limit l, and we write, we've already done this a couple of times, the limit of a sub n is l, or a sub n approaches l as n approaches infinity. We know how to take limits, it talks about taking limits here, so that's not new. We say the sequence converges or is convergent. If it doesn't converge, then we call it divergent. That's terminology we've already used. Now, how can it converge? And by the way, these are dots, these are not connected curves because we're only using integer values for n, right? So we don't use n equals a third, or two thirds, or 17 20ths, or anything else. So that's why it's a collection of dots, as opposed to a curve. So it is possible for our points to kind of cross through the limiting value and approach it from the other side. Basically, we don't care what happens over here, we care what happens eventually, as n approaches infinity. It is possible for it to oscillate on either side. This has a, what, a declining or decaying amplitude, so it could be a sine or cosine curve. This is possible as well, we kind of oscillate and approach zero from either side, or whatever the limiting value is. That's possible. Again, what we are concerned with is what happens way out here to the right, we don't really care how many times it crossed that limiting value, any of that stuff, we care what happens way out to the right. So that really technically is a definition, so we just kind of read it, accept it, and use it. Here's a theorem which requires some proof. It relates it to stuff that we've already done. So how our f of x and f of n, well if we say the description of the sequence is really some function of n, and we convert that to a function of x, that's really hard to do. Wherever there's an n, you put an x. And just like we used to take limits of functions, now how do functions differ? They connect the dots. So whether the dots are connected or not, we're still kind of rooting it or tying it to something we've already done. So if in fact the function has a limit as x approaches infinity, then the sequence, a sub n, has a limit as n approaches infinity. It's just that we are not connecting the dots. It's the integer values only on this diagram, not the connected curve. I don't know if I have a picture of that, yes I do. Occasionally we'll come up with a stubborn sequence that is difficult to analyze by itself, but we'll have some other ones, one that's bounded above by that one, another it's bounded below, and we'll be able to analyze those. Therefore, by the squeeze theorem, we'll be able to analyze what happens to the one in the middle. So again, these are dots. So here's the one kind of underneath the one in question. It seems to be convergent. The one that's bounding above is this one, which also appears to be convergent as we work our way out to the right. Here's the stubborn one, the b sub n, the one in the middle. We don't know what it's doing, but we know it's bounded by the other two. If the other two are convergent, and in fact convergent to the number L, then certainly the one in question in here that is doing some kind of unpredictable wavering, if it's bounded by those two, then it must also have the same limit as the other two. So that's the squeeze theorem for sequences. You don't need it very often, but occasionally you have a stubborn one in here that is bounded above and bounded below. If they have the same limit, we now know the limit to the one in the middle. So if the limit of a sub n and c sub n are both L, then the one in the middle must also be L. I guess this technically kind of results from the squeeze theorem. So if the limit of the absolute value of a function, not a function, a description of the nth term of the sequence is zero, then even without the absolute value, it should also be zero. So we're kind of closing in from either side, so certainly the one-sided version of that is also convergent. Let's say we have an nth term that is described as natural log of n over n, and we're trying to decide if it's convergent. Well, as n approaches infinity, doesn't the numerator get larger? Right? Infinitely large, so to speak. And as n approaches infinity, the denominator is just n, so certainly n approaches infinity. So if the numerator is getting infinitely large, I'm giving a big hint here, and the denominator gets infinitely large. Pardon? Because they increase at different rates. They increase at different rates. How do we handle infinity over infinity? Um... Lopitalz rule. Lopitalz rule. So anything that we've used to this point, we can also invoke with these as well. So this thing is headed toward infinity. This thing is headed toward infinity. So if we want the limit, as n approaches infinity, we can take the derivative of the top. What's the derivative of? And I know it may look awkward with the n's, if it bothers you just put an x there. The derivative of the natural log of n would be... What do you think is the case with this particular version of infinity over infinity? Which is faster? You think it's going to converge? Taking the natural