 Welcome back. We're at lecture 40. My guess is that they'll probably be about somewhere in the low 60-62 class sessions in this particular course before all is said and done, so we're Figure it out. I mean it is a math class after all. We're two-thirds of the way there almost We have a test this week we will cover Chapter 8 section 3. It's got a couple of different Tests to see if something is a series is convergent or divergent So we're kind of continuing in that and that reminds me that I forgot something today. I'm going to give you a chart kind of a pecking order chart that if you're Testing a series and you really don't have any Clue what's going to work is in an alternating series, which we're not there yet A comparison test we're going to do today limit comparison test. I think we'll get to today P series maybe today probably tomorrow That test for divergence that we talked about last time if the limit of the nth term does not approach zero then We know for a fact that it diverges So we're going to have a pretty long list of convergence and divergence tests So I want to give you a chart the chart has kind of a Decision making hierarchy on it I didn't make the chart Marilyn McCollum who is a math instructor in our department made it and with her permission We're going to continue to use it because it's very good She recently retired and Spending more time at the beach and that sound kind of pleasant Let's just go let's close it up and go to the beach so I will try to bring that tomorrow and Hopefully that'll be helpful. It's not something you can use on a test But I think it will kind of organize your thinking Help you organize your thinking for a test that you'll try this it takes a few seconds If it works, it's good if it doesn't then let's progress on to some of these other ones So I'm take a long time. So you don't want to try them for example a ratio test which we'll get at the end of the chapter near the end of the chapter you don't want to you know Kind of blindly try the ratio test and all it didn't work You know, we got a limit equal to one therefore the test fails Well, you've just wasted seven or eight minutes So we want to save that one till we've ruled out some of the ones that take a matter of a few seconds. I Think today we'll be able to get the integral test and the comparison test and Possibly the limit comparison test. So we'll have three more Tests for convergence and divergence near the end of this section Probably tomorrow possibly Wednesday will Using some of these approximating techniques of a series that converges We'll try to see how close our answer actually is or if we wanted a certain level of accuracy How many terms would we have to add together in order to get that level of accuracy? so do we add together 10 terms or 30 terms or 52 terms to get us, you know within 1 1 millionth of the actual sum All right, so 8.3 Just kind of put a star beside the new tests as we come to them and when you look back at your notes you'll be able to pick off the Kind of the different tests and the basics about how to do that the integral test is a is one of the ones Lower in the hierarchy. So you can ask yourself the question Basically, could I integrate that function and get a solution if you can then we're going to be able to make a decision We'll do some of the examples about half of the examples that I have are ones that are actually in the book because they're Just really good ones really classic ones 1 over x 1 over x squared And then we'll make those into kind of Since we already know something about them we'll be able to use those because we know what those integrals do and consequently hopefully their series that goes along with them does the same thing so we know there is a relationship between this s and this s and We've seen When integrals converge and diverge when you get an answer finite answer it converges and when you don't you get something that does not exist and it diverges so we've seen that and we have seen the interaction between these two Summation symbols this is really for a finite number of Of things that we're adding together and if we go to n and then let n go to infinity that becomes an infinite number of things and We then can switch we've The probably the first tie that we saw between these two symbols was this one That if we are adding the areas of skinny little rectangles together the Riemann some Concept and we wanted there to be in rectangles and we added their areas together and then Not just in but we wanted to let the number of rectangles approach zero that became by definition the value of the definite integral from A to B whatever the starting value is to whatever the ending value is So we've already seen kind of how these things are related We can't really get to an integral let alone a definite integral without this concept of adding together an infinite number of pieces in this case skinny little rectangles To get the value of the definite integral so what we want Is a summation that's what the answer that we really want But we're going to use an integral to help us get that answer So there's the series we're dealing with it's an infinite series It's not geometric because we're not multiplying by the same thing as we progress from one term to the next But the terms are clearly positive and they are decreasing so it has a chance of Converging we can't do this. We can't say that the limit of one over in we could say it better be wrong So we really probably shouldn't say it that the limit of one over n squared as In approaches infinity is zero therefore it converges with that's not going to work for us It is true that the limit of one over n squared is zero, but that doesn't guarantee convergence What's a good counter example of what I just said? Where it's a series that does the nth term does get closer and closer