log of a number, although it increases, it doesn't increase as fast as the number itself increases. Right? If you go from the natural log of 2 to the natural log of 3, yes, it increases, but not as much as you increase the denominator in going from 2 to 3. So the numerator is what? 1 over n. Denominator is 1. So we took the derivative of the top, derivative of the bottom. We reload. The limit of this is 0. Therefore, this sequence, the value of the terms as we work our way out to the right gets closer and closer to 0. Got to get a little factorial in there today. Factorial with sequences and series. In fact, we're going to deal with both of these later in chapter 8 on a pretty regular basis. What do you think? Which one? They both get fairly large fairly quickly. n factorial and n to the n. Which one gets larger faster? n to the n. n factorial. n to the n. Or n factorial. Well, let's write it out. The numerator is going to be something like 1 times 2 times 3 times 4. And it's going to go all the way up to n. Right? What's the denominator going to look like? So if it be 5 to the 5th, we'd have 5, 5 times. Right? So in terms of n, we'd have n used as a factor n times. Does that make sense? Because we're going to have a matching n underneath of it. So we're going to have factors of n, n times, and up here we have 1 times 2 times 3 all the way up to n. n to the n is bigger. Right? At a more rapid rate than n factorial. Now this follows in the textbook. It follows the squeeze theorem. Can you think of a series, not a series, a sequence that this is smaller than, isn't it smaller than 1 over n? This particular series is smaller than 1 over n. And even though we might have difficulty, I think we see this now when we have n's down here. Well, they only match up in the last place. Every other comparative value in the numerator is smaller than its matching value in the denominator. So clearly, the numerator is larger than this. But if you'll take kind of the rest of this other than 1 over n, I think it's pretty clear that the 1 in question, a sub n, is less than 1 over n. 1 over n already converges to 0. Therefore, this one must also converge to 0. Does that make sense? So this is one way that we could use the squeeze theorem. This one converges to 0. This one's even smaller. It must also converge to 0. So it's kind of being squeezed down by the one that is larger than it's bounded above by 1 over n. Therefore, this one must also converges to 0. Some of these things we really don't need right now. I'm trying to pick and choose as we go through this section. As it progressed out to the right, it was some number, R, and we do want to talk about what kind of R values are going to make sense in this problem. But the first term would be R to the 1, then R squared, R cubed, R to the fourth, and so on. It may not look like this ever has a chance to converge, but it does. What type of sequence is this where you multiply by R as you progress from one term to the next? We add anything like that? I'm trying to think if we've had something like that at this point in time. It's a geometric progression. So if you multiply by something as you go, so a geometric progression, and later in this chapter we're going to have geometric series. So we're going to want to add these things together. Right now we're just kind of analyzing what happens to the value of this term Well, let's take a real simple value. Let's say R is one. Well, if R is one, one, one squared, one cubed, one to the fourth, that's certainly convergent, right? I'm not talking about add them. I'm talking about the value of the term. Aren't we converging on the value one as we go? In fact, every one of them is one. So one ought to work if we ever expect this thing to converge. How about negative one? Negative one? This would be negative one. This would be positive one. This would be negative one. We're not converging to one specific number. They're alternating between one and negative one. So I don't think we want that one. So let's take the one category that the limit is one. If R itself is one, isn't it kind of absurd to examine values that cause this to get larger, like two or three or five, two squared, two cubed, two to the fourth? There's no way that's going to converge. So what types of numbers we looked at one? We looked at negative one. We think numbers like two, three, and four are a problem. In like fashion, negative two, negative three, and negative four are also problems. Don't they get larger in magnitude? Shouldn't we be examining values between negative one and one? So what do you think? Let's say our ratio is a half. So the first term is a half. Then we've got a half squared. Then we've got a half cubed. Think those would be converging on something? What? Zero. Let's take negative two-thirds. If R is negative two-thirds and you keep taking negative two-thirds to higher and higher powers, which one wins? Negative two to the 170th power or three to the 170th power? Three. Doesn't the denominator win? Numerators pretty large in magnitude