to zero But it does not One over in or the harmonic series, so we've got a counter example to that So we can't just say that because the limit of the nth term goes to zero therefore its convergence not going to work What do you think about this one? Think it's getting small fast enough in order for this to converge One over one one over two one over three one over four that did not Converge because that although they got small they didn't get small fast enough What about this one? Most of you are saying that you think it's probably going to converge But we need to be able to prove that so we're going to use the integral test Let's see on a diagram where this series is Numerically an actual picture of it on the diagram see if we can use an integral to make the decision for us and in general Can we use the value of a definite integral to help us decide on what's happening on the series? So let's take a look at We want the curve This is one over n squared, so we want the curve to be One over x squared So let's see if we can find on this curve the values of This thing once one over one squared plus one over two squared and so on there's our first term There's one over one squared isn't when I come up from one and the curve is one over x squared The height of that curve is one over one squared the width is one Therefore the area of this is one and that happens to be the value of our first term now the question is can we continue that? Well if I come up from two And if I'm on the curve one over y equals one over x squared when x is two, what is that height? one over two squared right because there's the curve and One over two squared is the value of the Second term of this series now. Let's see if we can continue with that come up here to three How tall is that box that is one unit wide by the way all these are going to be one unit wide? That'd be one ninth That's the value of the third term and we could do the same thing I don't know that I can actually do this But we could continue this pattern on out to the right that rectangle its area would be what? 116 which is the value of the fourth term, so we can relate the curve. We don't really have the curve we have these Rectangles that are under the curve, but we can relate the curve to the to the series that we had what we really want is the sum of all these areas This is a little problematic the first one One but as we mentioned last class or possibly the class before that what happens initially in a series We'll never change what's going to happen eventually in that series. So why don't we start our? discussion in our Let's kind of leave that first rectangle out there by it by its side Not include it in our discussion, but then let's restart the discussion on the right of that rectangle Knowing that whatever we get we're going to get One added to it this first rectangle So I want to start at one and let this thing go to infinity and I want to find the area under One over x squared so the area under one over x squared is really all of this Now if we could determine that this Converges to something. I don't know. Let's say five five's not right, but let's say it converged to five And then we'd add one to it. So we say we want the one plus the area under this curve if this converges Doesn't the sum of all these little boxes Have to also converge Does that make sense? So if the area under I guess it did or you wouldn't have given me that look Maybe that was a Monday morning look and not a I don't understand that So if we're able to find the area under this curve from one All the way out to infinity Doesn't it make sense that we're also going to be able to find the sum of the areas of these blocks It's not even the same as the area under the curve and we can find that Because we're missing all these pieces these little so-called little triangular pieces. So if this converges Then we're in business the area in this particular series converges It's even smaller Than the area under the curve itself So if this is small enough to converge certainly anything smaller than that should converge We're going to see that a couple times today with the comparison test as well Now that doesn't mean that the sum of one fourth plus one ninth Plus one sixteenth is exactly the same as the area under the curve. That's not what we're saying We're just making a decision about convergence That if we can find the area under the curve certainly we can find the area of the boxes that are Underneath this curve from one to infinity. So if the integral converges So does the series that's associated with? Well, let's see what happens to one over x squared Kind of looks like an improper integral to me What do we do with an improper integral? Kind of save that so-called bad value the infinity To the end of the problem, so we'll go from one to a we'll actually evaluate from one to a and then the last step in the problem We'll let a approach infinity So that's x to the negative second. What's the integral of x to the negative second? Good negative x to the negative one so that's the same thing as negative one over x We'll evaluate that From one to a so we'll start off at a negative one over a minus negative one over one We like to see the negative one over a as a approaches infinity because what happens to negative one over a as a gets larger and larger and larger This disappears doesn't it that goes to zero as a gets larger and we're left with negative of negative one which is One so the area under one over x squared from one to infinity is one So let's go back to this one We had this one out here by itself, which is this guy right here We're going to add that in because that's the first term of the series We just found that the area under this curve from one to infinity is also one Certainly if we can add this one block this one by one block to the entire area under the curve from one to infinity We're going to be able to find the sum of these terms now It won't be two it'll be less than two is that correct if I add one plus one fourth plus one ninth Plus one sixteen all the way out to infinity. We're not going to get to But it's certainly bounded above by two so we're going to get something less than two in fact It's about one point six five roughly if we could add that up again That's an approximation, but certainly it's bounded above by two Therefore by the integral test because this integral converges from one to infinity the series associated with it also converges So how would you write that down? You would say that the integral converges Therefore and if you're asked to kind of check a block, what's the reason the reason is the integral test? This is small enough to converge Therefore anything that finishes underneath of it even if we're adding a finite number of terms to it will also converge Can we make a similar decision with something that we know? diverges In fact, this is another validation or verification that the harmonic series diverges just if you didn't happen to like the first version where we were grouping a certain amount of terms together first term second term the next two terms the next four terms the next eight terms if you didn't like that that those Kind of force this thing to diverge here's another validation for The harmonic series we already know this diverges. Here's a good validation why it diverges They're positive. They're decreasing. They're getting smaller the nth term does approach zero But just because the nth term approaches zero does not guarantee convergence. This is the classic counter example So we already know This diverges Let's see if we can validate this another way and also have a picture that goes along with it So if we wanted the integral that goes along with this particular series, what would the integral be for 1 over n it'd be 1 over x so there's our curve 1 over x it does Kind of decrease asymptotically to the x-axis just not quite as quickly as the one we looked at first today So our first one is one So instead of coming back over here and drawing by one by one block There's a method for that or a reason for what I'm doing. I'm going to come over here to the right. That's one by one So there's our first block at two We want one half. So there's that one at three. We bring this guy over to here That'd be one third At four we do the same thing so we generate kind of the same Process, but we don't have one over there to the left standing by itself Again, we want to start our discussion right here at one We want to integrate so we want the area under the curve from one to infinity What's the integral of one over x integrated with respect to x? natural on Absolute value, but we don't have to pay attention to absolute value with this one because we're going from one to infinity See if I can resurrect my Saving of the bad value to the end of the problem That's what it should look like. I'm not liking the natural log of a I like it a whole lot better when the a's and the denominator Then it can increase without bound and that term disappears. I don't think we're going to get quite so lucky here natural log of one is Zero that's the power you'd raise e to get one so that disappears so we don't need that anymore So what happens to the natural log of a as a is allowed to increase without bound to disappear? natural log of ten natural log of a hundred natural log of a Million natural log of six hundred trillion don't those values keep getting larger and larger So we don't have anything disappearing in fact just the opposite this limit does not exist Keeps getting larger and larger and larger So we knew that we've already done that earlier in in this course Maybe even in 141 that we knew we couldn't get an answer to the area under 1 over x from 1 to infinity Doesn't exist Therefore the integral diverges well if the integral diverges look at our blocks here Don't they go above the integral? Don't you think that this particular sum is it in even worse shape? Then the area under the curve 1 over x from 1 to infinity not only does it have all the area that's under the curve It's got some extra the area under the curve is already in trouble when you add some extra It's not going to help it. It's going to make it worse We're already falling down with all this luggage on our backs. Well. Let's add in some more that's going to help things Yeah, that's going to make it easier to carry. Isn't it? No, that's going to make it worse It's already in trouble. We add more to it. It is even worse in trouble. Yes The curve would Be I think I know where you're going with it wasn't diverging then If it wouldn't be conclusive, okay? I'm not let me see if I can get to the What I thought you were going to ask and you tell me if that's really what you just asked me So let's say we determined that this was convergent Okay, then my headed down the right path here and Then we have this that is a little bit larger than something that's convergent. Does that mean it also converges? It does not It's got to be underneath the convergent It's got to be above the divergent if it's any mixture of that we can't make a decision so one way to look at that is this net is Not that good. This net is not big enough to catch That particular situation we'll have to get another net Okay, and we do have some other nets that are a little bigger So if it's bigger Then the divergent we're in business if it's smaller than the convergent We're in business any mixture of that we cannot make a decision. We'll have to use another test Is that what is that where you're headed? Okay, so what would we write down because this? integral diverges Therefore the series associated with that because it actually goes above that curve Sorry, that's not what