but the denominator's much larger because it's a larger number that's raised to the 170th power. So even though these alternate in this area, aren't all of these values causing this particular geometric progression to converge? Is that right? And what would we say the limit would be? Zero? The value of the n-th term gets closer and closer and closer to zero as long as R is chosen anywhere in here and the other kind of strange value is when R is one, not a very interesting problem at all. It just stays at one, so the limit is one, but we also know that that is convergent. So if R is any other value, we would say this is a... this R to the n would be divergent. If R is any other value outside of here. So we could put them together and say that... but this will come in handy later in the chapter. Win is an infinite geometric series We're not there yet. Win is an infinite geometric series convergent. Right now it's just the sequence. If you set foot on our planet today for the very first day, you would probably know how to do these. So let's go through these real quickly. This is not anything of Earth-shattering magnitude. If you have a limit of a sum, it's the sum of the limits. The limit of a difference is the difference of the limits. We've already seen these, right? With functions, all these limit properties. If you have a constant times a function, you can farm the constant out in front. The limit of a product is the product of the limits. Ooh, there's one we almost forgot. It's a good thing we picked that one up. The limit of a constant is, in fact, that constant. Glad we got that one. The limit of a quotient is the quotient of the limits as long as the denominator doesn't do anything wacky. And the limit of something to a power is the limit of that function first, then raise that limit to a power. These are all things that we've done with functions. We now have the right to do all of those things with nth term descriptions of sequences. First part of this, we did the second part. Just as we describe functions that are increasing and decreasing, we can do the same thing with sequences. We did this already. This is where we decided when r to the n is convergent. So when is the sequence increasing? If the nth term is smaller than its successor, that's earth shattering for all n greater than or equal to 1. Just as a function is an increasing function, same thing with these terms in a sequence. When is it decreasing? If the nth term is larger than its predecessor for any term that you look at in this series. It is called, now that may be a new term, monotonic, is that a term that has come up before? Other than talking about your calculus professor that he talks in a monotone, I'm sorry about that. He's monotonic. If the whole function is either increasing or decreasing, it's called monotonic. It always is doing the same thing. So it doesn't increase for a while, then decrease for a while. That's not monotonic. If it's always increasing or always decreasing, this is a term that is used to describe it. So why is that word important? We'll put that monotonic together with this term bounded and we'll be able to come up with a conclusion. So what does bounded mean? A sequence is bounded above if there is a number m and there might be a lot of numbers m, but there just has to be one number m for this to be true, such that m is greater than or equal to any value of the sequence that you come up with at any point in time. It's bounded below, and again there might be a lot of different numbers, lower case m, but there's at least one number where m is less than or equal to every value of the sequence that we come up with. So for the entire sequence in both cases, if it's bounded above and bounded below, that actually happens some, then it is called a bounded sequence. Now let's put those two words together, bounded and monotonic. For the monotonic sequence theorem, every bounded monotonic sequence is convergent. So if you think about that, here's our capital M. We've got an increasing sequence. So it's monotonic because it's always increasing. This one is bounded above, doesn't it have to be convergent? Aren't we getting closer and closer and closer to this value capital M? Now aren't there other values that bound this sequence above? Yeah, there's a whole bunch of other values. There's another value that clearly is larger than every value that we come up with, but there only has to be one for it to be bounded above, and conveniently if it's bounded above by m, as in this picture, and its limit is also m, then it is a convergent sequence. All right, so that puts us to series, we should start tomorrow. Is there anything we can do in two or three minutes on that problem? I mean you have the c1 and c2 values, right? That worked and you tried them and they worked, and we can do that after class if necessary to see where that error is. All right, so that finishes 8-1. We will switch horses tomorrow and start with series, which now we're going to be adding terms up, which is a different kind of animal.