I want One over in and again Justification would be the integral test So what did what was true about both of these curves and what do we want to be true about any? Curve that has a chance of being decided by the integral test. We want it to be a positive terms We want it to be each term to be positive We'll deal with later. What happens when they're alternating, but right now we want each term to be positive We want it to be decreasing If it's increasing it doesn't have a chance of converging and if we're going to be able to integrate it It has to be a continuous curve So when we convert it from the series to the integral, it's got to be a continuous curve you can kind of Flavor that just a little bit it technically it doesn't have to be decreasing Every term decreases from its predecessor, but it has to ultimately be decreasing So you could have a few terms that got a little bit larger a little bit larger But then beyond some point moving all the way out to the right all the way to infinity. It does have to be decreasing ultimately Decreasing another thing that both of these did that doesn't have to be true. These both started at 1 They do not need to start at n equals 1 The nice thing about starting at n equals 1 and a couple of people have asked me about this in either office hours or email or after class The nice thing about starting at 1 is when you put 1 in it generates the first term. That's convenient You put 2 in it generates the second term 3 generates the third term that that just We like that visually we like that, but that doesn't have to be true. You could take a series let's say an infinite geometric series a R to the n minus 1 and if we started at 1 What's the first term? Not 0, but the power is 0 right a r to the 0 which is just a And the next term is a r to the 1 and a r squared But we can write that another way Let's say we want to start at n equals 0 Isn't that the same series? Well put in 0. What do you get? a r to the 0 which is a Put in 1 a r to the 1 which is a r And so on so you can kind of adjust and make them start wherever you want to but this is Probably the the more convenient because 1 generates the first term doesn't have to be the case Alright, so what do we have we have? Kind of the harmonic series itself we've now validated that two different ways that it diverges You're not going to be asked to validate the harmonic Anymore, I mean we've said it's divergent two different ways, so we kind of use that If the harmonic series diverges, let's say that we have this is the series in question Is it permissible to take that three halves out in front? And we're going to encounter this a lot, so let's go ahead and decide now what we're going to do with this if we have any Number any constant any coefficient out here Times a series that we already know what it does Is that going to change what it does No, it's not going to change convergence or divergence. It will change the value to which it converges But it's not going to change our decision So any constant times a converging is also converging We don't have to make a big deal about that when we're trying to make a decision Any constant times a an existing divergent series is going to also diverge So we can still call this Harmonic it is harmonic. It's three halves times our One that we traditionally call harmonic, but it is also a harmonic All right, one of our First ones that will kind of go through will go through the kind of the test for divergence That's quick and easy We'll probably then see ask ourselves. Can I integrate this thing if I can integrate it and get an answer? Then I'll be able to use the integral test. The next thing is this p-series. So what does a p-series look like? It looks like this one over into the p So for certain values of p, this is going to diverge. In fact, we already know one. We just did one when p is one One over into the one is just one over in which is harmonic, which does what? Diverges We already did another one today our first example One over n squared. We decided that converged right one over n squared If p is Zero That's not very interesting if p is zero. That's just one added to itself That one would diverge. Let's let's see if we can somehow categorize this Let's leave this value out since we already know what it does and let's say that P cannot be one For this particular discussion So for all other values of p, how do we And this is why we need the integral test first because we're going to use the integral test to make the decisions on this p-series For p not equal to one. This is one over into the p is into the what negative p. I'm not liking that. I Don't like working in terms of the ends when we're integrating. So let's call this x to the negative p So we're converting the function the nth term description into a function So a sub n has to be convertible To some function, I don't think that's too big of a stretch to say the ends where the letters the variables We're now changing them to x's So we can integrate with respect to x We're going to rewrite that at the top of the sheet So we're throwing out we're throwing aside the natural log answer So we know we cannot get a natural log answer. So if p is one, this is x to the negative one Which is one over x, which is the natural log? We're not dealing with that. We already know what happens when p is one. We know it diverges So how do you integrate x to the negative p if we throw out the possibility that it could be a natural log? It's a power series. What do you do typically when you integrate x to a power? What if it were x to the negative third? What would you do? Add one to the power divide by the new power right? Remember the only case you can't do that there's when the power is negative one So we've already said it can't be negative one So we get one added to the power divided by That new power Let's go ahead and convert this so we save the bad value To the end of the problem. All right. We're supposed to put in the upper limit of integration x to the negative p plus one Sorry So that replaces the x and then we're supposed to put in one Why don't we factor out that denominator? So we don't have to kind of think about that So let's take that out in front. What is that numerator? one to Any power we choose is going to be one right? Here's the one that we need to isolate on and we need to decide for certain values of p This term will disappear. We want it to disappear, right? What what do we know causes this term? As a approaches infinity Where do we want the a to ultimately end up? We want it to end up in the denominator, right? We can somehow get that a to be in the denominator I don't care what power it is in the denominator if it's in the denominator and a approaches infinity Won't that term eventually disappear? So what value for p? Would send the a down to the denominator p has to be More than one Everybody okay with that if p is greater than one Then we've got a negative exponent That's what we want. We want the exponent to be negative that sends the a to the denominator Then if a approaches infinity and it's in the denominator the value of that term disappears So we want p to be greater than one So let's say p is greater than one when p is greater than one this term approaches zero and Whatever else is in the problem. Don't we get an answer when this term disappears? So our decision would be what convergent and if this term did not disappear When would it not disappear? When p is Right, there we go. That was my next question. Thank you for answering it before I ask it We already know what happens when p is one it already diverges Also when p is less than one The a stays in the numerator The exponent is positive therefore the a stays in the numerator and as a gets larger that whole thing gets larger We don't get a decision or well. We do get a decision. It doesn't disappear like we want it to so this means that It's going to converge When the exponent is positive the a stays in the numerator as a gets larger that term gets larger Nothing is disappearing so it diverges so P series if p is greater than one We can automatically say because we've already validated it right here that that series is convergent If p is one or less We can go right to the conclusion that particular series is divergent Here's an example of each Don't want the integral. Sorry That's not an example Now we're not going to have one quite this dumb kind of got a dumb example here One over into the 1.4. It's going to be three halves or something else but because We already know what to do with one over in to the p which we classify now as a p series We know if p is what? greater than one it converges and If p is less than or equal to one it diverges. Well, what does this do? Converges, so what do you write down? It's a p series And p is greater than one that'll do it might want to at least think about that as instead of square root of in in to a power so that's one over in to the one half Decision diverges it's a p series and what p is less than or equal to one Those are pretty quick and easy So you'll want to check those kind of right in that category with the integral test Can I integrate it and get a solution then I'll use the integral test is it a p series? It is okay. That's a quick decision as long as we justify That it is a p series and what Category p is in we can get a quick and easy decision Let me write this out, and then we'll continue that thought Until tomorrow. This is the beginning of a new test Let's say that we're trying to kind of Categorize this Would this be a candidate for an integral test? Is that look like something? That you would want to integrate I can tell you it doesn't look like anything that I would want to integrate So I would probably rule out the integral test pretty quickly in my Hierarchy as far as I don't want to do that Any guesses about this? Convergent divergent 50-50 chance right? How about this one? How about if the five were not there? You would say convergent. Why would you say convergent? One is right because I mean if it was just like one over Okay, you're right. The top is gonna be one and the bottom is gonna increase by by a factor of three right? We do know what this is. So let's write a couple of them out first one would be what one third? Second term is one ninth One twenty seventh kind of series is that Geometric don't we multiply by something as we go? What's the ratio? One third so there's our category. It's an infinite geometric series What is what is cement? What is something else that we need to decide? Help us decide if it converges or not the ratio itself We want the absolute value of the ratio to be Less than one Well, is that true? Yes So what Noah said is correct it converges and in fact in this particular problem We could say that it not only does it converge we could find the value to which it converged which would be what? first term over one minus the ratio first term is one third One minus the ratio is also one third So one third over two-thirds if we could add all these terms together. What would we get? one half So it is convergent. We actually in this case know the value to which it converges back to this problem Does that help you decide? If the five is not there it converges. What do you think it's going to do when the five is there in the denominator? It's also going to converge and somebody said a nice word there. It converges even faster, right? Isn't it even smaller? so We'll talk about what this test is Using something that we already know what it does and something that is Close to that will begin class with that test tomorrow. So is it okay for me to say test tomorrow? This test this particular test is tomorrow